# Given matrix $A \in \Bbb R^{n \times n}$ such that $A^2=-I$, find $\det(A)$

Given an $$n \times n$$ matrix $$A$$ with real entries such that $$A^2=-I$$, find the $$\det (A)$$.

My Attempt:

$$|A|^2=(-1)^n\implies|A|=(-1)^{\frac n2}$$

The answer given is $$|A|=1$$

Eigenvalues are not syllabus. So, can this question be explained without the concept of eigenvalues? Thanks.

Perhaps the condition that the entries are real has something to do with $$|A|=1$$, but not sure how.

A similar question exists here. As per this, $$n$$ is even. Does that mean the question I posted is incomplete?

• A real matrix has a real determinant. The square of a real value cannot be negative. Can you finish? May 19, 2022 at 13:41
• @DatBoi if $A$ is real, and $det(A)=-1$ then $|A|^2=1$. So, $n$ must be even. But that still doesn't confirm us about $det(A)$, does it? May 19, 2022 at 13:50
• May 19, 2022 at 14:23

Let $$p(t)=\det (t-A)$$. Then $$p(t)p(-t)=\det (-I-t^2)=(-1)^n(t^2+1)^n\text{.}$$ If $$n=0$$ then $$p(t)=1$$. If $$n=1$$ then $$t^2 + 1$$ splits into linear factors, a contradiction since it is irreducible.

For other $$n$$ note that $$t^2+1$$ is prime, so it divides either $$p(t)$$ or $$p(-t)$$, and thus divides both. Let $$p(t)=(t^2+1)q(t)$$. Then we must have

$$q(t)q(-t)=(-1)^{n-2}(t^2+1)^{n-2}\text{.}$$ Repeating the argument $$\lfloor n/2\rfloor$$ times gives a contradiction if $$n$$ is odd; for $$n$$ even it gives $$p(t)=(t^2+1)^{n/2}$$ whence $$\det A =1$$ (and not $$-1$$). No complex numbers, no eigenvalues.

Eigenvalues are not syllabus. So, can this question be explained without the concept of eigenvalues?

Here's a proof that does not need the notion of characteristic polynomials or eigenvalues. It does need the notion of a fixed point and some sophistication with orthogonality.

lemma: $$A$$ has no non-zero fixed points
proof: $$A\mathbf x = \mathbf x\implies -\mathbf x=-I \mathbf x = A^2\mathbf x = A\big(A\mathbf x\big) = A \big(\mathbf x\big) = \mathbf x \implies 2\mathbf x = \mathbf 0\implies \mathbf x =\mathbf 0$$
(note this can be interpreted in terms of eigenvalue 1 but such interpretation isn't needed; a fixed point is really a general and very useful concept that shows up in topology and analysis.)

special case: Orthogonal A
Then $$A$$ is skew symmetric because $$A = A^{-1}A^2 =A^TA^2 =A^T(-I) = -A^T$$
checking the determinant
$$0\neq \det\big(A\big)=\det\big(A^T\big)=\det\big(-I_n A\big)=\det\big(-I_n\big)\det\big( A\big)=(-1)^n\det\big(A\big)\implies \text{n is even}$$

Since $$A$$ is orthogonal it may be decomposed into $$m$$ Householder reflection matrices $$H_k$$, each of the form $$H_k:=1 -2\mathbf x_k\mathbf x_k^T$$ for some length one $$\mathbf x_k$$, for some $$m\in\big\{1,2,\dots, n-1,n\big\}$$.
Since $$\det\big(H_k\big)=-1$$ (e.g. by matrix determinant lemma) this us that

$$\det\big(A\big)=-1\implies m \text{ is odd}$$ and $$\det\big(A\big)=1\implies m \text{ is even}$$

now suppose for contradiction that $$\det\big(A\big)=-1$$. We have

$$\det\big(A\big)=-1\implies m \text{ is odd}\implies m\lt n$$ since $$n$$ is even. Thus write
$$A =H_1H_2\cdots H_{m-1}H_m =\prod_{k=1}^m H_k$$
and collect $$M:= \bigg[\begin{array}{c|c|c|c|c|c|c} \mathbf x_1 &\mathbf x_2 & \cdots & \mathbf x_m \end{array}\bigg]$$

then there is some $$\mathbf y\neq \mathbf 0$$ orthogonal to all these $$\mathbf x_k$$
proof: apply rank-nullity to $$M^T$$. Now
$$H_k\mathbf y =\big(I -2\mathbf x_k\mathbf x_k^T\big)\mathbf y = \mathbf y -2\mathbf x_k(\mathbf x_k^T\mathbf y) = \mathbf y -\mathbf 0=\mathbf y$$
so $$\mathbf y$$ is a fixed point for these $$H_k$$ thus

$$A\mathbf y = \prod_{k=1}^m H_k\mathbf y= \prod_{k=1}^{m-1} H_k\mathbf y=\cdots = \prod_{k=1}^2 H_k\mathbf y = H_1\mathbf y =\mathbf y$$
so $$A$$ has a non-zero fixed point, which is a contradiction.

conclude m is even and $$\det\big(A\big)=1$$

general case
$$A$$ is similar to orthogonal matrix $$A'$$, so $$S^{-1}AS= A'$$
proof: $A$ is real matrix and for some $k\geq 2,A^{k}$ is similar to an orthogonal matrix,how to prove $A$ is also similar to an orthogonal matrix?

where of course $$(A')^2=(S^{-1}AS)^2=S^{-1}A^2S= S^{-1}(-I)S = -I$$ so the above argument proves $$\det\big(A'\big)=1$$ and similar matrices have the same determinant so $$\det\big(A\big)=1$$.

$$(\det A)^2=(-1)^n$$ implies that there is such a matrix only if $$n$$ is even. For $$n$$ even, consider the following construction. Pick a non-zero vector $$v_1$$ and call $$v_2=Av_1$$. Then, pick a vector $$v_3\notin \operatorname{span}(v_1,v_2)$$ and $$v_4=Av_3$$. Keep selecting $$v_{2m+1}\notin \operatorname{span}(v_1,\cdots,v_{2m})$$ and $$v_{2m+2}=Av_{2m+1}$$ until $$\operatorname{span}(v_1,\cdots,v_{2m})=\Bbb R^n$$. I claim that $$\dim\operatorname{span}(v_1,\cdots, v_j)=j$$ for all $$j\le n$$. This implies that the procedure terminates at $$n$$ and that $$v_1,\cdots, v_{n}$$ is a basis. In fact, let $$k\le j$$ be the first index such that $$\operatorname{span}(v_1,\cdots, v_{k-1})\ni v_k$$. By construction, $$k$$ can't be odd, therefore $$k=2m+2$$. Notice that $$\operatorname{span}(v_1,\cdots,v_{2m})$$ is invariant for both $$A$$ and $$A^{-1}$$. Now, let $$w\in\operatorname{span}(v_1,\cdots, v_{2m})$$ such that $$v_{2m+2}=\alpha v_{2m+1}+w$$. Then \begin{align} &v_{2m+2}-\alpha v_{2m+1}=w\\ &-A(v_{2m+2}-\alpha v_{2m+1})=\alpha v_{2m+2}+v_{2m+1}=-Aw\end{align}

Solving those two equations for $$v_{2m+1}$$ and $$v_{2m+2}$$ (say, with Cramer) we obtain that $$v_{2m+1}=-\frac1{\alpha^2+1}Aw-\frac\alpha{\alpha^2+1}w\in\operatorname{span}(v_1,\cdots, v_{2m})$$, against minimality of $$2m+2$$. Hence, $$\dim\operatorname{span} (v_1,\cdots, v_j)=j$$.

By the cosinderations above we have a basis $$v_1,\cdots, v_n$$ where the $$v_{2m}=Av_{2m-1}$$ for all $$1\le m\le \frac n2$$. If we make the corresponding change of basis we obtain that $$PAP^{-1}$$ is a block-diagonal matrix where each block is the $$2\times 2$$ matrix $$\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}$$, whose determinant is $$1$$. Therefore $$\det A=\det (PAP^{-1})=1^{n/2}=1$$. The form of $$PAP^{-1}$$ is indeed a matrix such that $$A^2+I=0$$ of size $$n$$ for any even $$n>0$$.

First of all, if $$n$$ is odd, one cannot have a real matrix $$A$$ such that $$A^2=-I$$. Indeed, the characteristic polynomial would be a polynomial of odd degree which means that at least one root is real and there is no real number whose squared is equal to minus one.

If $$n$$ is even, then we need that all the eigenvalues be $$-i$$ or $$i$$ and since the matrix is real, the roots of the polynomial must be complex conjugate, therefore, we have $$n/2$$ roots equal to $$i$$ and $$n/2$$ roots equal to $$-i$$. This means that the determinant of $$A$$ is given by $$(i)^{n/2}(-i)^{n/2}=(1)^{n/2}=1.$$

$$\det(A^2) = \det(-I)$$
$$\det(A) * \det(A) = \det (-I)$$
$$\det(A) * \det(A) = (-1)^n$$
$$\det(A) = \sqrt{(-1)^n}$$

if $$n$$ is a multiple of 2 then $$\det(A)= \pm 1$$, else it doesn't have solution in the real space

• Can you find a $2\times 2$ real matrix such that $A^2+I=0$ and $\det A=-1$? May 19, 2022 at 13:03
• Pick $A=[0\ -1;1\ 0]$?
– KBS
May 19, 2022 at 13:35
• @SassatelliGiulio I know... See my answer.
– KBS
May 19, 2022 at 13:38
• @SassatelliGiulio I can't because it doesn't exist. Can you prove it? May 19, 2022 at 14:00
• You could try to improve your answer (the basic idea is correct and simpler than the other answers): add explanations, logic, do not take a square root before thinking about its validity.
– daw
May 19, 2022 at 14:20