Given these conditions on a matrix, prove (or provide a counterexample) that the matrix is invertible

Question

Give a proof if the statement is true, or give a counterexample if it is false:

a) If $A=A^*$, then $A+iI$ is invertible.

b) If $U$ is unitary, $U + 3/4 I$ is invertible.

c) If a matrix $A$ is real, $A - iI$ is invertible.

So for a), I was thinking this is true. If $A=A*$ and $A$ is $n × n$, then $A$ has $n$ real eigenvalues. So then $A+iI$ will have $n$ complex eigenvalues, and none will be zero. So then $A+iI$ will be invertible. Is this correct?

b) This one I am unsure. It makes sense when I try with matrices so I think it is true, but I am unsure how to go about proving it.

c) I think it is true. My proof would be by contradiction. So let $A-Ii$ not be invertible. So there is a solution to $(A-iI)x=0 $such that $x\neq 0$. So $Ax-ix=0$. So then $Ax=ix$. So then $i$ is an eigenvalue of $A$. But $A$ is real so it has to have real eigenvalues. Contradiction. Therefore $A-iI$ is invertible.

Please let me know if I am on the right track with these!

Answer 1

Much of your reasoning is very good. Let's consider $(b)$.

Suppose that $U + \frac{3}{4}I$ is not invertible. Then there is some vector $x$ such that $(U + \frac{3}{4}I)x = 0$, or equivalently that $Ux = -\frac{3}{4} I x = - \frac{3}{4}x$.

This would mean that $-\frac{3}{4}$ is an eigenvalue of $U$. But as $U$ is unitary, all of its eigenvalues are of magnitude $1$.

Answer 2

For $(c)$, take counterexample as $A=\begin{pmatrix}0&1\\-1&0\\\end{pmatrix}$