When the problem says "repeated according to multiplicities" does that mean there is one value of lambda that is repeated $n$ times? I'm confused on why the determinant of $A$ must be the product of the $n$ eigenvalues of $A$. Couldn't lambda be any value? I only see how the determinant of $A$ would equal the $n$ eigenvalues if lambda is 0. Why is this result true when complex eigenvalues are considered?

Prove that the determinant of an $n × n$ matrix A is the product of the eigenvalues (counted according to their algebraic multiplicities). Hint: Write the characteristic polynomial as $p(\lambda) = (\lambda_1 − \lambda)(\lambda_2 − \lambda)· · ·(\lambda_n − \lambda)$.

Solution: If the eigenvalues of $A$ are $\lambda_1, . . . , \lambda_n$ (counted with algebraic multiplicity), then as the hint says, the characteristic polynomial of $A$ is $det(A − \lambda I) = (\lambda_1 − > \lambda)(\lambda_2 − \lambda)...(\lambda_n − \lambda)$. Plugging in $\lambda = 0$ yields $det A = \lambda_1\lambda_2... \lambda_n$.

  • $\begingroup$ Hint: Jordan decomposition/diagonalization $\endgroup$ – Jacky Chong Mar 30 '17 at 5:50
  • $\begingroup$ Meaning set $A$ equal to $PDP^{-1}$? $\endgroup$ – stumped Mar 30 '17 at 6:09

Repeated according to multiplicity

This simply means that we want a copy of $(\lambda_r-\lambda)$ for each time that $\lambda_r$ pops up as an eigenvalue. For an example, consider $A=\begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix}$. $3$ is an eigenvalue with (algebraic) multiplicity two. We want $\det(A-\lambda I) = (3- \lambda)(3-\lambda) = (3-\lambda)^2$ and NOT simply $(3-\lambda)$.

$\det(A- \lambda I)$ is a polynomial

You are absolutely correct, $\lambda$ can have any value, just as $x$ can have any value in $f(x) = x^2 - 4 = (x+2)(x-2)$. What makes eigenvalues special is that they are the roots of the polynomial $\det(A-\lambda I)$, i.e., the special values of lambda that give $\det(A-\lambda I) =0$. When factoring this polynomial as given in your question, it should be clear what the roots are, namely $\{ \lambda_1, \lambda_2, \dots, \lambda_n \}$.

$\det(A)$ = product of eigenvalues

What happens when we evaluate this polynomial at $\lambda=0$, just as we might evaluate the polynomial $f(x) = x^2-4$ at $x=0$ to get $f(0) = -4$? \begin{align*} \det(A - 0 I) &= (\lambda_1 - 0) (\lambda_2 - 0) \cdots (\lambda_n - 0) \\ &= \lambda_1 \cdot \lambda_2 \cdots \lambda_n \end{align*} However, $\det(A-0I) = \det(A - 0) = \det(A)$, so we have $$\det(A) = \lambda_1 \cdot \lambda_2 \cdots \lambda_n.$$

Again, notice the importance of listing repeated eigenvalues multiple times. For example, $\det \left( \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} \right) = 9$ and this matrix has a repeated eigenvalue of $3$. We see that $3\cdot 3 =9$, but if we only listed this eigenvalue once we would have $3 \neq 9$.

  • $\begingroup$ Thank you for your answer! But what if in the case we didn't make $\lambda=0$? If we instead had $\lambda=4$? So then we have $(\lambda_1− 4)(\lambda_2 − 4)...(\lambda_n-4)$. I'm confused on how setting $\lambda=0$ solves the problem because doesn't this only work for that one case where $\lambda=0$? $\endgroup$ – stumped May 15 '17 at 4:24
  • $\begingroup$ Think of it like a good old quadratic $f(x) = x^2 + b x + c = (x-r_1)(x-r_2)$. The $y$-intercept of $f$ is defined to be $f(0) = c = (-r_1)(-r_2)$. If we evaluated this polynomial at $x=4$, then we have $f(4) = 4^2 + b * 4 + c = (4-r_1)(4-r_2)$ which is just...$f(4)$; it's not still the $y$-intercept. Similarly, if we evaluate $\det(A-\lambda I)$ at $\lambda = 4$, we just have this polynomial evaluated at some random input and no longer get $\det(A)$ as the output. $\endgroup$ – erfink May 15 '17 at 4:33
  • $\begingroup$ For another explanation, $A = A-0 = A- 0 I$, but $A \neq A - 4I$. As such $\det(A) = \det(A-0I)$, but $\det(A) \neq \det(A - 4I)$. $\endgroup$ – erfink May 15 '17 at 4:36

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