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)
