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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?

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    $\begingroup$ A real matrix has a real determinant. The square of a real value cannot be negative. Can you finish? $\endgroup$
    – DatBoi
    May 19, 2022 at 13:41
  • $\begingroup$ @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? $\endgroup$
    – aarbee
    May 19, 2022 at 13:50
  • $\begingroup$ Related: example of complex structure with negative determinant $\endgroup$
    – K B Dave
    May 19, 2022 at 14:23

5 Answers 5

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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.

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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 $\det\big(A\big)=1$.

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$(\det A)^2=(-1)^n$ 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 $n>0$.

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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 $$(i)^{n/2}(-i)^{n/2}=(1)^{n/2}=1.$$

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

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

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  • $\begingroup$ Can you find a $2\times 2$ real matrix such that $A^2+I=0$ and $\det A=-1$? $\endgroup$ May 19, 2022 at 13:03
  • $\begingroup$ Pick $A=[0\ -1;1\ 0]$? $\endgroup$
    – KBS
    May 19, 2022 at 13:35
  • $\begingroup$ @SassatelliGiulio I know... See my answer. $\endgroup$
    – KBS
    May 19, 2022 at 13:38
  • $\begingroup$ @SassatelliGiulio I can't because it doesn't exist. Can you prove it? $\endgroup$
    – Ander Cruz
    May 19, 2022 at 14:00
  • $\begingroup$ 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. $\endgroup$
    – daw
    May 19, 2022 at 14:20

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