What is the rank of $M- \lambda U U^*$ Let $M$ be an $n×n$ Hermitian matrix of rank $k,k \neq n$.If $\lambda \neq 0$ is an eigenvalue of M with corresponding unit column vector $U$. Then prove that,
$rank(M- \lambda U U^*)=k-1$.
Now I have ,
$MU=\lambda U$, so
$M-(\lambda U) U^*=M-MUU^*=M(I-UU^*).
$
I can get only this far, after this I have tried my best to get that rank but nothing significant I have achieved.THANK YOU.
 A: Since $M$ is Hermitian, it admits an orthonormal eigenbasis $B = \{v_1, v_2, \cdots, v_n\}$. Since $M$ is of rank $k$, we know that $n - k$ of these eigenvectors have eigenvalue $0$. Let us suppose without loss of generality that $U = \vec{v}_1$. Then for any $\vec{v} \in B$ with $M \vec{v} = \vec{0}$, we have $(M - \lambda U U^{*})(\vec{v}) = M\vec{v} - \lambda \langle \vec{v}_1, \vec{v} \rangle \vec{v}_1$. But since by construction of the basis $B$ $\vec{v} \perp \vec{v}_1$, this expression is just the zero vector. Furthermore, it should be clear that $M \vec{v}_1 = \vec{0}$ as well, so we have found $n - k + 1$ linearly independent vectors in the null space of $M$.
Now we can also check that for any $\vec{v} \in B$ with $M \vec{v} \neq \vec{0}$ and $\vec{v} \neq \vec{v}_1$, we have
$$(M - \lambda \vec{v} \vec{v}^*)\vec{v} = M \vec{v} - \lambda \langle \vec{v}_1, \vec{v} \rangle \vec{v}_1 = \mu \vec{v}$$
where $\mu$ is the eigenvalue associated with $\vec{v}$. There are $k - 1$ such vectors, so indeed we have verified there are $k - 1$ linearly independent vectors in the column space of $M$.
These two facts are enough to show that $\text{rank}(M) = k - 1$. $\square$
Remark: In fact, the spectral decomposition of $M$ can also be written as
$$M = \sum_{k = 1}^n \lambda_k \vec{v}_k \vec{v}_k^*$$
where $\vec{v}_k$ is an eigenvector of $M$ with eigenvalue $\lambda_k$. Note that each term of the summation is a rank one matrix. Often times, we approximate these Hermitian matrices by truncating this sum to a fewer number of terms (usually the terms with the largest eigenvalues). The rank of this approximation is then exactly the number of terms which have non-zero eigenvalues.
In your problem, we are effectively removing the term of the summation with $\lambda_k = \lambda$ and $\vec{v}_k = U$. Since the eigenvalue corresponding to this term is nonzero, we would be reducing the rank by one, exactly as claimed.
