Maximum value of $A-\lambda P_v \succeq 0$, where $A$ is a positive symmetric matrix and $P_v$ is a projection onto a unit vector

Question

For a unit vector $v \in \Bbb R^n$, consider the projection $P_v$ on $v$, defined by $P_v(x)=(x \cdot v)v$. Let $A$ be a positive symmetric matrix. Find the maximal value of $\lambda \in \Bbb R$ so that $A - \lambda P_v \succeq 0$.

(Hint: If $c_1,...,c_n$ are the eigenvalues of $A-\lambda P_v$ with $c_1 \geq \cdots \geq c_n$, then we must have $c_n=0$.)

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(For a symmetric matrix $A$, we write $A \succ 0$ if all the eigenvalues of $A$ are positive, and we write $A \succeq 0$ if all the eigenvalues of $A$ are nonnegative.)

I have two questions.

1. I think that we should regard $A-\lambda P_v$ as an $n \times n$ matrix, where $P_v$ is represented by the standard basis. Then is it true that $P_v$ is symmetric?

2. I cannot understand the hint. Why should we have $c_n =0$?

Answer 1

$P_v$ is always symmetric. It can be written as $P_v = v\cdot v^T = P_v^T$. The difference of a symmetric matrix is always symmetric, i.e we can build a normal basis of eigenvectors with real eigenvalues.

The hint gives you the following idea: Let $c_1$ to $c_n$ be the EVs of $A-\lambda P_v$ and let the smallest eigenvalue $c_n$ be positive. Let $v$ be in some form associated with the eigenvector to $c_n$, the smallest eigenvalue. So a change in $\lambda$ will have an influence on $c_n$. It will only decrease $c_n$, if you increase $\lambda$. So we can reach a better target value (maximum of $\lambda$) while still having a positive semi-definitve matrix.