How many $3 \times 3$ non-symmetric and non-singular matrices $A$ are there such that $A^{T}=A^2-I$?

How many $$3 \times 3$$ non-symmetric and non-singular matrices $$A$$ are there such that $$A^{T}=A^2-I$$?

Note: $$I$$ denotes the identity matrix of size $$3 \times3$$ and $$A^{T}$$ represents the transpose of the matrix $$A$$.

I took transpose on both sides in given equation to get $$A=(A^T)^2-I$$ and then I put value of $$A^T$$ in this equation using $$A^{T}=A^2-I$$ to get $$A^3-2A-I=0$$ which ultimately gave $$(A^T-A)(A+I)=0$$ . How to deal the problem from here? And is there any better approach to tackle this problem? The given answer is $$0$$.

• Well, don't you have your answer? From your matrix equation you get the fact that the only matrices satisfying this equation are either symmetric matrices or $-I$, so no non-symmetric matrices exist which satisfy the matrix equation Commented Apr 24, 2018 at 20:17
• @junkquill The product of two matrices is 0 doesn't imply that one matrix is zero. Commented Apr 24, 2018 at 20:20
• Ah! My bad, completely slipped my mind Commented Apr 24, 2018 at 20:20
• Did you get this from the GRE? Sounds familiar
– BCLC
Commented Apr 24, 2018 at 20:20
• @dan_fulea In this context that's a pretty pedantic question, it is apparent that it's either $\mathbb{R}$ or $\mathbb{C}$ and the answer doesn't depend on which.
– Ian
Commented Apr 24, 2018 at 20:31

We have $$(A^T - A)(A + I)=0$$

Since $A$ is nonsingular, $\det A \neq 0 \implies \det A^T \neq 0 \implies \det (A^2 - I) \neq 0 \implies \det(A+I)\cdot\det(A-I)\neq 0$ $$\implies \det(A+I) \neq 0 \neq \det(A-I)$$

Therefore, $$(A^T - A)(A + I)=0 \implies A^T - A = 0$$ And we're done

Here's something else that I noticed. You arrive at the following equation: $$A^3 - 2A - I=0 \implies A^3 - A = A+I \implies A(A^2 - I) = A+I \implies AA^T = A +I \\ \implies A = AA^T - I$$ Therefore, $A$ is symmetric

• If $A=I$ then $\det(A^T-A)=0$. Commented Apr 24, 2018 at 20:57
• $A = I$ doesn't satisfy the original matrix equation, so that case will never arise Commented Apr 24, 2018 at 20:58

I would like to say why the above dan_fulea's comment is good and Ian's comment (upvoted!) is inappropriate and also why it is assumed that $A$ is invertible. Note also that the hypothesis $n=3$ has nothing to do here. Assume that $A\in M_n(K)$.

Case 1. $K=\mathbb{R}$. Since $AA^T=A^TA$, $A$ is normal and therefore unitarily diagonalizable; then we may assume that $A=diag(\lambda_i),A^T=diag(\overline{\lambda_i})$ where $\overline{\lambda_i}=\lambda_i^2-1$. The previous equation has only real solutions and, consequently, $A=A^T$. Note that the hypothesis "$A$ invertible" is useless.

Case 2. $K=\mathbb{C}$. Note that $A(A+I)(A^2-A-I)=0$ and $A$ is diagonalizable over $\mathbb{R}$.

If $\det(A)\not= 0$, then $A(A^2-I)$ is invertible and consequently $A(A+I)$ too; we deduce easily that $A^2-A-I=0$ and $A=A^T$.

Assume that $\det(A)=0$ and $A$ non-symmetric; then $0,0^2-1=-1\in spectrum(A)$ and $A^2-A-I\not= 0$.

For $n=2$ a solution is $A_2=1/2\begin{pmatrix}-1&-i\\i&-1\end{pmatrix}$ where $spectrum(A_2)=\{0,-1\}$.

For $n=3$, a solution is $A_3=diag(A_2,(1+\sqrt{5})/2)$ where $spectrum(A_3)=\{0,-1,(1+\sqrt{5})/2\}$.

• For $K=\mathbb{R}$, correct me if I'm wrong, if $A$ wasn't invertible, then there would be at least one $\lambda_i=0$ and then it wouldn't satisfy the original matrix equation. So I'm guessing, it's useful Commented Apr 26, 2018 at 19:36
• @junkquill , Yes, when $K=\mathbb{R}$, all the solutions (necessarily symmetric) are invertible.
– user91684
Commented Apr 26, 2018 at 20:45

Your idea is very good: $A=(A^T)^2-I$, so $$A=(A^2-I)^2-I$$ which becomes $A=A^4-2A^2+I-I$ and, owing to $A$ nonsingular, $A^3-2A-I=0$. This can be rewritten as $A(A^2-I)-(A+I)=0$ or $(A+I)(A^2-A-I)=0$, hence $(A+I)(A^T-A)=0$.

You have to exclude that $-1$ is an eigenvalue of $A$, so ensuring $A+I$ is invertible, which would lead to $A^T=A$.

If $Av=-v$, with $v\ne0$, then $A^Tv=A^2v-v=0$. This is not possible because $A^T$ is nonsingular. The assumption the matrices are $3\times3$ is not required.

Since $A^3-2A-I=0$ , any eigenvalue of $A$ satisfies $$\lambda^3-2\lambda-1=0 \Rightarrow \lambda=-1, \frac{1 \pm \sqrt{5}}{2}$$

Let $\lambda_{1,2,3}$ be the eigenvalues of $A$. Then $$tr(A^T)=tr(A^2-I) \Rightarrow \lambda_1+\lambda_2+\lambda_3 = \lambda_1^2+\lambda_2^2+\lambda_3^2-3 \\ \Rightarrow \sum_{j=1}^3(\lambda_j^2 -\lambda_j -1) =0 \qquad (*)$$

Note that $\lambda= \frac{1 \pm \sqrt{5}}{2}$ satisfy $\lambda^2 -\lambda-1=0$, while $\lambda=-1$ satisfies $\lambda^2 -\lambda-1>0$

It follows from $(*)$ that $-1$ is NOT an eigenvalue of $A$.

Therefore, the minimal polynomial of $A$ is $$\mu_A(x)=(x- \frac{1 +\sqrt{5}}{2})^\alpha (x- \frac{1 -\sqrt{5}}{2})^\beta$$

Moreover, since $A^3-2A-I=0$ we have $\mu_A(x) | X^3-2X-1$ and hence $\alpha, \beta \leq 1$.

Therefore, we have three choices left: $$\mu_A(x)=(x- \frac{1 +\sqrt{5}}{2})\\ \mu_A(x)= (x- \frac{1 -\sqrt{5}}{2})\\ \mu_A(x)=(x- \frac{1 +\sqrt{5}}{2}) (x- \frac{1 -\sqrt{5}}{2})$$

The first two lead to the solutions $$A=\lambda I \qquad \lambda= \frac{1 \pm +\sqrt{5}}{2}$$ While the last gives $$A^2-A-I =0$$ and hence $$A=A^T$$