If matrix $A$ is a square matrix. And $A$'s characteristic polynomial is $p(t) = t^2 + 1$. It's not necessary true that $A$ is non-singular right? Because, the eigenvalues are $i,-i$. and if $\dim(A)= 2$ then $A$ is a $2 \times 2$ matrix. Thus, I can put in its non diagonal values that will give a result that will make $A$ a singular matrix, or even better, the zero matrix which is pure singular (Cuz its determinant is zero)

Or I'm wrong and if the eigenvalues are $+i , -i$ then its determinant is $-i\cdot i = 1$?

  • $\begingroup$ The last sentence is correct. I can't understand your argument in the first paragraph. And what is $\dim (A)$? $\endgroup$
    – Git Gud
    Jun 21 '14 at 10:23
  • $\begingroup$ @GitGud I meant dimA=4 to represent a 2x2 matrix. anyways, thx. $\endgroup$ Jun 21 '14 at 10:24

The determinant is the product of all eigenvalues, hence the determinant is $1$, which is nonzero, so we can conclude that the matrix is invertible.

  • $\begingroup$ Will accept ur answer in 5min, that is what I figured out myself. wasn't sure about it. $\endgroup$ Jun 21 '14 at 10:31
  • $\begingroup$ I don't think that if the determinant of a matrix is non-zero it will be invertible. $\endgroup$ Jun 21 '14 at 12:13
  • $\begingroup$ @user3628041 note that the determinant is zero iff one of the eigenvalues is zero and use Surb answer to complete the argument that indeed the implication I made holds $\endgroup$
    – Belgi
    Jun 21 '14 at 12:24

More generally, a matrix $A$ is non-singular if and only $0$ is not an eigenvalue of it. This is easily shown as follow:

  • if $0$ is an eigenvalue of $A$ then there is some $v\neq 0$ such that $Av= 0v = 0$ and thus $0 \neq v \in \ker(A)$ which shows that $A$ is singular.

  • if $A$ is singular then there $\ker(A) \neq \{0\}$ and there exists some $v \in \ker(A), v \neq 0$ such that $Av = 0 = 0v$, it follows that $(0,v)$ is an eigenpair of $A$.

In particular, if your matrix has only $i$ and $-i$ as eigenvalues, it must be non-singular. This is the case when the associated characteristic polynomial is $p(t)=t^2+1$.

Edit: Here is a "determinant" approach of the problem. We know that $\lambda$ is an eigenvalue of $A$ if and only if $\det(A-\lambda I)=0$, where $I$ is the identity matrix. This is the reason why we look for the roots of the characteristic polynomial. We also know that $A$ is singular if and only if $\det(A)=0$. It is now clear that $0$ is an eigenvalue of $A$ if and only if $0=\det(A-0I)=det(A)$.

To make a connection with the first argument, note that $\det(A-\lambda I)=0 \iff \ker(A-\lambda I)\neq \{0\}$ and $\ker(A-\lambda I)$ is the eigenspace associated to the eigenvalue $\lambda$.


From Cayley-Hamilton theorem you know that $p(A)=0$, i.e., in this case:
$$ \begin{align} A^2+I&=0\\ -A^2&=I\\ A^{-1}&=-A \end{align} $$ Since $A$ has an inverse, it is not singular. (And, additionally, you know how to express the inverse matrix using $A$.)

  • $\begingroup$ @ilan from this you should be convinced that the matrix is invertible exactly when the constant term of the characteristic polynomial is nonzero. $\endgroup$
    – rschwieb
    Jun 21 '14 at 12:35

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