Prove that the trace of a matrix A is equal to the sum of its eigenvalues?

Answer 1

Given a matrix A=([a,b],[c,d]), the trace of A is a+d, and the det of A is ad-bc.

By using the characteristic equation, and representing the eigenvalues with x, we have the equation

x2-(a+d)x+(ad-bc)=0

Which, using the formula for quadratic equations, gives us the eigenvalues as,

x1=[(a+d)+√((a+d)2-4(ad-bc))]/2

x2=[(a+d)-√((a+d)2-4(ad-bc))]/2

now by adding the two eigenvalues together we get:

x1+x2=(a+d)/2+[√((a+d)2-4(ad-bc))]/2+ (a+d)/2-[√((a+d)2-4(ad-bc))]/2

The square roots cancel each other out being the same value with opposite signs, leaving us with:

x1+x2=(a+d)/2+(a+d)/2

x1+x2= 2(a+d)/2

x1+x2=(a+d)

x1+x2=trace(A)

Q.E.D.

General proofThe above answer only works for 2x2 matrices. I'm going to answer it for nxn matrices. (Not mxn; the question only makes sense when the matrix is square.)

The proof uses the following ingredients:

(1) Every nxn matrix is conjugate to an upper-triangular matrix

(2) If A is upper-triangular, then tr(A) is the sum of the eigenvalues of A

(3) If A and B are conjugate, then tr(A) = tr(B)

(4) If A and B are conjugate, then A and B have the same characteristic polynomial (and hence the same sum-of-eigenvalues)

If these are all true, then we can do the following: Given a matrix A, find an upper-triangular matrix U conjugate to A; then (letting s(A) denote the sum of the eigenvalues of A) s(A) = s(U) = tr(U) = tr(A).

Now to prove (1), (2), (3) and (4):

(1) This is an inductive process. First you prove that your matrix is conjugate to one with a 0 in the bottom-left corner. Then you prove that this, in turn, is conjugate to one with 0s at the bottom-left and the one above it. And so on. Eventually you get a matrix with no nonzero entries below the leading diagonal, i.e. an upper-triangular matrix.

(2) Suppose A is upper-triangular, with elements a1, a2, ... , an along the leading diagonal. Let f(t) be the characteristic polynomial of A. So f(t) = det(tI-A). Note that tI-A is also upper-triangular. Therefore its determinant is simply the product of the elements in its leading diagonal. So f(t) = det(tI-A) = (t-a1) * ... * (t-an). And its eigenvalues are a1, ... , an. So the sum of the eigenvalues is a1 + ... + an, which is the sum of the diagonal elements in A.

(3) This is best proved using Summation Convention. Summation convention is a strange but rather useful trick. Basically, the calculations I've written below aren't true as they're written. For each expression, you need to sum over all possible values of the subscripts. For example, where it says b_ii, it really means b11 + b22 + ... + bnn. Where it says bil deltali, it means (b11 delta11 + ... + b1n deltan1) + ... + (bn1 delta1n + ... + bnn deltann). Oh, and deltakj=1 if k=j, and 0 otherwise.

Suppose B = PAP-1. Let's say the element in row j and column k of A is ajk. Similarly, say the (i,j) element of P is pij, the (k,l) element of P-1 is p*kl, and the (i,l) element of B is bil. Then:

bil = pij ajk p*kl

And the trace of B is given by:

tr(B) = bii

= bil deltali

= p*kl deltali pij ajk

= p*kl plj ajk

= deltakj ajk (since p and p* are inverses)

= ajj

= tr(A)

(4) Again, suppose B = PAP-1. Then, for any scalar t, we have tI-B = P(tI-A)P-1. Hence det(tI-B) = det(P).det(tI-A).det(P-1). Since det(P).det(P-1)=det(PP-1)=det(I)=1, we have det(tI-B) = det(tI-A).