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So if I have an $n\times n$ matrix $A$ that is diagonalizable then how do I determine $\mbox{tr}(A)$ and $\det(A)$ as functions of the eigenvalues of the matrix?

For $\det(A)$, I know the formula for eigenvalues is $\det(A−tI)$ which would be $$(-1)^n(t-\lambda_1)\dots (t-\lambda_n)$$ where $\lambda_i$ are the eigenvalues of $A$.

For $\mbox{tr}(A)$ could I also do something similar to that? I'm just not sure how this question works.

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  • $\begingroup$ The trace is the sum of the eigenvalues. $\endgroup$ – Eduardo Longa Nov 30 '16 at 17:26
  • $\begingroup$ If $A$ is diagonalisable, then $A$ is similar to a diagonal matrix with eigenvalues on the main diagonal. What does that mean for the determinant and the trace? If $A$ is not diagonalisable, there is always the Jordan decomposition, but that is killing a bug with a shotgun. EDIT: You can also bring out the formula for trace and determinant directly from the characteristic polynomial, but that is 'ugly'. $\endgroup$ – user305860 Nov 30 '16 at 17:28
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    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – GDumphart Nov 30 '16 at 17:29
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If we write the Jordan Normal Form for the matrix $A$ we get:

$$A=P^{-1}JP$$

and $J$ has all eigenvalues on the diagonal but $J$ not necessearily is a diagonal matrix.

So $\det(A)=\det(J)$ and $\det(J)$ is the product of the eigenvalues.

Now using the known relation $\mbox{tr} (AB)=\mbox{tr} (BA)$ we have:

$$\mbox{tr} (A)=\mbox{tr} (P^{-1}JP)=\mbox{tr} (JPP^{-1})=\mbox{tr} (J)$$

and $\mbox{tr} (J)$ is the sum of eigenvalues.

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If A is diagonalizable, $A=UDU^{-1}$, with $D$ diagonal matrix with the eigenvalues in the diagonal ($D=diag(\{\lambda\}_{i=1}^{n})$), and $U$ invertible. The determinant does not vary when multiplying by invertible matrices.

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  • $\begingroup$ You should probably say something about the trace as well. $\endgroup$ – Semiclassical Nov 30 '16 at 17:33
  • $\begingroup$ Ok so then I know that tr(A) is the sum of all eigenvalues. Does that depend on whether or not A is diagonalizable? $\endgroup$ – david mah Nov 30 '16 at 17:33
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It is not difficult to show that the coefficient of the term $\lambda^{n-1}$ of the characteristic equation $(A-\lambda I)=0$ of an $n \times n$ matrix $A$ is exactliy $ a_{n-1}=\mbox{tr} (A)$.

Now remember that the eigenvalues are the roots $\lambda_i$ of this polynomial, so that $$(A-\lambda I)=0 \quad \iff \quad (\lambda -\lambda_1)(\lambda- \lambda_2)\cdots(\lambda-\lambda_n)=0$$ so, from the Vieta's formulas, the coefficient of $\lambda^{n-1}$ is the sum of the $n$ eigenvalues (counted with their multiplicity).

Note that this result does not depend on the fact that the matrix $A$ is diagonalizable or not.

In an analogous way we can proof that the term $a_0$ is the determinant of the matrix $A$ and it is the product of the eigenvalues.

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