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I am currently trying to solve this exercise:

Let A be a diagonalisable matrix with eigenvalues $\lambda_1,\lambda_2...\lambda_n.$ Show that $\det A = \prod_{n=1}^n\lambda_i $

I tried solved it for a $3x3$ matrix:

A diagonalisable matrix can be written as:

$$ \begin{pmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{pmatrix} $$

If we compute the determinant by the Rule of Sarrus

we get

$$\lambda_1\times\lambda_2\times\lambda_3 + 0 + 0 - 0 - 0 - 0$$

Therefore we can write:

$A = \prod_{n=1}^3\lambda_i $

My problem is how to show this generally?

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    $\begingroup$ Diagonalizable does not mean diagonal. $\endgroup$ – Julien Oct 29 '13 at 15:42
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    $\begingroup$ If $\lambda_1$ is an eigenvalue of $A$, what does it imply? $\endgroup$ – freak_warrior Oct 29 '13 at 15:45
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    $\begingroup$ @freako This implies: "For any set of distinct eigenvalues of a matrix A, a set of associated eigenvectors, one per eigenvalue, is linearly independent." $\endgroup$ – Kare Oct 29 '13 at 15:52
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    $\begingroup$ Note that your computation of det is only valid for $n=3$. The claim holds for general $n$. $\endgroup$ – Hagen von Eitzen Oct 29 '13 at 15:55
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    $\begingroup$ @user5414 try to use the eigenvalues and eigenvectors to get a diagonalization of the matrix $A$. $\endgroup$ – freak_warrior Oct 29 '13 at 16:03
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If $A$ is diagonalisable (i.e. it has distinct eigenvalues), then we can write: \begin{align*} Av_1 &= \lambda_1v_1 \\ Av_2 &= \lambda_2v_2 \\ & \vdots \\ Av_n &= \lambda_nv_n \end{align*} for distinct eigenvalues $\lambda_1,...,\lambda_n$ and linearly independent eigenvectors $v_1,...,v_n$.

Define a matrix $P = \bigg[v_1|...|v_n\bigg]$, where the columns of $P$ are the eigenvectors. Then \begin{align*} AP = A\bigg[v_1|...|v_n\bigg] = \bigg[\lambda_1 v_1|...|\lambda_nv_n\bigg] = \bigg[v_1|...|v_n\bigg]\left[\begin{matrix} \lambda_1 & & \\ & \ddots & \\ && \lambda_n\end{matrix}\right] \end{align*} i.e. \begin{align*} AP = PD \end{align*} where $D$ is a diagonal matrix with eigenvalues on its diagonal. Then $\det{A}\det{P} = \det{P}\det{D} \Rightarrow \det{A} = \det{D}$

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In general, $$\det(A) = \sum_{\sigma\in\Sigma_n} \prod_{k=1}^n{\rm sgn}(\sigma)A_{k,\sigma(k)} = \sum_{\sigma\in\Sigma_n}{\rm sgn}(\sigma)\prod_{k=1}^n \lambda_k\delta_{k, \sigma(k)},$$ where ${\rm sgn}(\sigma)$ is the parity of permutation $\sigma$ and $\delta$ is the kronecker delta function.

If for some $k$, $\sigma(k) \not = k$, the kronecker delta evaluates to zero. Hence the only term in the sum that is not zero corresponds to the identity permutation.

Consquently the determinant is $$\prod_{k=1}^n\lambda_k.$$

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If your matrix is diagonalizable, then there is a matrix $Q$ so that

$$ A = QDQ^{-1} $$

where $D$ is a diagonal matrix with the eigenvalues. You've already shown that $\det(D) = \Pi_{i=1}^n \lambda_i$. Use the fact that $$ \det(BC) = \det(B)\det(C) $$ for any square matrices $B$ and $C$ to conclude.

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If $A$ is diagonalizable, then there exists $P$ a invertible matrix such that $$A= P^{-1}\begin{pmatrix} \lambda_1 & & & \\ & \lambda_2 & & \\ & & \ddots & \\ &&& \lambda_n \end{pmatrix}P, $$ then use the fact that $\det(BC)=\det(B)\det(C)$ and $\det(P^{-1})=(\det P)^{-1}$. To find the determinant of a diagonal matrix, just develop the determinant with respect to the first row.

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You would need a formula for computing the determinant of a larger matrix. This is usually done using the Laplace or cofactor expansion, which can be found here: Laplace Expansion on Wikipedia. Have a look at the example there. Note that this works for square matrices of any size and allows you to express the determinant in terms of smaller determinants.

In your case, dealing with a diagonal matrix, the Laplace expansion amounts to the statement that the determinant of the matrix $diag(\lambda_1, \dots, \lambda_n)$ equals $\lambda_n$ times the determinant of $diag(\lambda_1, \dots, \lambda_{n-1})$. Now, since you already know that for small matrices, the determinant equals $$ det \left(diag(\lambda_1, \dots, \lambda_{n-1})\right) = \Pi_{i = 1}^{n-1} \lambda_i $$ you can conclude by induction that for any $n$

$$ det \left(diag(\lambda_1, \dots, \lambda_{n})\right) = \Pi_{i = 1}^{n} \lambda_i $$

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