# Inverse of a real matrix plus identity times i

How would you proof that given a real square matrix $A$ then the inverse of the matrix ( $A + i I$) exists?

• Cheating: the claim is false. For example, take $\;A=-iI\;$ ... Oct 14 '13 at 18:46
• Perhaps your matrix is supposed to be symmetric? Oct 14 '13 at 18:46
• @DonAntonio: did you miss the "real"? Oct 14 '13 at 18:47
• No, I didn't @RobertIsrael. I'm using "i" in the same sense the OP seems to be using it, meaning $\;i\in\Bbb R\;$ . If the OP meant $\;i=\sqrt{-1}\;$ then the claim's still false but my example doesn't work. Oct 14 '13 at 18:49
• Unless it's being used as a "dummy variable", e.g. in a sum or product, the standard mathematical meaning of $i$ is $\sqrt{-1}$. There's no $\sum$ or $\prod$ here. Oct 14 '13 at 19:47

$$A = \left(\begin{array}{cc}0 & 1 \\ -1 & 0 \end{array} \right)$$ then $$A + iI = \left(\begin{array}{cc}i & 1 \\ -1 & i \end{array} \right)$$ has determinant $i^2 + 1 = 0$.
There are conditions that you can place on $A$ to ensure $A + iI$ is invertible.
HINT: fill in the blank:"$A - \lambda I$ is non-invertible if and only if $\lambda$ is an ____ of $A$".