# Is a matrix with characteristic polynomial $t^2 +1$ invertible?

Given that $A$ is a square matrix with characteristic polynomial $t^2+1$, is $A$ invertible?

I'm not sure, but this question seems to depend on whether $A$ is over $\mathbb{R}$ or over $\mathbb{C}$. My reasoning is that if $A$ is over $\mathbb{C}$ then $A$ has two distinct eigenvalues $-i$ and $i$ and is diagonalizable. Since it's diagonalization is invertible, $A$ is also invertible.

However if $A$ is over $\mathbb{R}$ then $A$ has no eigenvalues and therefore... I don't know where to go from there.

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A matrix is invertible iff its determinant is nonzero. Can you tell the determinant from the characteristic polynomial? –  Chris Eagle Jul 3 '12 at 17:33
If you have already learned Cayley-Hamilton theorem, then you know that $A^2+I=0$, i.e., $A^2=-I$. –  Martin Sleziak Jul 3 '12 at 17:36
The constant coefficient of the characteristic polynomial = determinant of $A.$ A non-zero determinant implies the matrix is invertible. –  user2468 Jul 3 '12 at 18:38
The invertibility property does not depend on the base field. So if a matrix with real coefficients has a complex inverse, this inverse is in fact real. –  Lierre Jul 3 '12 at 19:52

The eigenvalues of the matrix are all roots of the characteristic polynomial.

A square matrix is invertible if and only if $0$ is not an eigenvalue of the matrix.

Therefore, a square matrix is invertible if and only the constant term of its characteristic polynomial is <fill in the blank>

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not zero. I get it. It was the fact that the matrix has no real eigenvalues that was throwing me for a loop. But I guess that doesn't matter as long as zero isn't an eigenvalue. –  Robert S. Barnes Jul 3 '12 at 18:00
@RobertS.Barnes: Exactly. –  Arturo Magidin Jul 3 '12 at 18:00

Yes it is. In fact $A^{-1}=-A$.

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$A$ is a square matrix and $\det A=1$ so it is invertible.
Hint: $\pmatrix{ \hphantom{-}0&1\\ -1&0 }$ is an example of a real matrix with characteristic polynomial $t^2+1$ which is invertible. Its inverse is $\pmatrix{ 0&-1\\ 1&\hphantom{-}0 }$, as you can easily check.
I would take some objection to your final paragraph; eigenvalues are defined in terms of eigenvectors, and if you view the matrix as acting on $\mathbb{R}^2$, then it has no eigenvectors (and hence no eigenvalues). –  Arturo Magidin Jul 3 '12 at 17:39