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1 Introduction

2 OrthogonalMatrices

3 OrthogonallyDiagonalizableMatrices

4 The SpectralTheorem

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Chapter 8 Orthogonality8.3 Diagonalizing Symmetric Matrices and QR

Factorization

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Remembering Back ...

What did it mean for vectors to be orthogonal?

How did we test to see if vectors are orthogonal?

What did it mean for vectors to be orthonormal?

Anyone want to take a shot at defining anorthogonal matrix? It is not a trick question, but not asstraight forward as we may want ...

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

DefinitionA square matrix whose columns form an orthonormal basis iscalled an orthogonal matrix.

Example

Is A =

[47

√33√

337 −4

]orthogonal?

Is the dot product of the columns equal to 0? Even so, thesecond vector is not a unit vector, so the matrix is notorthogonal. Is there anything we could do to this matrix toproduce an orthogonal matrix? Divide both entries of thesecond column by the norm, which is 7.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

Example

Is A =

[47

√33√

337 −4

]orthogonal?

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

Example

Is A =

[47

√33√

337 −4

]orthogonal?

Is the dot product of the columns equal to 0?

Even so, thesecond vector is not a unit vector, so the matrix is notorthogonal. Is there anything we could do to this matrix toproduce an orthogonal matrix? Divide both entries of thesecond column by the norm, which is 7.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

Example

Is A =

[47

√33√

337 −4

]orthogonal?

Is the dot product of the columns equal to 0? Even so, thesecond vector is not a unit vector, so the matrix is notorthogonal.

Is there anything we could do to this matrix toproduce an orthogonal matrix? Divide both entries of thesecond column by the norm, which is 7.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

Example

Is A =

[47

√33√

337 −4

]orthogonal?

Is the dot product of the columns equal to 0? Even so, thesecond vector is not a unit vector, so the matrix is notorthogonal. Is there anything we could do to this matrix toproduce an orthogonal matrix?

Divide both entries of thesecond column by the norm, which is 7.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

Example

Is A =

[47

√33√

337 −4

]orthogonal?

Is the dot product of the columns equal to 0? Even so, thesecond vector is not a unit vector, so the matrix is notorthogonal. Is there anything we could do to this matrix toproduce an orthogonal matrix? Divide both entries of thesecond column by the norm, which is

7.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

Example

Is A =

[47

√33√

337 −4

]orthogonal?

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

Example

Is A =

[12

√32

2√5− 1√

5

]orthogonal?

Are the columns unit vectors?

Is the dot product of the vectors 0?

The columns do not have a dot product of 0, so we do nothave an orthogonal matrix.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

Example

Is A =

[12

√32

2√5− 1√

5

]orthogonal?

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

Example

Is A =

[12

√32

2√5− 1√

5

]orthogonal?

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

Example

Is A =

[12

√32

2√5− 1√

5

]orthogonal?

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

Example

Is A =

3√11

√211

−√

211

3√11

orthogonal?

Is the dot product of the vectors equal to 0?

Since both are satisfied, the matrix is indeed orthogonal.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

Example

Is A =

3√11

√211

−√

211

3√11

orthogonal?

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

Example

Is A =

3√11

√211

−√

211

3√11

orthogonal?

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

Example

Is A =

3√11

√211

−√

211

3√11

orthogonal?

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Why We Care

We looked in chapter 6 at diagonalizing a matrix - when isthat possible?

Here, we want to see if we can orthogonally diagonalize amatrix. That is, see if we can find an orthogonal matrix Pand a diagonal matrix D such that A = PDP−1.

This is desirable because of what it does for us if we have anorthogonal matrix. Thinking back, though, to finding thediagonalization of a matrix larger than 2× 2, what was thehardest part? Finding the inverse ...

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Why We Care

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Why We Care

This is desirable because of what it does for us if we have anorthogonal matrix. Thinking back, though, to finding thediagonalization of a matrix larger than 2× 2, what was thehardest part?

Finding the inverse ...

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Why We Care

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

TheoremIf P is an orthogonal matrix, then P−1 = PT .

Proof.Let P be an orthogonal matrix and consider PTP.

I How would we find this product of matrices? It wouldbe the dot products of the columns of P.

I If we take the dot product of the i th column of P withthe i th column of P, what will be the result? 1 - why?Because of normality ...

I What will be the product of the i th column of P withthe j th column of P? 0 - why? Because the columns areorthogonal.

I So what is PTP equal to? The identity matrix.

So, PTP = In = P−1P ⇒ PT = P−1.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

So, PTP = In = P−1P ⇒ PT = P−1.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

I How would we find this product of matrices?

It wouldbe the dot products of the columns of P.

So, PTP = In = P−1P ⇒ PT = P−1.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

So, PTP = In = P−1P ⇒ PT = P−1.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

I If we take the dot product of the i th column of P withthe i th column of P, what will be the result?

1 - why?Because of normality ...

So, PTP = In = P−1P ⇒ PT = P−1.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

I If we take the dot product of the i th column of P withthe i th column of P, what will be the result? 1 - why?

Because of normality ...

So, PTP = In = P−1P ⇒ PT = P−1.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

So, PTP = In = P−1P ⇒ PT = P−1.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

I What will be the product of the i th column of P withthe j th column of P?

0 - why? Because the columns areorthogonal.

So, PTP = In = P−1P ⇒ PT = P−1.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

I What will be the product of the i th column of P withthe j th column of P? 0 - why?

Because the columns areorthogonal.

So, PTP = In = P−1P ⇒ PT = P−1.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

So, PTP = In = P−1P ⇒ PT = P−1.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

I So what is PTP equal to?

The identity matrix.

So, PTP = In = P−1P ⇒ PT = P−1.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

So, PTP = In = P−1P ⇒ PT = P−1.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonal Matrices

So, PTP = In = P−1P ⇒ PT = P−1.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Orthogonally Diagonalizable Matrices

Example

Find an orthogonal matrix P that orthogonally diagonalizes

A =

[−2 22 1

]

We begin as we did before when we were diagonalizing byfinding the eigenvalues and associated eigenvectors.

|A− λI | =

∣∣∣∣ −2− λ 22 1− λ

∣∣∣∣= (−2− λ)(1− λ)− 4 = 0

λ2 + λ− 6 = 0

λ = −3, 2

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example

A =

[−2 22 1

]

We begin as we did before when we were diagonalizing by

finding the eigenvalues and associated eigenvectors.

|A− λI | =

∣∣∣∣ −2− λ 22 1− λ

∣∣∣∣= (−2− λ)(1− λ)− 4 = 0

λ2 + λ− 6 = 0

λ = −3, 2

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example

A =

[−2 22 1

]

|A− λI | =

∣∣∣∣ −2− λ 22 1− λ

∣∣∣∣= (−2− λ)(1− λ)− 4 = 0

λ2 + λ− 6 = 0

λ = −3, 2

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example

A =

[−2 22 1

]

|A− λI | =

∣∣∣∣ −2− λ 22 1− λ

∣∣∣∣

= (−2− λ)(1− λ)− 4 = 0

λ2 + λ− 6 = 0

λ = −3, 2

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example

A =

[−2 22 1

]

|A− λI | =

∣∣∣∣ −2− λ 22 1− λ

∣∣∣∣= (−2− λ)(1− λ)− 4 = 0

λ2 + λ− 6 = 0

λ = −3, 2

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example

A =

[−2 22 1

]

|A− λI | =

∣∣∣∣ −2− λ 22 1− λ

∣∣∣∣= (−2− λ)(1− λ)− 4 = 0

λ2 + λ− 6 = 0

λ = −3, 2

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example

A =

[−2 22 1

]

|A− λI | =

∣∣∣∣ −2− λ 22 1− λ

∣∣∣∣= (−2− λ)(1− λ)− 4 = 0

λ2 + λ− 6 = 0

λ = −3, 2

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Now, we find the eigenvectors:

For λ = 2, we have

A− 2I =

[−4 22 −1

]∼[

1 −12

0 0

]

So the eigenvector associated with λ = 2 is

[12

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Now, we find the eigenvectors:For λ = 2, we have

A− 2I =

[−4 22 −1

]∼[

1 −12

0 0

]

[12

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

A− 2I =

[−4 22 −1

]

∼[

1 −12

0 0

]

[12

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

A− 2I =

[−4 22 −1

]∼[

1 −12

0 0

]

[12

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

A− 2I =

[−4 22 −1

]∼[

1 −12

0 0

]

[12

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

For λ = −3, we have

A + 3I =

[1 22 4

]∼[

1 20 0

]

So the eigenvector associated with λ = −3 is

[−21

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

A + 3I =

[1 22 4

]

∼[

1 20 0

]

[−21

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

A + 3I =

[1 22 4

]∼[

1 20 0

]

[−21

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

A + 3I =

[1 22 4

]∼[

1 20 0

]

[−21

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

So, we have {[12

],

[−21

]}What do we have?

The vectors form an orthogonal basis, sosince we have distinct eigenvalues, we could use this matrixto diagonalize A, but we aren’t done yet because the vectorsneed to be normalized.

And how do we normalize?

p1 =v1‖v1‖

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

So, we have {[12

],

[−21

]}What do we have? The vectors form an orthogonal basis, sosince we have distinct eigenvalues, we could use this matrixto diagonalize A, but we aren’t done yet because

the vectorsneed to be normalized.

p1 =v1‖v1‖

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

So, we have {[12

],

[−21

]}What do we have? The vectors form an orthogonal basis, sosince we have distinct eigenvalues, we could use this matrixto diagonalize A, but we aren’t done yet because the vectorsneed to be normalized.

p1 =v1‖v1‖

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

So, we have {[12

],

[−21

p1 =v1‖v1‖

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

So, we have {[12

],

[−21

p1 =v1‖v1‖

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

p1 =

1√5

[12

]p2 =

1√5

[−21

]So, we have

PTAP =1√5

[1 2−2 1

] [−2 22 1

]1√5

[1 −22 1

]=

[2 00 −3

]= D

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

p1 =1√5

[12

]

p2 =1√5

[−21

]So, we have

PTAP =1√5

[1 2−2 1

] [−2 22 1

]1√5

[1 −22 1

]=

[2 00 −3

]= D

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

p1 =1√5

[12

]p2 =

1√5

[−21

]So, we have

PTAP =1√5

[1 2−2 1

] [−2 22 1

]1√5

[1 −22 1

]=

[2 00 −3

]= D

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

p1 =1√5

[12

]p2 =

1√5

[−21

]

So, we have

PTAP =1√5

[1 2−2 1

] [−2 22 1

]1√5

[1 −22 1

]=

[2 00 −3

]= D

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

p1 =1√5

[12

]p2 =

1√5

[−21

]So, we have

PTAP =1√5

[1 2−2 1

] [−2 22 1

]1√5

[1 −22 1

]=

[2 00 −3

]= D

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

Example

A =

[13 66 −3

]

|A− λI | =

∣∣∣∣ 13− λ 66 −3− λ

∣∣∣∣= (13− λ)(−3− λ)− 36 = 0

λ2 − 10λ− 75 = 0

λ = −5, 15

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

Example

A =

[13 66 −3

]

We begin as we did before when we were diagonalizing by

finding the eigenvalues and associated eigenvectors.

|A− λI | =

∣∣∣∣ 13− λ 66 −3− λ

∣∣∣∣= (13− λ)(−3− λ)− 36 = 0

λ2 − 10λ− 75 = 0

λ = −5, 15

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

Example

A =

[13 66 −3

]

|A− λI | =

∣∣∣∣ 13− λ 66 −3− λ

∣∣∣∣= (13− λ)(−3− λ)− 36 = 0

λ2 − 10λ− 75 = 0

λ = −5, 15

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

Example

A =

[13 66 −3

]

|A− λI | =

∣∣∣∣ 13− λ 66 −3− λ

∣∣∣∣

= (13− λ)(−3− λ)− 36 = 0

λ2 − 10λ− 75 = 0

λ = −5, 15

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

Example

A =

[13 66 −3

]

|A− λI | =

∣∣∣∣ 13− λ 66 −3− λ

∣∣∣∣= (13− λ)(−3− λ)− 36 = 0

λ2 − 10λ− 75 = 0

λ = −5, 15

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

Example

A =

[13 66 −3

]

|A− λI | =

∣∣∣∣ 13− λ 66 −3− λ

∣∣∣∣= (13− λ)(−3− λ)− 36 = 0

λ2 − 10λ− 75 = 0

λ = −5, 15

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

Example

A =

[13 66 −3

]

|A− λI | =

∣∣∣∣ 13− λ 66 −3− λ

∣∣∣∣= (13− λ)(−3− λ)− 36 = 0

λ2 − 10λ− 75 = 0

λ = −5, 15

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

Now, the normalized vectors:

A− 15I =

[−2 66 −18

]∼[

1 −30 0

]

So, the eigenvector associated with λ = 15 is

[31

]and the

normalized vector is

[3√101√10

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

A− 15I =

[−2 66 −18

]

∼[

1 −30 0

]

[31

]and the

[3√101√10

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

A− 15I =

[−2 66 −18

]∼[

1 −30 0

]

[31

]and the

[3√101√10

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

A− 15I =

[−2 66 −18

]∼[

1 −30 0

]

[31

]

and the

[3√101√10

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

A− 15I =

[−2 66 −18

]∼[

1 −30 0

]

[31

]and the

[3√101√10

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

A− 15I =

[−2 66 −18

]∼[

1 −30 0

]

[31

]and the

[3√101√10

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

A + 5I =

[18 66 2

]

∼[

1 13

0 0

]

So, the eigenvector associated with λ = −5 is

[1−3

]and the

[1√10

− 3√10

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

A + 5I =

[18 66 2

]∼[

1 13

0 0

]

[1−3

]and the

[1√10

− 3√10

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

A + 5I =

[18 66 2

]∼[

1 13

0 0

]

[1−3

]

and the

[1√10

− 3√10

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

A + 5I =

[18 66 2

]∼[

1 13

0 0

]

[1−3

]and the

[1√10

− 3√10

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

A + 5I =

[18 66 2

]∼[

1 13

0 0

]

[1−3

]and the

[1√10

− 3√10

].

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

So, we have

D =

[−5 00 15

]P =

[1√10

3√10

− 3√10

1√10

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

So, we have

D =

[−5 00 15

]P =

[1√10

3√10

− 3√10

1√10

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Another Example

So, we have

D =

[−5 00 15

]P =

[1√10

3√10

− 3√10

1√10

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

Example

Find the orthogonal diagonalization of

A =

11 −1 −1−1 11 −1−1 −1 11

|A− λI | =

∣∣∣∣∣∣11− λ −1 −1−1 11− λ −1−1 −1 11− λ

∣∣∣∣∣∣= 1296− 360λ+ 33λ2 − λ3 = 0

λ = 9, 12, 12

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

Example

A =

11 −1 −1−1 11 −1−1 −1 11

|A− λI | =

∣∣∣∣∣∣11− λ −1 −1−1 11− λ −1−1 −1 11− λ

∣∣∣∣∣∣

= 1296− 360λ+ 33λ2 − λ3 = 0

λ = 9, 12, 12

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

Example

A =

11 −1 −1−1 11 −1−1 −1 11

|A− λI | =

∣∣∣∣∣∣11− λ −1 −1−1 11− λ −1−1 −1 11− λ

∣∣∣∣∣∣= 1296− 360λ+ 33λ2 − λ3 = 0

λ = 9, 12, 12

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5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

Example

A =

11 −1 −1−1 11 −1−1 −1 11

|A− λI | =

∣∣∣∣∣∣11− λ −1 −1−1 11− λ −1−1 −1 11− λ

∣∣∣∣∣∣= 1296− 360λ+ 33λ2 − λ3 = 0

λ = 9, 12, 12

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6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

Now the normalized vectors: For λ = 9

A− 9I =

2 −1 −1−1 2 −1−1 1 2

∼

1 0 −10 1 −10 0 0

111

and so the

1√31√31√3

.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

A− 9I =

2 −1 −1−1 2 −1−1 1 2

∼1 0 −1

0 1 −10 0 0

111

and so the

1√31√31√3

.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

A− 9I =

2 −1 −1−1 2 −1−1 1 2

∼1 0 −1

0 1 −10 0 0

111

and so the

1√31√31√3

.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

For λ = 12, what are we expecting?

2 vectors in the bases ...

A− 12I =

−1 −1 −1−1 −1 −1−1 −1 −1

∼1 1 1

0 0 00 0 0

The eigenvectors associated with λ = 12 are−1

10

,−1

01

and so the normalized vectors are−

1√2

0

,−

1√2

01√2

.

Are we good?

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

For λ = 12, what are we expecting? 2 vectors in the bases ...

A− 12I =

−1 −1 −1−1 −1 −1−1 −1 −1

∼

1 1 10 0 00 0 0

10

,−1

01

1√2

0

,−

1√2

01√2

.

Are we good?

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

A− 12I =

−1 −1 −1−1 −1 −1−1 −1 −1

∼1 1 1

0 0 00 0 0

10

,−1

01

1√2

0

,−

1√2

01√2

.

Are we good?

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

A− 12I =

−1 −1 −1−1 −1 −1−1 −1 −1

∼1 1 1

0 0 00 0 0

10

,−1

01

1√2

0

,−

1√2

01√2

.

Are we good?

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

A− 12I =

−1 −1 −1−1 −1 −1−1 −1 −1

∼1 1 1

0 0 00 0 0

10

,−1

01

1√2

0

,−

1√2

01√2

.

Are we good?

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

−

1√2

0

,−

1√2

01√2

We have a basis, but these vectors are not orthogonal. Sohow do we handle this?

Gram-Schmidt ...

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

−

1√2

0

,−

1√2

01√2

We have a basis, but these vectors are not orthogonal. Sohow do we handle this?

Gram-Schmidt ...

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

v1 =

−1√2

1√2

0

v2 =

−1√2

01√2

− projv1

−

1√2

01√2

=

−1√2

01√2

− 12

1

−1√2

1√2

0

=

−1

2√2

− 12√2

1√2

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

v1 =

−1√2

1√2

0

v2 =

−1√2

01√2

− projv1

−

1√2

01√2

=

−1√2

01√2

− 12

1

−1√2

1√2

0

=

−1

2√2

− 12√2

1√2

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

v1 =

−1√2

1√2

0

v2 =

−1√2

01√2

− projv1

−

1√2

01√2

=

−1√2

01√2

− 12

1

−1√2

1√2

0

=

−1

2√2

− 12√2

1√2

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

v1 =

−1√2

1√2

0

v2 =

−1√2

01√2

− projv1

−

1√2

01√2

=

−1√2

01√2

− 12

1

−1√2

1√2

0

=

−1

2√2

− 12√2

1√2

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

v1 =

−1√2

1√2

0

v2 =

−1√2

01√2

− projv1

−

1√2

01√2

=

−1√2

01√2

− 12

1

−1√2

1√2

0

=

−1

2√2

− 12√2

1√2

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

v1 =

−1√2

1√2

0

v2 =

−1√2

01√2

− projv1

−

1√2

01√2

=

−1√2

01√2

− 12

1

−1√2

1√2

0

=

−1

2√2

− 12√2

1√2

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

Now, we need to normalize this vector again.

1√34

−1

2√2

− 12√2

1√2

=2√3

−1

2√2

− 12√2

1√2

=

−1√6

− 1√6

2√6

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

Now, we need to normalize this vector again.

1√34

−1

2√2

− 12√2

1√2

=2√3

−1

2√2

− 12√2

1√2

=

−1√6

− 1√6

2√6

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

One More of These

So, we have

P =

1√3− 1√

2− 1√

61√3

1√2− 1√

61√3

0 2√6

D =

9 0 00 12 00 0 12

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

What Do You Notice ...

If we look at the matrices we used for the last 3 examples,what do they have in common?

[−2 22 1

] [13 66 −3

] 11 −1 −1−1 11 −1−1 −1 11

They are all symmetric. Is this a coincidence?

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

What Do You Notice ...

If we look at the matrices we used for the last 3 examples,what do they have in common?

[−2 22 1

] [13 66 −3

] 11 −1 −1−1 11 −1−1 −1 11

They are all symmetric. Is this a coincidence?

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

The Spectral Theorem

A matrix is orthogonally diagonalizable iff it is symmetric.

The proof of this is omitted in the text and is beyond thescope of an undergraduate class. If interested, you can lookat Schur’s Theorem, which deals with unitary matrices andupper triangular matrices with eigenvalues on the maindiagonal. A unitary matrix is a matrix U such that UUT = Ifor real matrices.

1 Introduction

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6 QR Example 1

7 QR Example 2

8 QR Example 3

The Spectral Theorem

A matrix is orthogonally diagonalizable iff it is symmetric.

The proof of this is omitted in the text and is beyond thescope of an undergraduate class. If interested, you can lookat Schur’s Theorem, which deals with unitary matrices andupper triangular matrices with eigenvalues on the maindiagonal. A unitary matrix is a matrix U such that UUT = Ifor real matrices.

1 Introduction

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6 QR Example 1

7 QR Example 2

8 QR Example 3

Consequences of the Spectral Theorem

I Eigenvalues of symmetric matrices are always real.

I The dimension of an eigenspace for a symmetric matrixis always equal to the multiplicity of the eigenvalue as aroot of the characteristic equation.

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6 QR Example 1

7 QR Example 2

8 QR Example 3

Consequences of the Spectral Theorem

I Eigenvalues of symmetric matrices are always real.

I The dimension of an eigenspace for a symmetric matrixis always equal to the multiplicity of the eigenvalue as aroot of the characteristic equation.

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6 QR Example 1

7 QR Example 2

8 QR Example 3

QR Factorization

We can think of diagonalization as a way to factor a matrix.But, we have seen that there are conditions for diagonalizinga matrix and even more for orthogonally diagonalizing amatrix. When these fail, we can look to Jordan form for afactorization. But all of these rely on one common fact: thematrix must be square.

What if the matrix is not square? Is there a way to factor it?

If we have an n ×m matrix A with linearly independentcolumns a1, . . . , am, then we can apply the Gram-Schmidtorthonormalization process to these vectors and thisamounts to factoring the matrix. Note that n is notnecessarily equal to m.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

QR Factorization

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

QR Factorization

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

QR Factorization

QR FactorizationSuppose A = [a1 a2 . . . am] is an n ×m matrix withlinearly independent columns. Then there exists an n ×mmatrix with orthonormal columns Q = [q1 q2 . . . qm] andan m ×m upper triangular matrix R whose diagonal entriesare positive such that A = QR.

So how do we find this Q and R?

The Process

I Apply the Gram-Schmidt process to the columns of Ato find the orthonormal basis {q1,q2, . . . ,qm}

I Take these as the columns of Q

I Solve for R using matrix multiplication in the equationR = QTA

Why are we not using Q−1?

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

QR Factorization

The Process

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

QR Factorization

The Process

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

QR Factorization

The Process

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

QR Factorization

The Process

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

QR Factorization

The Process

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

Example

Find the QR factorization of A =

[1 −12 1

].

First, we perform the Gram-Schmidt process on thesevectors.

v1 =

[12

]v2 = s2 − projv1s2

=

[−11

]− 1

5

[12

]=

1

5

[−63

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

Example

[1 −12 1

].

v1 =

[12

=

[−11

]− 1

5

[12

]=

1

5

[−63

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

Example

[1 −12 1

].

v1 =

[12

=

[−11

]− 1

5

[12

]=

1

5

[−63

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

Example

[1 −12 1

].

v1 =

[12

]

v2 = s2 − projv1s2

=

[−11

]− 1

5

[12

]=

1

5

[−63

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

Example

[1 −12 1

].

v1 =

[12

]v2 =

s2 − projv1s2

=

[−11

]− 1

5

[12

]=

1

5

[−63

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

Example

[1 −12 1

].

v1 =

[12

=

[−11

]− 1

5

[12

]=

1

5

[−63

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

Example

[1 −12 1

].

v1 =

[12

=

[−11

]−

1

5

[12

]=

1

5

[−63

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

Example

[1 −12 1

].

v1 =

[12

=

[−11

]− 1

5

[12

]

=1

5

[−63

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

Example

[1 −12 1

].

v1 =

[12

=

[−11

]− 1

5

[12

]=

1

5

[−63

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

So, we have Q =

[1 −6

52 3

5

]?

Why not?

v1‖v1‖

=1√5

[12

]=

[1√52√5

]v2‖v2‖

=1√95

[−6

535

]

=

[− 2√

51√5

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

So, we have Q =

[1 −6

52 3

5

]?

Why not?

v1‖v1‖

=1√5

[12

]=

[1√52√5

]v2‖v2‖

=1√95

[−6

535

]

=

[− 2√

51√5

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

So, we have Q =

[1 −6

52 3

5

]?

Why not?

v1‖v1‖

=

1√5

[12

]=

[1√52√5

]v2‖v2‖

=1√95

[−6

535

]

=

[− 2√

51√5

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

So, we have Q =

[1 −6

52 3

5

]?

Why not?

v1‖v1‖

=1√5

[12

]=

[1√52√5

]

v2‖v2‖

=1√95

[−6

535

]

=

[− 2√

51√5

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

So, we have Q =

[1 −6

52 3

5

]?

Why not?

v1‖v1‖

=1√5

[12

]=

[1√52√5

]v2‖v2‖

=

1√95

[−6

535

]

=

[− 2√

51√5

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

So, we have Q =

[1 −6

52 3

5

]?

Why not?

v1‖v1‖

=1√5

[12

]=

[1√52√5

]v2‖v2‖

=1√95

[−6

535

]

=

[− 2√

51√5

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

So, we have Q =

[1 −6

52 3

5

]?

Why not?

v1‖v1‖

=1√5

[12

]=

[1√52√5

]v2‖v2‖

=1√95

[−6

535

]

=

[− 2√

51√5

]

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

So, we have Q =

[1√5− 2√

52√5

1√5

]. Now what?

R = QTA =

[1√5

2√5

− 2√5

1√5

] [1 −12 1

]=

[5√5

1√5

0 3√5

]

So, we have

QR =

[1√5− 2√

52√5

1√5

][√5 1√

5

0 3√5

]=

[1 −12 1

]= A

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

So, we have Q =

[1√5− 2√

52√5

1√5

]. Now what?

R = QTA

=

[1√5

2√5

− 2√5

1√5

] [1 −12 1

]=

[5√5

1√5

0 3√5

]

So, we have

QR =

[1√5− 2√

52√5

1√5

][√5 1√

5

0 3√5

]=

[1 −12 1

]= A

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

So, we have Q =

[1√5− 2√

52√5

1√5

]. Now what?

R = QTA =

[1√5

2√5

− 2√5

1√5

] [1 −12 1

]

=

[5√5

1√5

0 3√5

]

So, we have

QR =

[1√5− 2√

52√5

1√5

][√5 1√

5

0 3√5

]=

[1 −12 1

]= A

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

So, we have Q =

[1√5− 2√

52√5

1√5

]. Now what?

R = QTA =

[1√5

2√5

− 2√5

1√5

] [1 −12 1

]=

[5√5

1√5

0 3√5

]

So, we have

QR =

[1√5− 2√

52√5

1√5

][√5 1√

5

0 3√5

]=

[1 −12 1

]= A

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 1

So, we have Q =

[1√5− 2√

52√5

1√5

]. Now what?

R = QTA =

[1√5

2√5

− 2√5

1√5

] [1 −12 1

]=

[5√5

1√5

0 3√5

]

So, we have

QR =

[1√5− 2√

52√5

1√5

][√5 1√

5

0 3√5

]=

[1 −12 1

]= A

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

Example

1 1 01 0 10 1 1

.

Remember the process:

1. Apply the Gram-Schmidt process to the columns

2. Normalize the vectors

3. Take these vectors as the columns of Q

4. Multiply QTA to find R

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

Example

1 1 01 0 10 1 1

.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

Example

1 1 01 0 10 1 1

.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

Example

1 1 01 0 10 1 1

.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

Example

1 1 01 0 10 1 1

.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

Example

1 1 01 0 10 1 1

.

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

v1 =

110

v2 =

101

− v1 · s2‖v1‖2

110

=

101

− 1

2

110

=

12−1

21

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

v1 =

110

v2 =

101

− v1 · s2‖v1‖2

110

=

101

− 1

2

110

=

12−1

21

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

v1 =

110

v2 =

101

− v1 · s2‖v1‖2

110

=

101

− 1

2

110

=

12−1

21

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

v1 =

110

v2 =

101

− v1 · s2‖v1‖2

110

=

101

− 1

2

110

=

12−1

21

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

v1 =

110

v2 =

101

− v1 · s2‖v1‖2

110

=

101

− 1

2

110

=

12−1

21

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

v1 =

110

v2 =

101

− v1 · s2‖v1‖2

110

=

101

− 1

2

110

=

12−1

21

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

v3 =

011

− s3 · v1‖v1‖2

110

− s3 · v2‖v2‖2

12−1

21

=

011

− 1

2

110

− 1232

12−1

21

=

011

−1

2120

− 1

6−1

613

=

−23

2323

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

v3 =

011

− s3 · v1‖v1‖2

110

− s3 · v2‖v2‖2

12−1

21

=

011

− 1

2

110

− 1232

12−1

21

=

011

−1

2120

− 1

6−1

613

=

−23

2323

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

v3 =

011

− s3 · v1‖v1‖2

110

− s3 · v2‖v2‖2

12−1

21

=

011

− 1

2

110

− 1232

12−1

21

=

011

−1

2120

− 1

6−1

613

=

−23

2323

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

v3 =

011

− s3 · v1‖v1‖2

110

− s3 · v2‖v2‖2

12−1

21

=

011

− 1

2

110

− 1232

12−1

21

=

011

−1

2120

− 1

6−1

613

=

−23

2323

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

v3 =

011

− s3 · v1‖v1‖2

110

− s3 · v2‖v2‖2

12−1

21

=

011

− 1

2

110

− 1232

12−1

21

=

011

−1

2120

− 1

6−1

613

=

−23

2323

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

Now, we normalize ...

q1 =

1√2

110

=

1√21√2

0

q2 =

1√32

12−1

21

=

1√6

− 1√6

2√6

q3 =

1√43

−23

2323

=

−1√3

1√31√3

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

q1 =1√2

110

=

1√21√2

0

q2 =

1√32

12−1

21

=

1√6

− 1√6

2√6

q3 =

1√43

−23

2323

=

−1√3

1√31√3

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

q1 =1√2

110

=

1√21√2

0

q2 =1√32

12−1

21

=

1√6

− 1√6

2√6

q3 =

1√43

−23

2323

=

−1√3

1√31√3

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

q1 =1√2

110

=

1√21√2

0

q2 =

1√32

12−1

21

=

1√6

− 1√6

2√6

q3 =

1√43

−23

2323

=

−1√3

1√31√3

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

q1 =1√2

110

=

1√21√2

0

q2 =

1√32

12−1

21

=

1√6

− 1√6

2√6

q3 =

1√43

−23

2323

=

−1√3

1√31√3

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

q1 =1√2

110

=

1√21√2

0

q2 =

1√32

12−1

21

=

1√6

− 1√6

2√6

q3 =1√43

−23

2323

=

−1√3

1√31√3

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

q1 =1√2

110

=

1√21√2

0

q2 =

1√32

12−1

21

=

1√6

− 1√6

2√6

q3 =

1√43

−23

2323

=

−1√3

1√31√3

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

q1 =1√2

110

=

1√21√2

0

q2 =

1√32

12−1

21

=

1√6

− 1√6

2√6

q3 =

1√43

−23

2323

=

−1√3

1√31√3

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

So, we have

Q =

1√2

1√6− 1√

31√2− 1√

61√3

0 2√6

1√3

and

R = QTA =

2√2

1√2

0 3√6

1√6

0 0 2√3

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 2

So, we have

Q =

1√2

1√6− 1√

31√2− 1√

61√3

0 2√6

1√3

and

R = QTA =

2√2

1√2

0 3√6

1√6

0 0 2√3

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 3

Example

Find the QR factorization of

A =

1 2 02 3 10 1 −11 4 4

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 3

Example

A =

1 2 02 3 10 1 −11 4 4

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 3

Example

A =

1 2 02 3 10 1 −11 4 4

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 3

Example

A =

1 2 02 3 10 1 −11 4 4

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 3

Example

A =

1 2 02 3 10 1 −11 4 4

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 3

Example

A =

1 2 02 3 10 1 −11 4 4

1 Introduction

5 QR Factorization

6 QR Example 1

7 QR Example 2

8 QR Example 3

Example 3

Jumping to the punchline ...

Q =

1√6

0 − 1√6√

23 − 1√

60

0 1√6−√

23

1√6

√23

1√6

R =

√

6 2√

6√

6

0√

6√

6

0 0√

6

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