# Hermitian Matrix and nondecreasing eigenvalues

I am studying for finals and looking at old exams. I found this question and am not sure how to proceed.

Let $A$ be an $n\times n$ Hermitian matrix with eigenvalues $\lambda_1 \leq \lambda_2 \leq \cdots \leq \lambda_n$. Prove that if $a_{ii}=\lambda_1$ for some $i$, then every other entry of row and column $i$ is zero.

Any help is greatly appreciated!

Let $\langle \cdot, \cdot \rangle$ be the standard inner product on $\mathbb{C}^n$. For the Hermitian matrix $A$, we will show that $\inf_{||v||=1} \langle Av,v \rangle = \lambda_1$ for $v \in \mathbb{C}^n$ having unit norm.
First note that $\langle Av,v \rangle$ is real, since $\langle Av,v \rangle = \langle v,A^*v \rangle = \langle v,Av \rangle = \overline{\langle Av,v \rangle}$. $A$ is Hermitian, hence Normal, so by the Spectral Theorem, $\mathbb{C}^n$ has an orthonormal basis $\{ v_i \}_{i=1}^{n}$ of characteristic vectors of $A$ whose eigenvalues respectively are $\{ \lambda_i \}_{i=1}^{n}$. If $v = \sum_{i=1}^{n} \alpha_iv_i$with norm $1$ (i.e, $\sum_{i=1}^{n} |\alpha_i|^2 = 1$), then we have, $$\langle Av,v \rangle = \langle \sum_{i=1}^{n} \alpha_i\lambda_iv_i, \alpha_iv_i \rangle = \sum_{i=1}^{n} |\alpha_i|^2\lambda_i \ge (\sum_{i=1}^{n} |\alpha_i|^2) \lambda_1 = \lambda_1$$ and $\langle Av_1, v_1 \rangle = \lambda_1$, which proves the assertion.
Since we also have $\langle Ae_i, e_i \rangle = \lambda_1$ by hypothesis, we have equality in the above computation. If $e_i = \sum_{i=1}^{n} \alpha_iv_i$, then we have $\sum_{i=2}^{n} |\alpha_i|^2 (\lambda_i - \lambda_1 ) = 0$ and hence all terms in the sum are zero. If $k$ is the largest index for which $\alpha_k \neq 0$ (such $k$ exists), then, $\lambda_k = \lambda_1 \implies \lambda_1 = \lambda_2 = \cdots = \lambda_k$
Thus, $$Ae_i = A(\sum_{i=1}^{k} \alpha_iv_i) = \lambda_1(\sum_{i=1}^{k} \alpha_iv_i) = \lambda_1e_i$$ which means the $i$-th column of $A$ is actually $\lambda_1e_i$, thus, except $a_{ii}$, all other entries in column $i$ of $A$ are zero. (and hence, follows for row $i$ too)