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Is it always true that product of eigenvalues is determinant of a matrix ? what if one of the eigenvalues are same and matrix is not diagonalizable ? Is this statement is still true ?

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  • $\begingroup$ Yes, if you count each eigenvalue with suitable multiplicity. $\endgroup$ – Angina Seng Aug 31 '17 at 8:32
  • $\begingroup$ Do you know what a Jordan form is? $\endgroup$ – 5xum Aug 31 '17 at 8:37
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    $\begingroup$ More precisely, determinant is the product of all zeroes of the characteristic polynomial, multiplicity included. These are not necessarily eigenvalues, as the underlying field does not have to be algebraically closed. The proof in general uses Vieta's formulas on the zeroes of the characteristic polynomial, not diagonalizing the matrix and then using the fact that the determinant is a similarity invariant (although that proof works for diagonalizable matrices). $\endgroup$ – mechanodroid Aug 31 '17 at 9:16
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$\DeclareMathOperator{\Tr}{Tr}$ The determinant is the product of the zeroes of the characteristic polynomial (counting with their multiplicity), and the trace is their sum, regardless of diagonalizability of the matrix. If the underlying field is algebraically closed (such as $\mathbb{C}$), then those zeroes will exactly be the eigenvalues.

Proof:

Let $k_A$ be the characteristic polynomial $k_A(\lambda)=\det(A-\lambda I)$ of the $n\times n$ matrix $A$. It's expanded form is $k_A(\lambda)=a_n\lambda^n + a_{n-1}\lambda^{n-1} + \cdots + a_1\lambda + a_0$, noting that the degree of the polynomial is $n$ because the determinant is a sum of products of $n$ matrix entries (each entry having at most one $\lambda$).

Observe that $k_A(0) = \det(A)$, so $a_0 = \det(A)$.

Notice that the only way to get the power $\lambda^n$ as a product of $n$ matrix entries, taking exactly one from every row and column, is to choose the entries on the main diagonal: $(a_{11} - \lambda)(a_{22} - \lambda)\cdots (a_{nn} - \lambda)$. This yields $(-1)^n\lambda^n$ when expanded, so $a_n = (-1)^n$.

Now, let's calculate $a_{n-1}$: the power $\lambda^{n-1}$ is yielded also only from the product of main diagonal entries $(a_{11} - \lambda)(a_{22} - \lambda)\cdots (a_{nn} - \lambda)$, when multiplying $n-1$ terms with $-\lambda$, and one $a_{ii}$ term. This is multiplied to $a_{ii}(-1)^{n-1}\lambda^{n-1}$. Doing this for every $i \in \{1, 2, \ldots, n\}$ and summing the results, we get $(a_{11} + a_{22} + \cdots + a_{nn})(-1)^{n-1}\lambda^{n-1} = \Tr(A)(-1)^{n-1}\lambda^{n-1}$, so $a_{n-1} = \Tr(A)(-1)^{n-1}$

So, $k_A(\lambda)=(-1)^n\lambda^n + \Tr(A)(-1)^{n-1}\lambda^{n-1} + \cdots + a_1\lambda + \det(A)$. Let's denote the roots of $k_A$ with $\lambda_1, \ldots, \lambda_n$ (not necessarily distinct).

Now, using Vieta's formulas we have:

$$\lambda_1\lambda_2\cdots\lambda_n = (-1)^n\frac{a_0}{a_n} = \det(A)$$

$$\lambda_1 + \lambda_2 + \cdots + \lambda_n = -\frac{a_{n-1}}{a_n} = \Tr(A)$$

As you probably know, if $A$ is diagonalizable, then we can write:

$$A = P^{-1}\begin{bmatrix} \lambda_1 & 0 & 0 & \ldots & 0 \\ 0 & \lambda_2 & 0 & \ldots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \ldots & \lambda_n \\ \end{bmatrix}P$$

for an invertible matrix $P$. Then, using the fact that $\det$ and $\Tr$ are similarity invariants, we get the desired identities.

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