Question

I want as simple or most intuitive proof as possible of the two things: That the determinant of a matrix is equal to the product of its eigenvalues and the trace of a determinant is equal to the sum of its eigenvalues.

Answer 1

For the determinant here is what I have:

A is an nxn matrix and has eigenvalues. We know these eigenvalues are the roots of the characteristic polynomial so we can write this equation: \[\Large \left| A - \lambda I \right|= \prod_{i=1}^n (\lambda_i-\lambda)\] Now we can set lambda = 0 since it is just a variable. \[\Large \left| A \right|= \prod_{i=1}^n \lambda_i\] So that seems to show it pretty well, but it almost seems too easy, like perhaps I have left something out.

As far as the trace of a matrix being the sum of its eigenvalues, that is not at all obvious to me except in the case where the matrix is a diagonal matrix.

Answer 2

only when its diagonal matrices , right ?

Answer 3

No, this is a true fact in general.

Answer 4

well
|A-lambda I | =0 to get lambda , right ?

Answer 5

That would be in general to solve for specific eigenvalues, but if we leave lambda as a variable, remember this is really just a polynomial. The polynomial has zeroes only when lambda is an eigenvalue, so that's why I wrote it out with that product thingy on the right.\[\Large \left| A - \lambda I \right|= p( \lambda)=\prod_{i=1}^n (\lambda_i-\lambda)\]

So you can see for n=2 that we'll have: \[\Large \prod_{i=1}^2 (\lambda_i-\lambda)=(\lambda_1-\lambda)(\lambda_2-\lambda) \] \[\\ \Large =\lambda ^2-(\lambda_1 + \lambda_2)\lambda + \lambda_1 \lambda_2=p(\lambda)\] So really this is just an equivalent way of saying \[\Large p(\lambda)=0\] is the polynomial you solve for the specific eigenvalues.

Answer 6

If I'm using like confusing looking math nonsense tell me so I can either explain the notation or explain it a different way.

Answer 7

im ok with is so far , for 2D seems ok

Answer 8

but im not sure about trace thingy its goes true for diagonal only

Answer 9

Maybe it helps to look at it as this polynomial:\[\Large \left| A-\lambda I \right| = (\lambda_1-\lambda) (\lambda_2-\lambda)... (\lambda_n-\lambda)\] We can plug in any value, we want, like let's choose the last eigenvalue: \[\Large \left| A-\lambda_n I \right| = (\lambda_1-\lambda_n) (\lambda_2-\lambda_n)... (\lambda_n-\lambda_n)\] Sure enough, that last term is 0 so we know that it is indeed an eigenvalue.\[\Large \left| A-\lambda_n I \right| = 0\] And of course this is true in general of any term in the product.

As far as the trace goes, I might have found something.

Answer 10

interesting ....
have u tried to look for counter example ?

Answer 11

Well that's kind of why I'm asking here on OS haha, I can't find a counter example, and it seems pretty solid.

Answer 12

hehe
i agree that
lambda ^2 - (trace ) lambda + det(A)=0
xD for 2x2
if that helps u lol

Answer 13

Here is the best proof I can find so far of the trace:

A is an nxn matrix, \[\Large tr(A)=tr (P^{-1}DP)=tr (DP^{-1}P)=tr(D)\] The only steps I am unsure about are:

What if A is not diagonalizable?

How do I prove this statement \[\Large tr(XY)=tr(YX)\]

Answer 14

ur first claim is lambda1.lambda2=det(A)
(i'll only do it for 2x2)
second claim
trace A= labda1+labda2

Answer 15

so , from general for of quadratic thingy lets find lambda1 and lambda2
to make sure , if it goes will with 2x2
then we can generalize maybe ?

Answer 16

That's a good idea. I guess let's solve for the eigenvalues in terms of the determinant an trace idk?

Answer 17

yes lets try

Answer 18

Looks like this is both of them: \[\Large \lambda_\pm = \frac{tr(A) \pm \sqrt{tr(A)^2-4 \det(A)}}{2}\]

Answer 19

yep lets see whats there product and sum :)

Answer 20

Well since I derived this from: \[\Large \det(A)=(\lambda_+) (\lambda_- )\\ \Large tr(A) = (\lambda_+) +(\lambda_- )\] it should work out haha

Answer 21

wow ! its true for determinant !

Answer 22

haha lol its true for both

Answer 23

amazing approach! i should learn that polynomial u represented !

Answer 24

The first time I saw a polynomial of this type sort of come up was in the proof of \[\Large \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}\]

Answer 25

aha , this can be proved with Fourier and some complex approach , whats ur prove ?
i would like to see !

Answer 26

Here is the proof I saw http://en.wikipedia.org/wiki/Basel_problem#Euler.27s_approach but I think there are others.

Answer 27

i only understand 2 of them

Answer 28

ok 4 of them lol