If matrix $A-I$ is positive semidefinite, does $\lambda_{\inf} \geq 1$ hold?

Question

If matrix $A-I$ is positive semidefinite, does the following hold?

$$\lambda_{\inf} \geq 1$$

where $\lambda_{\inf}$ is the infimum of the set of all eigenvalues of $A$. If so, why?

Thanks in advance.

Answer 1

HINT: Suppose you had an eigenvector $v$ with corresponding eigenvalue $\lambda$. What does $\langle (A-I)v,v\rangle\ge 0$ tell you?

Answer 2

Let $\lambda^\star$ be an eigenvalue of $A-I$. Because $A-I$ is positive semidefinite, $\lambda^\star\ge0$.

The characteristic equation of $A-I$ is $$|\lambda^\star-(A-I)|=|(\lambda^\star+1)I-A|=0$$ and the characteristic equation of $A$ is $$|\lambda I-A|=0$$.

We have $$0=|\lambda I -A |=|(\lambda^\star+1)I-A|$$ so the set of eigenvalues of $A$ are in one-to-one correspondence with the set of eigenvalues of $A-I$,

and $\forall j\in\{1 ,\cdots ,\text{number of columns of }A\}$, $$\lambda_j = \lambda_j^\star+1$$ Let $\lambda=\min \lambda_j$ and $\lambda^\star = \min \lambda^\star_j$ $$\lambda^\star \ge 0,$$ $$\lambda^\star +1 \ge 1,$$ $$\lambda \ge 1.$$ All of the eigenvalues of $A$ are greater or equal to $1.$