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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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    $\begingroup$ A matrix is invertible iff its determinant is nonzero. Can you tell the determinant from the characteristic polynomial? $\endgroup$
    – Chris Eagle
    Commented Jul 3, 2012 at 17:33
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    $\begingroup$ If you have already learned Cayley-Hamilton theorem, then you know that $A^2+I=0$, i.e., $A^2=-I$. $\endgroup$
    – Martin Sleziak
    Commented Jul 3, 2012 at 17:36
  • $\begingroup$ The constant coefficient of the characteristic polynomial = determinant of $A.$ A non-zero determinant implies the matrix is invertible. $\endgroup$
    – user2468
    Commented Jul 3, 2012 at 18:38
  • $\begingroup$ 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. $\endgroup$
    – Lierre
    Commented Jul 3, 2012 at 19:52

5 Answers 5

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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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  • $\begingroup$ 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. $\endgroup$
    – Robert S. Barnes
    Commented Jul 3, 2012 at 18:00
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    $\begingroup$ @RobertS.Barnes: Exactly. $\endgroup$
    – Arturo Magidin
    Commented Jul 3, 2012 at 18:00
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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.

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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.

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  • $\begingroup$ 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). $\endgroup$
    – Arturo Magidin
    Commented Jul 3, 2012 at 17:39
  • $\begingroup$ I agree. I have deleted the objectionable paragraph. Thank you. $\endgroup$
    – MJD
    Commented Jul 3, 2012 at 17:40
  • $\begingroup$ It's not misguided, with the appropriate caveats, and it can be made very precise (e.g., you can "complexify" the vector space to obtain eigenvalues), or you can talk about roots of the characteristic polynomial over an algebraic closure of the field of definition. I just thought that such a categorical statement might be confusing given how eigenvalues are usually introduced. (For that matter, my own phrasing in my question could likewise be misconstrued, so I have just fixed it...) $\endgroup$
    – Arturo Magidin
    Commented Jul 3, 2012 at 17:47
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One can use the Cayley-Hamilton Theorem, which states that $f(A) = 0$, where $f(x)$ is the characteristic polynomial of $A$, to determine that $A^{2} + I = 0 \implies A^{2} = -I \implies \det(A)^{2} = (-1)^{2} \implies \det(A) = \pm 1 \ne 0 \implies A$ is invertible.
Note that since the characteristic polynomial has degree $2$ we may assume we are working with $2 \times 2$ matrices.

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