# Eigenvalues

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• December 2nd 2008, 08:49 PM
Brokescholar
Eigenvalues
I'm having trouble with these two problems. Thanks guys!

1) Let A be an n x n matrix. Show that det(A) is the product of all the roots of the characteristic polynomial of A.

I get if A is a triangular matrix, how the determinant would equal the roots of the char. polynomial. I don't understand how I would prove that though. And I don't understand how I would figure it out if A is not a triangular matrix.

I know that from det( λIn - A) you can figure out the char. poly. equation and get the roots from that. But how do you tie that in with det(A)? I've been looking through the equations in my book and through the properties of determinants and I can't figure out how to tie these two together.

2) If L:V -> V is a linear transformation, show that L is not one-to-one if and only if 0 is an eigenvalue of L.

What I have so far is that

L(x) = λx and if λ= 0 then
L(x) = 0x
Now to prove that L is one to one we need to know if L(u) = L(v) and u = v
So I plugged in L(u) = 0u and L(v) = 0v which would give
0u = 0v but unless u = v then L is not a one to one function.

I'm just wondering if there's an easier way to do this. Thanks!
• December 3rd 2008, 06:38 AM
ThePerfectHacker
Quote:

Originally Posted by Brokescholar
1) Let A be an n x n matrix. Show that det(A) is the product of all the roots of the characteristic polynomial of A.

I get if A is a triangular matrix, how the determinant would equal the roots of the char. polynomial. I don't understand how I would prove that though. And I don't understand how I would figure it out if A is not a triangular matrix.

I know that from det( λIn - A) you can figure out the char. poly. equation and get the roots from that. But how do you tie that in with det(A)? I've been looking through the equations in my book and through the properties of determinants and I can't figure out how to tie these two together.

You need to remember a factor about polynomials. Let $f(x)$ be a monic non-constant real polynomial with (possibly comples) zeros $z_1,z_2,...,z_n$ (counting multiplicity) then $f(x)=(x-z_1)(x-z_2)...(x-z_n)$. Let $x=0$ and we get $f(0) = (-1)^nz_1z_2...z_n$. We see that the constant term of $f(x)$ is the product of the roots of $f(x)$ (with a plus or minus sign depending on the parity of $n$). Now let $g(x) = \det (xI - A)$. The product of the roots of $g(x)$ is $(-1)^n g(0)$ by the above reasoning. Thus, the product of the roots is $(-1)\cdot \det( - A) = (-1)^n (-1)^n \det (A) = \det (A)$.

Quote:

2) If L:V -> V is a linear transformation, show that L is not one-to-one if and only if 0 is an eigenvalue of L.
If $L$ is not one-to-one then $L(x) = L(y)$ for $x\not = y$. But then $L(x-y) = 0\cdot (x-y)$. Thus, $x-y\not = 0$ is an eigenvalue. The converse is similar.
• December 3rd 2008, 11:35 AM
Brokescholar
Wow I would not have figured out that first part on my own. Thank you so much!