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TERM PAPER 

SUB:- ENGINEERING MATEHMATICS

TOPIC:- For a given matrix A, prove that trace of A

equals the sum of its eigen values. What will

be the Eigen values of A-kI ? Also give examples.

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SUBMITED TO:-SUBMITED BY:-

Mr.GURPREET SINGH BHATIANAME:-MAAN

ROLL.NO:-RF4002A50

CERTIFICATE

THIS IS TO CERTIFIED THAT THE TERM PAPER HAS BEEN PREPARED BY “Maan ” {ROLLNO:RF4002A50} UNDER OUR SUPERVISION AND GUIDANCE. HIS EFFORT IS

SATISFACTORY AND WE APPRECIATE IT.

Sig.of teacher:

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ACKNOWLEDGEMENT 

IT IS THE MATTER OF GREAT PLEASURE FOR METO ACKNOWLEDGE MYINDEBTNESS TO MY TEACHER Lect.Mr.GURPREET SINGH BHATIA (DEPTT. OFMATHEMATICS), LOVELY PROFESSIONAL UNIVERSITY, PHAGWARA(PB).

FOR LENDING HIS HELPING HAND IN THECOMPLITION OF MY TERM PAPER. HE INSPIREDME THROUGH OUT THE ACADEMIC SEMSTER 

 AND HELPED ME IN FINDING THE SUITABLETERM PAPER FOR ME. VALUABLE GUIDENESS,ENCOURAGEMENT AND THE FACILITIES

PROVIDED TO ME DURING THE PREPRATION OFMY TERM PAPER ARE VERY HELPFUL TO ME.

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Introduction to

Eigen Values-:

In mathematics, and in particular in linear algebra, an important tool for describing eigenvalues of square matrices is the characteristic polynomial: saying that λ  is an eigenvalue of A is equivalent tostating that the system of linear equations (A - λ I) v =

0 (where I is the identity matrix) has a non-zerosolution v (namely an eigenvector ), and so it isequivalent to the determinant det (A - λ I) being zero.The function p(λ ) = det (A - λ I) is a  polynomial in λ  since determinants are defined as sums of products.

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This is the characteristic polynomial of A: theeigenvalues of a matrix are the zeros of its characteristic

 polynomial. 

It follows that we can compute all the eigenvalues of amatrix A by solving the equation p A(λ ) =0.If A is an n-by-n matrix, then p A has degree n and Acan therefore have at most n eigenvalues. Conversely,

 the  fundamental theorem of algebra says that this

equation has exactly n roots (zeroes), counted withmultiplicity. All real polynomials of odd degree have areal number as a root, so for odd n, every real matrixhas at least one real eigenvalue. In the case of a realmatrix, for even and odd n, the non-real eigenvaluescome in conjugate pairs.

 An example of a matrix with no real eigenvalues is the

  90-degree rotation

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whose characteristic polynomial is x2 + 1 and so itseigenvalues are the pair of complex conjugates i, -i.

The Cayley–Hamilton theorem states that every squarematrix satisfies its own characteristic polynomial, thatis, p A(A) = 0.

Types-:

Eigenvalues of 2×2 matrices

 An analytic solution for the eigenvalues of 2×2matrices can be obtained directly from the quadratic

 formula: if 

 then the characteristic polynomial is

so the solutions are-

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Notice that the characteristic polynomial of a 2×2matrix can be written in terms of the trace tr(A) = a +d and determinant det(A) = ad − bc as

where I2 is the 2×2 identity matrix. The solutions for  the eigenvalues of a 2×2 matrix can thus be written as

Thus, for the very special case where the 2×2 matrix has zero determinant, but non-zero trace, the eigenvaluesare zero and the trace (corresponding to the negativeand positive roots, respectively). For example, theeigenvalues of the following matrix are 0 and (a2 + b2):

It cannot be stressed enough that this formula holds for only a 2×2 matrix.

Eigenvalues of 3×3 matrices-:

If 

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 then the characteristic polynomial of A is

 Alternatively the characteristic polynomial of a 3×3matrix can be written in terms of the  trace tr(A) anddeterminant det(A) as

where I3 is the 3×3 identity matrix. 

The eigenvalues of the matrix are the roots of this polynomial, which can be found using the method for solving cubic equations.

 A formula for the eigenvalues of a 4×4 matrix could bederived in an analogous way, using the formulae for  the solutions of the quartic equation.

Trace (linear algebra)-:

In linear algebra, the trace of an n-by-n square matrix

 A is defined to be the sum of the elements on the main

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diagonal (the diagonal from the upper left to the lower 

right) of A, i.e.,

where aii represents the entry on the ith row and ithcolumn of A. Equivalently, the trace of a matrix is thesum of its eigenvalues, making it an invariant withrespect to a change of basis. This characterization can

be used to define the trace for a linear operator ingeneral. Note that the trace is only defined for a squarematrix (i.e. n×n).

Examples

Let T be a linear operator represented by the matrix

Then tr(T ) = −2 + 1 − 1 = −2. 

The trace of the identity matrix is the dimension of the

space; this leads to generalizations of dimension using trace. The trace of a projection (i.e., P2 = P) is the rankof the projection. The trace of a nilpotent matrix is zero.

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The product of a symmetric matrix and a skew-symmetric matrix has zero trace.

More generally, if f(x) = (x − λ 1)d1···(x − λ k)dk is thecharacteristic polynomial of a matrix A, then

If A and B are  positive semi-definite matrices of thesame order then

Properties-:

The trace is a linear map. That is,

 tr(A + B) = tr(A) + tr(B),

 for all square matrices A and B, and all scalars c.

If A is an m×n matrix and B is an n×m matrix, then

 tr(AB) = tr(BA).Conversely, the above properties characterize the tracecompletely in the sense as follows. Let f be a linear 

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 functional on the space of square matrices satisfying f(xy) = f(yx). Then f and tr are proportional.

The trace is similarity-invariant, which means that Aand P−1 AP have the same trace. This is because

 tr(P − 1 AP) = tr(P − 1(AP)) = tr((AP)P − 1) =

 tr(A).

 A matrix and its transpose have the same trace:

 tr(A) = tr(AT ).

Let A be a symmetric matrix, and B an anti-symmetric matrix. Then

 tr(AB) = 0.

When both A and B are n by n, the trace of the (ring- theoretic) commutator  of A and B vanishes: tr([A, B]) = 0; one can state this as "the trace is a mapof Lie algebras from operators to scalars", as thecommutator of scalars is trivial (it is an abelian Liealgebra). In particular, using similarity invariance, it

 follows that the identity matrix is never similar to thecommutator of any pair of matrices.

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Conversely, any square matrix with zero trace is thecommutator of some pair of matrices. Moreover, anysquare matrix with zero trace is unitarily equivalent to

a square matrix with diagonal consisting of all zeros.

The trace of any power of a nilpotent matrix is zero.When the characteristic of the base field is zero, theconverse also holds: if tr(xk) = 0 for all k, then x isnilpotent.

Note that order does matter in taking traces: in general,

In other words, we can only interchange the two halvesof the expression, albeit repeatedly. This means that the

 trace is invariant under cyclic permutations, i.e.,

 tr(ABCD) = tr(BCDA) = tr(CDAB) = tr(DABC).

However, if products of three symmetric (or, moregenerally, Hermitian) matrices are considered, any

 permutation is allowed. (Proof: tr(ABC) = tr(AT  BT  CT )

= tr((CBA)T 

) = tr(CBA).) For more than three factors this is not true. This is known as the cyclic property.

Unlike the determinant, the trace of the product is not the product of traces. What is true is that the trace of the

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 tensor product of two matrices is the product of their  traces:

The trace of a product can be rewritten as the sum of allelements from a Hadamard product (entry-wise

 product):

.This should be more computationally efficient, since thematrix product of an matrix with an one(first and last dimensions must match to give a squarematrix for the trace) has mn2 multiplications and n +(m − 1)n2 additions, whereas the computation of the

Hadamard version (entry-wise product) requires onlynm multiplications followed by nm additions.

The exponential trace-:

Expressions like exp(tr(A)), where A is a square matrix,occur so often in some fields (e.g. multivariatestatistical theory), that a shorthand notation hasbecome common:

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This is sometimes referred to as the exponential trace function.

Trace of a linear operator-:

Given some linear map f : V → V (V is a finite-dimensional vector space) generally, we can define the

 trace of this map by considering the trace of  matrixrepresentation of f, that is, choosing a basis  for V anddescribing f as a matrix relative to this basis, and

 taking the trace of this square matrix. The result willnot depend on the basis chosen, since different baseswill give rise to similar matrices, allowing for the

 possibility of a basis independent definition for the trace of a linear map.

Such a definition can be given using the canonicalisomorphism between the space End(V) of linear mapson V and V⊗V*, where V* is the dual space of V. Let v bein V and let f be in V*. Then the trace of thedecomposable element v⊗ f is defined to be f(v); the

 trace of a general element is defined by linearity.

Using an explicit basis for V and the correspondingdual basis for V*, one can show that this gives the samedefinition of the trace as given above.

Eigen value relationships-:

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If A is a square n-by-n matrix with real or  complexentries and if  λ 1,...,λ n are the (complex and distinct)eigenvalues of A (listed according to their  algebraicmultiplicities), then

This follows from the fact that A is always similar toits  Jordan form, an upper  triangular matrix havingλ 

1,...,λ 

n on the main diagonal. In contrast, thedeterminant of A is the product of its eigenvalues; i.e.,

More generally,

Derivatives-:

The trace is the derivative of the determinant: it is theLie algebra analog of the (Lie group) map of thedeterminant. This is made precise in  Jacobi's formula

 for the derivative of the determinant (see under determinant). As a particular case, : the trace is

 the derivative of the determinant at the identity. From this (or from the connection between the trace and theeigenvalues), one can derive a connection between the

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 trace function, the exponential map between a Liealgebra and its Lie group (or concretely, the matrixexponential function), and the determinant: 

det(exp(A)) = exp(tr(A)).

For example, consider the one-parameter family of linear transformations given by rotation through angleθ,

These transformations all have determinant 1, so they preserve area. The derivative of this family at θ = 0 is the antisymmetric matrix

which clearly has trace zero, indicating that this matrixrepresents an infinitesimal transformation which

 preserves area.

 A related characterization of the trace applies to linear vector fields. Given a matrix A, define a vector field Fon R n by F(x) = Ax. The components of this vector fieldare linear functions (given by the rows of A). Thedivergence div F is a constant function, whose value is

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equal to tr(A). By the divergence theorem, one caninterpret this in terms of flows: if F(x) represents thevelocity of a fluid at the location x, and U is a region in

R n, the net flow of the fluid out of U is given by tr(A)·vol(U), where vol(U) is the volume of U.

The trace is a linear operator, hence its derivative isconstant:

Examples-:

1.  Any matrix A, with eigen values ai is similar to its jordan form which is an upper triangular matrix-:

a1 b1 c1

0 a2 c2

0 0 a3

and since similar matrices have same trace value, the

 trace of matrix A = sum of the eigen value.

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