# Is there an Odd Matrix Whose Square is $-I$? ## Problem 316

Let $n$ be an odd positive integer.
Determine whether there exists an $n \times n$ real matrix $A$ such that
$A^2+I=O,$ where $I$ is the $n \times n$ identity matrix and $O$ is the $n \times n$ zero matrix.

If such a matrix $A$ exists, find an example. If not, prove that there is no such $A$.

How about when $n$ is an even positive number? Add to solve later

## Hint.

The key technique to solve this problem very easily is determinant.
Recall the following properties of the determinant.
Let $A, B$ be $n\times n$ matrices and $c$ be a scalar Then we have

• $\det(AB)=\det(A)\det(B)$
• $\det(cA)=c^n\det(A)$.

## Solution.

### When $n$ is odd.

When $n$ is odd, we prove that there is no $n \times n$ real matrix $A$ such that $A^2+I=O$.
Seeking a contradiction, assume that we have $A$ such that $A^2+I=O$.
Since we have $A^2=-I$, we have
$\det(A^2)=\det(-I).$

Using the properties of determinant, we obtain
$\det(A)^2=(-1)^n\det(I)=-1$ because $n$ is odd and $\det(I)=1$.

Since $A$ is a real matrix, the determinant of $A$ is also real.
Thus, $\det(A)^2=-1$ is impossible. Hence there is no such $A$.

### When $n$ is even.

On the other hand, if $n$ is even there is $A$ such that $A^2+I=O$.
For example, consider
$A=\begin{bmatrix} 0 & -1\\ 1& 0 \end{bmatrix}.$ Then a direct computation shows that $A^2=-I$, hence $A^2+I=O$.

## Comment.

Recall that the imaginary number $i$ is the number whose square is $-1$.
Similarly, we found above that the square of the matrix $A=\begin{bmatrix} 0 & -1\\ 1& 0 \end{bmatrix}$ is $-I$.

But when $n$ is odd, there is no such matrix $A$ as we showed. Add to solve later

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