2
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ Cheating: the claim is false. For example, take $\;A=-iI\;$ ... $\endgroup$
    – DonAntonio
    Oct 14 '13 at 18:46
  • $\begingroup$ Perhaps your matrix is supposed to be symmetric? $\endgroup$ Oct 14 '13 at 18:46
  • 2
    $\begingroup$ @DonAntonio: did you miss the "real"? $\endgroup$ Oct 14 '13 at 18:47
  • 1
    $\begingroup$ 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. $\endgroup$
    – DonAntonio
    Oct 14 '13 at 18:49
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 14 '13 at 19:47
1
$\begingroup$

It won't necessarily exist. For instance, if

$$ A = \left(\begin{array}{cc}0 & 1 \\ -1 & 0 \end{array} \right) $$ 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 $A$".

$\endgroup$
1
  • 1
    $\begingroup$ eigenvalue.. of course! Thank you very much $\endgroup$ Oct 15 '13 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.