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I'm solving a problem where I am given the eigenvalues of a matrix $A$ and need to solve for the determinant of $A$. I know that if my matrix is diagonalizable I can find the determinant of $A$ by multiplying the eigenvalues together. However, I am given these eigenvalues: $-2, 3, -\frac23, 4, 4, 4, \frac12, \frac12$. As you can see, there are multiple repeats. I am given the fact that $A$ is an $8\times8$ matrix. I was just wondering that if I have repeated eigenvalues does that mean that I don't have $8$ distinct eigenvalues but instead $5$ eigenvalues?

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    $\begingroup$ The determinant is still the product of all eigenvalues, with multiplicities. $\endgroup$ – Amitai Yuval May 13 '15 at 21:28
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    $\begingroup$ Even if the matrix is not diagonalisable, it can be put intriangular form, with eigenvalues on the diagonal. The theorem you mention on computing the determinant for a diagonal matrix, is also valid for a triangular matrix. $\endgroup$ – Bernard May 13 '15 at 21:33
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Since your matrix is diagonalizable we have $$ A=P\Lambda P^{-1} $$ where $$ \Lambda= \scriptsize{ \begin{bmatrix} -2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -\frac23 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac12 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac12 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 4 \\ \end{bmatrix}} $$ It follows that \begin{align*} \det{A} &= \det(P\Lambda P^{-1}) \\ &= \det(P)\det(\Lambda)\det(P^-1) \\ &= \det(P)\det(P^{-1})\det(\Lambda) \\ &= \det(PP^{-1})\det(\Lambda) \\ &= \det(I)\det(\Lambda) \\ &= \det(I\Lambda) \\ &= \det(\Lambda) \\ &= (-2)\left(-\frac{2}{3}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)(3)(4)(4)(4) \\ &= 64 \end{align*} so the multiplicity of each eigenvalue does matter!

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