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Let $A$ be an $n \times n$ matrix with eigenvalues $\lambda_{1},...\lambda_{n}$. How do I prove that tr$(A^k) = \sum_{i=1}^{n}\lambda_{i}^{k}$?

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    $\begingroup$ Hmm, the easiest argument I can come up with is that it is clear when the matrix is diagonalizable, and then use an argument by denseness and continuity. $\endgroup$ – Tobias Kildetoft Dec 18 '15 at 10:26
  • $\begingroup$ The problem states $A$ has $n$ eigenvalues.... $\endgroup$ – David Wheeler Dec 18 '15 at 10:45
  • $\begingroup$ @DavidWheeler But not that they are distinct. $\endgroup$ – Tobias Kildetoft Dec 18 '15 at 10:46
  • $\begingroup$ True enough. Ir's possible some duplications were intended. $\endgroup$ – David Wheeler Dec 18 '15 at 10:49
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  1. $tr(A)=\sum \lambda_i$
  2. If $\lambda_i$ is eigenvalue of $A$, then $\lambda_i^k$ is eigenvalue of $A^k$. This mapping preserves multiplicities.

The first one is a classic result, easily deduced from the characteristic polynomial. The second one is a little trickier (if there are repeated eigenvalues).

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  • $\begingroup$ Ah, I see. Thank you for the answer! :) $\endgroup$ – Train Heartnet Dec 20 '15 at 13:03
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Sum of roots of a polynomial is given by $-\frac{b}{a}$, if the polynomial looks like $a\lambda^N+b\lambda^{N-1}+\dots$. Now, eigenvalues are roots of polynomial $\det(A-\lambda I)$. Try to prove that $-\frac{b}{a}$ in this case is $\mathrm{trace}(A)$. See that $A^kv=A^{k-1}(Av)=A^{k-1}(\lambda_i v)=\lambda_i^k v$ (where $\lambda$ is a eigenvalue). Thus $\lambda_i^k$ are eigenvalues of $A^k$. Thus $\lambda_i^k$ are roots of polynomial $\det (A^k-\lambda I)$. Thus, their sum should equal $\mathrm{trace}(A^k)$.

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  • $\begingroup$ sum of roots is from Vieta's formulas $\endgroup$ – qwr Nov 1 '19 at 18:46
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The trace is similarity-invariant. Supposing the multiplicity of each eigenvalue to be 1. Let $A^k = U^{-1}\Sigma^k U,$ where the matrix $U$ is composed by the eigenvectors of $A$ and $\Sigma$ is the diagonal matrix with the eigenvalues of $A$. So we have: $$\mathrm{tr}(A^k) = \mathrm{tr}(U^{-1}\Sigma^k U) = \mathrm{tr}(U^{-1}(\Sigma^k U)) = \mathrm{tr}((\Sigma^k U)U^{-1}) = \mathrm{tr}(\Sigma^k (U U^{-1})) = \mathrm{tr}(\Sigma^k I) = \mathrm{tr}(\Sigma^k).$$

Thus as $\Sigma = \left(\lambda_i \mathbb{1}_{\{i=j\}} \;;\; (i,j) \in [0,n]^2\right) $, we have $\Sigma^k = \left(\lambda_j^k \mathbb{1}_{\{i=j\}} \;;\; (i,j) \in , [0,n]^2\right) $ and, therefore, $$\mathrm{tr}(\Sigma^k) = \sum_{i=1}^n \lambda_i^k$$

Also, this argument may be extended easily to any matrix A n-by-n using the Jordan form as in wiki

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    $\begingroup$ This is missing the details of how to do this when the matrix is not diagonalizable. $\endgroup$ – Tobias Kildetoft Dec 18 '15 at 10:41
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    $\begingroup$ $n$ eigenvalues does not imply diagonalizable. $n$ distinct ones does. $\endgroup$ – Tobias Kildetoft Dec 18 '15 at 10:54
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    $\begingroup$ Ahh, using the Jordan form is definitely simpler than my argument from the comment on the question. $\endgroup$ – Tobias Kildetoft Dec 18 '15 at 11:00
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    $\begingroup$ Yes, the JNF is the way to go-assuming similarity invariance we can write $A = PJP^{-1}$, where $J = D+N$ and $D$ is diagonal and $N$ is nilpotent with the only non-zero entries on the super-diagonal. It's an easy induction proof to show that $\text{tr}(J^k) = \text{tr}(D^k)$. $\endgroup$ – David Wheeler Dec 18 '15 at 11:25
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    $\begingroup$ I was hoping to not resort to the Jordan Normal Form. But I understand that it's way more elegant that way. Thank you for all your inputs, it was really helpful. :) $\endgroup$ – Train Heartnet Dec 20 '15 at 13:10

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