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Prove $tr(A)=\sum\limits_{\lambda\in Spec(A)} m_\lambda\lambda$ where $m_\lambda$ is the algebraica multipicity.

My work

We know, The trace of a matrix coincides with the sum of all the eigenvalues ​​of the matrix.
Then
$tr(A)=\sum\limits_{i=1}^n a_{ii}=\lambda_1+...+\lambda_n=\sum\limits_{i=1}^n\lambda_i$

Here, i'm stuck. Can someone help me?

Note: $Spec(A)$ is the set of eigenvalues of $A$

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  • $\begingroup$ What is $tr(\cdot)$? $\endgroup$ – Mr Pie Feb 18 '18 at 12:21
  • $\begingroup$ @user477343 Trace of a matrix. $\endgroup$ – Wojowu Feb 18 '18 at 12:23
  • $\begingroup$ Given $i$, how many times does $\lambda_i$ appear in the sum ? $\endgroup$ – Maxime Ramzi Feb 18 '18 at 12:24
  • $\begingroup$ @Wojowu thank you for telling me :) As they say, you learn something new every day... It’s just that this is the second time I had seen $tr(\cdot)$ on the MSE and I was starting to become curious... $\endgroup$ – Mr Pie Feb 18 '18 at 12:24
  • $\begingroup$ depend of the multiplicity of the eigenvalue? @Max $\endgroup$ – Bvss12 Feb 18 '18 at 12:25
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Put $A$ into the form $A = PDP^{-1}$ where $D$ denotes the Jordan canonical form. Then, $$\textrm{Tr} (A) = \textrm{Tr} ( P D P^{-1})$$ Recall that the trace is invariable under cyclic permutations of the inside matrices, so that $$\textrm{Tr} (PDP^{-1} ) = \textrm{Tr} (D P^{-1} P ) = \textrm{Tr} (D)$$ And the trace of the Jordan block matrix is precisely $\sum_{\lambda \in \textrm{Spec} (A)} m_{\lambda} \lambda$.

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  • $\begingroup$ Thanks for you answer... I never see the theory of canonical jordan form. I hope this exercise have other way of prove. $\endgroup$ – Bvss12 Feb 18 '18 at 12:41
  • $\begingroup$ This is a bit harder to prove without Jordan form - what relevant theorems on diagonalizable matrices was covered before this exercise? $\endgroup$ – Rellek Feb 18 '18 at 12:47
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Write $\displaystyle\sum_{i=1}^n \lambda_i = \displaystyle\sum_{\lambda\in Sp(A)} \sum_{1\leq i \leq n, \lambda_i =\lambda} \lambda_i = \displaystyle\sum_{\lambda\in Sp(A)}\lambda \sum_{1\leq i \leq n, \lambda_i =\lambda} 1$.

Now there are exactly $m_\lambda$ $i$'s such that $\lambda_i = \lambda$, by definition, so this gives $\displaystyle\sum_{\lambda\in Sp(A)} m_\lambda \lambda$, which is the expected result

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