# Symmetric matrices with spectrum in the diagonal

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 $$n$$?

• What is the question exactly ? I read : "Let A be a symmetric matrix. [...] is A symmetric ?". – TheSilverDoe Feb 26 at 14:05
• What about $\begin{pmatrix} 1 & -1 \\ 0 & 2\end{pmatrix}$ (for the $n=2$ case)? – JJC94 Feb 26 at 14:08
• Should the conclusion be "... then is $A$ diagonal?" – Dave Feb 26 at 14:27

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.
• @J.Salieri Because $A$ is Hermitian (actually real symmetric in this case) here. – user1551 Feb 26 at 15:00
• @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.) – user1551 Feb 26 at 15:10
• @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 $\|A\|_2=\sigma_1=|\lambda_1|=\rho(A)$. – user1551 Feb 26 at 15:16