# Problem related to a square matrix

Let $A$ be an $n\times n$ matrix with real entries such that $A^{2}+I=\mathbf{0}$. Then:

(A) $n$ is an odd integer.
(B) $n$ is an even integer.
(C) $n$ has to be $2$
(D) $n$ could be any positive integer.

I was thinking about the problem.I noticed for a $2\times 2$ matrix $A$ of the form $$\begin{pmatrix} 1 &-2 \\ 1& -1 \end{pmatrix},$$ the given condition holds good.So option (C) is a possibility.But I am not sure about other options. Is there any convenient way to tackle it?With regards..

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@amWhy: Don't you think when we are supposed to ruled out any options in a multiple choices like above, we need just an example for doing this? Do you think, we have to prove this theoretically? Thanks. – Babak S. Jan 21 at 14:08
Hint: Think about how one multiplies block matrices. – peoplepower Jan 21 at 14:11

## 3 Answers

Note: $n=1\,$ is ruled out, since e.g., $A$ consists of the single scalar entry $1$: $A = [1],\; I = I_1,\;, A^2 + I = 2.\;$

Indeed there is no real scalar $\,k\,\neq 0\,$ (in the case $n = 1, A = k\,$) such that $\,k^2 = -1.\,$

So option (A) is ruled out, since $\,n = 1\,$ is odd, and option (D) is ruled out, since $\,n = 1 >0\, n \in \mathbb{Z}^+$.

What remains is to decide between (B) and (C).

You know for $n = 2\,$, the equality is satisfied (hence "(C)" is in the "running") but $n = 2\,$ is also even: so "(B)" has a chance. You can rule out (C) if there exists any $\,n= 2k,\; k\in \mathbb{Z}^+$, $n > 2,\,$ such that $\,A_{n\times n}^2 + I_n\, = 0$.

• Hint: try $n = 4$: construct a $4\times 4$ matrix made of $4$-square block $2 \times 2$ matrices, using your matrix for each of the two block entries on the diagonal, and zero blocks off the diagonal.)

Let $A = \begin{pmatrix} 1 &-2 \\ 1& -1 \end{pmatrix}. \quad$ So using $A^2 + I = 0$, construct $A_{4 \times 4} = \begin{pmatrix} A & 0 \\ 0 & A \end{pmatrix}$

$$A_{4\times 4}^2 + I_4 = \begin{pmatrix} A &0 \\ 0& A \end{pmatrix} \cdot \begin{pmatrix} A & 0 \\ 0 & A \end{pmatrix} + I_4 = \begin{pmatrix} A^2 & 0 \\ 0& A^2 \end{pmatrix} + \begin{pmatrix} I_2 & 0\\ 0 & I_2 \end{pmatrix}$$ $$= \begin{pmatrix} A^2 + I_2 & 0\\ 0 & A^2 + I_2 \end{pmatrix}= 0$$

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You did rule out those poor A and D. :D & +1 – Babak S. Jan 21 at 14:23
@amWhy so from your calculation,i see that $(B)$ is the right choice..Thanks a lot sir for the detailed clarification. – user33640 Jan 21 at 15:45
user33640: Yes indeed! B is correct: Your welcome! – amWhy Jan 21 at 15:50

As $A$ is real, so is $\det A$. Yet $A^2+I=0$ implies that $(\det A)^2=\det(-I)=(-1)^n$. Therefore $n$ has to be an even integer and (A), (D) are incorrect. Now, if there is a $2\times 2$ matrix $A$ such that $A^2+I=0$, then $\tilde{A}=\begin{pmatrix}A\\&A\end{pmatrix}$ would be a $4\times4$ matrix such that $\tilde{A}^2+I=0$. Therefore (C) is incorrect too. To verify that (B) is indeed the correct answer, consider the aforementioned $\tilde{A}$ with $A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$.

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Hint: Let's $n>1$ an interger positive and even. Set the $n\times n$ matrix $$J_{n\times n}=\begin{pmatrix} 0 & \ldots & 0 & \ldots &-1 \\ \vdots & & \vdots & & \vdots \\ 0 & \ldots & -1 & \ldots & 0 \\ \vdots & & \vdots & & \vdots \\ -1 & \ldots & 0 & \ldots &0 \\ \end{pmatrix}.$$ Note that $I_{n\times n}=J_{n\times n}^2$ for if $n$ is even and $$A^2+I_{n\times n}= ( A+J_{n\times n})(A-J_{n\times n})=0.$$

Then we have $A=\pm J$. If for n=1 we are a conter exemple. It's exclude (D). Then the answer is (C).

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Not n = 1!.............. – amWhy Jan 21 at 14:46

1
"Then n could be any positive integer." Not true: you have not established this is true for $n = 1$. Indeed, $A^2 + I = 0 \implies n \neq 1$. Hence $n$ cannot be any positive integer, nor can it be any odd integer, as $n = 1 > 0; \, n \text{ is odd.}$ – amWhy Jan 21 at 14:52
@amWhy, Tank's for you coments – Elias Jan 21 at 14:53
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