Let $A\in M_n(\mathbb R)$ be a non-zero non-identity matrix such that $A^t=A^2$. Then is it possible to find the minimal polynomial of $A$?

My try is: the given condition implies that $A^4=A$, hence the minimal polynomial $p(t)$ of $A$ divides $t^4-t=t(t-1)(t^2+t+1)$. So $p(t)$ can be some factor of $t(t-1)(t^2+t+1)$.

Also if $\lambda$ is an eigenvalue of $A$ and $x$ is the corresponding eigenvector (considered as a column vector), then $Ax=\lambda x \implies \lambda x^t=x^t A^t\implies \lambda x^t=x^tA^2\implies \lambda(x^t x)=x^t(A^2x)\implies \lambda(x^t x)=\lambda^2(x^tx)\implies (\lambda^2-\lambda)(x^tx)=0\implies\lambda^2-\lambda=0,\text{ as } x^tx \text{ is non-zero}.$

Hence $0$ and $1$ are the only eigenvalues of $A$. So the minimal polynomial is of the form $p(t)=t(t-1).$ Is this correct or did I misunderstood something?

P.S. $A^t$ denotes the transpose of the matrix $A$.

  • $\begingroup$ Looks correct to me $\endgroup$
    – marwalix
    Feb 6 '15 at 10:21
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    $\begingroup$ @fvel: I believe $t$ indicates the transpose... $\endgroup$
    – Fabian
    Feb 6 '15 at 10:22
  • $\begingroup$ @fvel $t$ denotes transposition and not a power $\endgroup$
    – marwalix
    Feb 6 '15 at 10:23
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    $\begingroup$ This isn't right. There are $2 \times 2$ matrices $A$ of order $3$ with $A^{t} = A^{2}$, for example $\left( \begin{array}{clcr} \cos \frac{2 \pi}{3} & \sin \frac{2 \pi}{3}\\ -\sin\frac{2 \pi}{3} & \cos \frac{2 \pi}{3} \end{array} \right) $. You have overlooked the fact that a real matrix can have a pair of complex conjugate eigenvalues. Note that this example has characteristic polynomial $x^{2}+x + 1.$ There are examples where the min poly really is $x^{4} - x.$ $\endgroup$ Feb 6 '15 at 10:50
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    $\begingroup$ If you have complex numbers then it is possible that $x^tx=0$ for nonzero vectors $x$. The eigenvectors of rotation matrices are $[1,i]^t$ and $[1,-i]^t$ $\endgroup$
    – Empy2
    Feb 6 '15 at 12:19

Since $$ A^4=\left(A^2\right)^2=\left(A^T\right)^2=\left(A^2\right)^T=\left(A^T\right)^T=A $$ the minimal polynomial must divide $x^4-x=x(x-1)(x^2+x+1)$.

For example, $$ A=\begin{bmatrix} -\frac12&\frac{\sqrt3}2\\ -\frac{\sqrt3}2&-\frac12 \end{bmatrix} $$ has minimal polynomial $x^2+x+1$ and $A^2=A^T$. $$ B=\begin{bmatrix} -\frac12&\frac{\sqrt3}2&0\\ -\frac{\sqrt3}2&-\frac12&0\\ 0&0&1 \end{bmatrix} $$ has minimal polynomial $x^3-1$ and $B^2=B^T$. $$ C=\begin{bmatrix} -\frac12&\frac{\sqrt3}2&0\\ -\frac{\sqrt3}2&-\frac12&0\\ 0&0&0 \end{bmatrix} $$ has minimal polynomial $x^3+x^2+x$ and $C^2=C^T$. $$ D=\begin{bmatrix} -\frac12&\frac{\sqrt3}2&0&0\\ -\frac{\sqrt3}2&-\frac12&0&0\\ 0&0&1&0\\ 0&0&0&0 \end{bmatrix} $$ has minimal polynomial $x^4-x$ and $D^2=D^T$.

  • $\begingroup$ So if I assume that $A$ is a non-zero non-identity matrix whose characteristic polynomial splits over $\mathbb{R}$, then is it true that $p(t)=t(t-1)$? I am curious to know $\endgroup$
    – user149418
    Feb 6 '15 at 13:25
  • $\begingroup$ @user149418: the minimal polynomial, $P$, of a matrix, $M$, is the monic polynomial of minimal degree so that $P(M)=0$. $\endgroup$
    – robjohn
    Feb 6 '15 at 14:29

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