User:IssaRice/Linear algebra/Determinant of matrix equals product of eigenvalues

This is chapter 4 exercise 1.10 in Linear Algebra Done Wrong (p. 105).

Theorem. Let $A$ be a matrix. Then $\det A=\lambda _{1}\cdots \lambda _{n}$ , where $\lambda _{1},\ldots ,\lambda _{n}$ are the eigenvalues of $A$ (counting multiplicities).

Proof. We know that $\lambda$ is an eigenvalue of $A$ if and only if $\det(A-\lambda I)=0$ . But $\det(A-\lambda I)$ is a polynomial in $\lambda$ by equation 4.2 in the book. This means that $\lambda$ is a root of the polynomial $p(z)=\det(A-zI)$ if and only if $\lambda$ is an eigenvalue of $A$ .

Let's look at equation 4.2 from the book (p. 89) more closely:

$\det A=\sum _{\sigma \in \mathrm {Perm} (n)}a_{\sigma (1),1}a_{\sigma (2),2}\cdots a_{\sigma (n),n}\mathrm {sign} (\sigma )$

The matrix $A-\lambda I$ has the form:

${\begin{pmatrix}a_{1,1}-\lambda &&*\\&\ddots &\\{*}&&a_{n,n}-\lambda \end{pmatrix}}$

where the $*$ are entries containing just numbers. Thus when we use equation 4.2 to compute $\det(A-\lambda I)$ , we obtain the highest degree term (in terms of $\lambda$ ) when we use the identity permutation (i.e. just go down the diagonal of the matrix), getting a degree $n$ term when multiplied out; in fact, the degree $n$ term will have coefficient $(-1)^{n}$ . Every other permutation will take some of the diagonal entries and also some of the number-only entries, resulting in a term that, when multiplied out, gives a degree less than $n$ . This shows not only that $\det(A-\lambda I)$ is a polynomial in $\lambda$ , but in fact a degree $n$ polynomial in $\lambda$ .

So $p(z)=\det(A-zI)$ is an $n$ th degree polynomial, and its roots are exactly the eigenvalues of $A$ . By the fundamental theorem of algebra, $p$ has exactly $n$ complex roots, $\lambda _{1},\ldots ,\lambda _{n}$ , which are also all the eigenvalues of $A$ (counting multiplicities). By the polynomial remainder theorem, this means that $\det(A-\lambda I)$ can be represented as $\det(A-\lambda I)=a_{n}(\lambda _{1}-\lambda )\cdots (\lambda _{n}-\lambda )$ . We want to show that $a_{n}=1$ .

Let's check the coefficient on the $\lambda ^{n}$ term. We already saw above that this equals $(-1)^{n}$ . Multiplying out $a_{n}(\lambda _{1}-\lambda )\cdots (\lambda _{n}-\lambda )$ we get $a_{n}(-1)^{n}$ for the $\lambda ^{n}$ term. Thus the equation $(-1)^{n}=a_{n}(-1)^{n}$ shows that $a_{n}=1$ as desired.

(What if we had written instead $\det(A-\lambda I)=b_{n}(\lambda -\lambda _{1})\cdots (\lambda -\lambda _{n})$ ? Then we would have gotten $b_{n}$ on the $\lambda ^{n}$ term. So we have the equation $b_{n}=(-1)^{n}$ , since we already know that the coefficient on the $\lambda ^{n}$ term has to be $(-1)^{n}$ . Yikes, are we in trouble? Thankfully not! We can take each of the n factors in $(-1)^{n}$ and distribute it to each $(\lambda -\lambda _{j})$ factor to get $(\lambda _{j}-\lambda )$ , so that we have $\det(A-\lambda I)=(\lambda _{1}-\lambda )\cdots (\lambda _{n}-\lambda )$ , which is just the same as before!)

So finally we get that $\det(A-\lambda I)=(\lambda _{1}-\lambda )\cdots (\lambda _{n}-\lambda )$ . Plugging in $\lambda =0$ we get $\det(A)=\lambda _{1}\cdots \lambda _{n}$ , which completes the proof. Alternatively, [TODO: this part I'm not sure about. idk what Treil had in mind here when he says to compare the terms without lambda. In particular, idk how to compute the constant term in det(A - lambda I).]

See also