There is a circle of ideas related to linear independence, span, and basis for vector spaces, which it is good to try and understand carefully. That's what we'll do today. These ideas generalize (and use) ideas we've already explored more concretely, and facts we already know to be true for the vector spaces n. (A vector space that does not have a basis with a finite number of elements is said to be infinite dimensional. For example the space of all polynomials of arbitrarily high degree is an infinite dimensional vector space. We often study finite dimensional subspaces of infinite dimensional vector spaces.)

Theorem 1 (constructing a basis from a spanning set): Let V be a vector space of dimension at least one, and let span v1, v2, ... vp = V. Then a subset of the spanning set is a basis for V. (We followed a procedure like this to extract bases for Col A.)

Theorem 2 Let V be a vector space, with basis = b1, b2, ... bn . Then any set in V containing more than n elements must be linearly dependent. (We used reduced row echelon form to understand this in n.)

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Theorem 3 Let V be a vector space, with basis = b1, b2, ... bn . Then no set = a1, a2, ... ap with p n vectors can span V. (We know this for n.)

Theorem 4 Let V be a vector space, with basis = b1, b2, ... bn . Let = a1, a2, ... ap be a set of

independent vectors that don't span V. Then p n, and additional vectors can be added to the set to create a basis a1, a2, ... ap, ... an (We followed a procedure like this when we figured out all the subspaces of 3.)

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Theorem 5 Let Let V be a vector space, with basis = b1, b2, ... bn . Then every basis for V has exactlyn vectors. (We know this for n.)

Theorem 6 Let Let V be a vector space, with basis = b1, b2, ... bn . If = a1, a2, ... an is

another collection of exactly n vectors in V, and if span a1, a2, ... an = V, then the set is automaticallylinearly independent and a basis. Conversely, if the set a1, a2, ... an is linearly independent, then

span a1, a2, ... an = V is guaranteed, and is a basis. (We know all these facts for n from reduced row echelon form considerations.)

Corollary Let Let V be a vector space of dimension n. Then the subspaces of V have dimensions 0, 1, 2,...n 1, n. (We know this for n.)

Remark We used the coordinate transformation isomorphism between a vector space V with basis = b1, b2, ... bn for Theorem 2, but argued more abstractly for the other theorems. An alternate

(quicker) approach is to just note that because the coordinate transformation is an isomorphism it preservessets of independent vectors, and maps spans of vectors to spans of the image vectors, so maps subspaces to subspaces. Then every one of the theorems above follows from their special cases in n, which we've already proven. But this shortcut shortchanges the conceptual ideas to some extent, which is why we've discussed the proofs more abstractly.

Fri Oct 19 4.6 The four subspaces associated with a matrix. the rank of a matrix.

Announcements:

Warm-up Exercise:

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Math 2270-002Friday October 19

Big example of four fundamental subspaces associated to each matrix. Here is a matrix and it's reduced row echelon form, from quiz 7, and related matrices

A

1 1 0 1 52 3 1 4 82 2 0 2 21 2 3 5 1

row reduces to

1 0 1 1 00 1 1 2 00 0 0 0 10 0 0 0 0

AT =

1 2 2 11 3 2 20 1 0 31 4 2 55 8 2 1

row reduces to

1 0 092

0 1 0 3

0 0 154

0 0 0 00 0 0 0

(A column reduces to

1 0 0 0 00 1 0 0 00 0 1 0 092 3

54 0 0

.)

What are dim Col A , dim Nul A , dim Row A , dim Nul AT ? Can you find bases for each subspace? How could you find these dimensions in general ?

ai ai ai aiai

dim weA 3 of pivotcools in rref A because correspondingcols in A Cai5,51

din NatA 2 of non pivot ols are a prettygood basisof freevariables in for wl Asoltns I to AI Iofbasisvectors in Nnl A

dim row A 3 of pivot rows in rref A same as pivotales

dim Nul ATI I of non pivoturls in rref ATi e pivots

T x AT A is in xn

T IRS 1124 T Pi Rm Tk A xC U UO si Rm Rn Slg Atg

RowA NaeA CRA NulAT

colAT S g ATgs 1124 IRS

We figured out that in general A has m rows n wlums

dim Col A dim RowA this is called the rantofthematrixdim Cola t din Nal A h ofcolumns dimof domain 112dim RowA t din Nal AT m of rows dim ofcodomain