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Let $A$ be a symmetric matrix of order $n$. If $\lambda_1,\ldots, \lambda_n$ are its eigenvalues and the main diagonal of $A$ is $\lambda_1, \ldots, \lambda_n$ then is $A$ diagonal?

If $n=2$, you can use the determinant to ensure that the nondiagonal entries are zero. If $n=3$ you can use the square of the trace. Is it possible to give a clean argument for a general $n$?

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    $\begingroup$ What is the question exactly ? I read : "Let A be a symmetric matrix. [...] is A symmetric ?". $\endgroup$ – TheSilverDoe Feb 26 at 14:05
  • $\begingroup$ What about $\begin{pmatrix} 1 & -1 \\ 0 & 2\end{pmatrix}$ (for the $n=2$ case)? $\endgroup$ – JJC94 Feb 26 at 14:08
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    $\begingroup$ Should the conclusion be "... then is $A$ diagonal?" $\endgroup$ – Dave Feb 26 at 14:27
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By a simultaneous permutation of rows and columns of $A$ if necessary, we may assume that $|\lambda_1|\ge|\lambda_2|\ge\cdots\ge|\lambda_n|$. Since $$ \|A\|_2\ge\|(1,0,\ldots,0)A\|_2=\|(\lambda_1,a_{12},\ldots,a_{1n})\|_2\ge|\lambda_1|=\rho(A)=\|A\|_2, $$ we must have $a_{12}=\cdots=a_{1n}=0$ by the squeezing principle. As $A$ is symmetric, this means $A=\pmatrix{\lambda_1&0\\ 0&B}$ for some $(n-1)\times(n-1)$ symmetric matrix $B$. Obviously, the eigenvalues and diagonal entries are $\lambda_2,\lambda_3,\ldots,\lambda_n$. Proceed recursively, we conclude that $A$ is a diagonal matrix.

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  • $\begingroup$ How do you prove that the spectral radius is equal to the norm 2 of the matrix? $\endgroup$ – J. Salieri Feb 26 at 14:35
  • $\begingroup$ @J.Salieri Because $A$ is Hermitian (actually real symmetric in this case) here. $\endgroup$ – user1551 Feb 26 at 15:00
  • $\begingroup$ Sorry but I don't know much. Why the norm 2 of an hermitian matrix coincides with the spectral radius ? Is it difficult to prove? $\endgroup$ – J. Salieri Feb 26 at 15:02
  • $\begingroup$ @J.Salieri $A$ is unitarily diagonalisable. So, every eigenvector of $A$ is automatically a singular value of $A$. It follows that the moduli of the eigenvalues of $A$ are the singular values of $A$. Hence the spectral radius of $A$ is the operator norm of $A$. (This explains why the operator norm is also called the spectral norm despite the fact that the operator norm in general has little to do with the spectrum of a matrix.) $\endgroup$ – user1551 Feb 26 at 15:10
  • $\begingroup$ @J.Salieri Alternatively, in matrix language, if $A=V\Lambda V^\ast$ is a unitary diagonalisation where the eigenvalues are arranged in an order of decreasing magnitude, let $\lambda_k=|\lambda_k|e^{i\theta_k}$ for each $k$ and define $\Sigma=\operatorname{diag}(|\lambda_1|,\ldots,|\lambda_n|), D=\operatorname{diag}(e^{i\theta_1},\ldots,e^{i\theta_n})$ and $U=VD$. Then $A=U\Sigma V^\ast$ is a singular value decomposition. It is now clear that $\|A\|_2=\sigma_1=|\lambda_1|=\rho(A)$. $\endgroup$ – user1551 Feb 26 at 15:16

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