How to obtain sum of square of eigenvalues without finding eigenvalues of a matrix?
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The eigenvalues of $A^2$ are the squares of the eigenvalues of $A$. The sum of the eigenvalues of any matrix (with algebraic multiplicity) is the trace. So the sum of the squares of the eigenvalues of $A$ (with algebraic multiplicity) is the trace of $A^2$.
Ian
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Thanks. But why are the eigenvalues of $A^2$ are the squares of the eigenvalues of $A$? – KaWa Yip Oct 04 '15 at 02:12

1@Elessarr If $\lambda$ is an eigenvalue of $A$ with eigenvector $x$ then $A^2x=A(Ax)=A(\lambda x)=\lambda Ax=\lambda^2 x$. Ensuring the multiplicities agree is a somewhat more difficult problem. – Ian Oct 04 '15 at 02:15
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Trace of a Matrix is the sum of its Eigenvalues
Proof that the Trace of a Matrix is the sum of its Eigenvalues
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For any $A\in \mathbb{R}^{n\times n}$, we have $$\sum_{i=1}^n \lambda_i^2 = \min_{\det(S)\neq 0} \S^{1}AS\_F$$ $\lambda_i$ are the eigenvalues of $A$.
Henry Davii
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