Let $A∈M_n$ ($\mathbb R$). If $A^2=-I$ then,

Question

(a) $n$ is even.

(b) $A=±I$

(c) all the eigen values of $A$ are in $\mathbb R$

(d) $A$ is a diagonal matrix.

Now I multiplied both sides by $A^{-1}$ and got the equation $A+A^{-1}=0$ but how do I check the given options using the derived equation? I think I'm missing something very obvious.

Answer 1

Observe that as $A^2 = -I$, if $\lambda$ is an eigenvalue of $A$ and let $x \neq 0$ be the corresponding eigenvector of $A$ then $A^2x = -x$ so $\lambda^2 x = -x$ fro which we see that eigenvalues are $\{i,-i\}$.So option (c) is ruled out as eigenvalues are complex numbers.

Next, since eigenvalues are complex and are the roots of characteristic polynomial so they must exist in pairs(complex number and its conjugate) and hence the number of them would be even meaning that the order of the matrix is also even.so $n$ is even and hence option $(a)$ is correct.

Now $A \neq \pm I$ as eigenvalues of $A$ then will be $\pm 1$.So it is discarded.So option $(b)$ is wrong.

Can we eliminate option $d$? like we try to see if that is true for a non-diagonal matrix that is we take a non-diagonal $2 \times 2$ matrix which looks like - $\begin{bmatrix} 0 & x \\ y & 0\end{bmatrix}$ and we compute $A^2 = \begin{bmatrix} xy & 0 \\ 0 & xy\end{bmatrix}$ and equating to $\begin{bmatrix} -1 & 0 \\ 0 & -1\end{bmatrix}$, we observe $xy =-1$ and we can take $x = 1,y=-1$ or many combination of $x$ and $y$ such that $xy =-1$.So there exists non-diagonal matrices such that $A^2 = -I$ so $(d)$ is not necessarily correct.

Answer 2

Guide:

• Consider $A=\begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}$

• Consider taking determinant on both sides.

Answer 3

Answer. (a)

The example that follows $$ A=\left(\begin{array}{rr}0& -1\\ 1 & 0\end{array}\right) $$ shows that (b), (c) and (d) do not necessarily hold.

Now, why (a) is the answer?

Since, $A^2=-I$, implies that all the eigenvalues of $A$ are in the set $\{i,-i\}$, and they have the same multiplicity, since the characteristic polynomial of $A$ has real coefficients. Hence the degree of the polynomial, i.e., $n$, is even.

Answer 4

Assume $A^2 = -I$. Taking the determinant on both sides gives $$0 \le (\det A)^2 = \det A^2 = \det (-I) = (-1)^n$$

so $n$ has to be even.

To show the other statements are false, consider examples such as

$$A = \begin{pmatrix} 0 & 1 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0\\ \end{pmatrix}$$ with $\sigma(A) = \{-i, i\}$.