Matrix with invariant subspace

Question

Find the 5×5 real matrix $ A$ such that:

1. $ A(1\ 1\ 1\ 1\ 1)^T =(1\ 1\ 1\ 1\ 1)^T $
2. $ A^{-1} = A^T $
3. $ A^6 = A $
4. $ A $ has distinct eigenvalues.
5. If $ W $ is the subspace generated by $ (1\ -1\ 0\ 0\ 0)^T $ and $ (0\ 2\ -1\ -1\ 0)^T $, then for any $ w\in W $, $ Aw\in W $.

This is what I tried: First, $A$ is invertible by $2.$ So $ A^5=I $ by 3. and $ A=UDU^{-1} $ with some diagonal matrix $D,$ and so $ D^5 =I $. Thus the eigenvalues of $ A $ are $1, \zeta, \zeta^2, \zeta^3, \zeta^4$ with $\zeta =e^{2\pi i \over 5} $. Now the characteristic polynomial of $A$ is $ t^5 -1 $. So $A$ is similar to $ \begin{bmatrix} 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{bmatrix} $. Now I think we should find $U$ but I don't know how to find it.

Answer 1

Let $v_1=(1\ 1\ 1\ 1\ 1)^T$, $v_2=(1\ -1\ 0\ 0\ 0)^T$, and $v_3=(0\ 2\ -1\ -1\ 0)^T$. Now, take two non-null vectors $v_4$ and $v_5$ which are orthogonal to each other and such that each of them is orthogonal to $v_1$, $v_2$, and $v_3$; I shall take $v_4=(0\ 0\ 1\ -1\ 0)^T$, and $v_5=(1\ 1\ 1\ 1\ -4)^T$.

Now, let $\Omega=\{w_1,w_2,w_3,w_4,w_5\}$ be what we get from applying the Gram-Schmidt process to $\{v_1,v_2,v_3,v_4,v_5\}$ (which is quite easy to apply in this case, since the only two distinct $v_i$'s which are not orthogonal are $v_2$ and $v_3$); we get:

• $w_1=\left(\frac1{\sqrt5}\ \frac1{\sqrt5}\ \frac1{\sqrt5}\ \frac1{\sqrt5}\ \frac1{\sqrt5}\right)^T$;
• $w_2=\left(\frac1{\sqrt{2}}\ -\frac1{\sqrt{2}}\ 0\ 0\ 0\right)^T$;
• $w_3=\left(\frac12\ \frac12\ -\frac12\ -\frac12\ 0\right)^T$;
• $w_4=\left(0\ 0\ \frac1{\sqrt2}\ \frac1{\sqrt2}\ 0\right)^T$;
• $w_5=\left(\frac1{\sqrt{20}}\ \frac1{\sqrt{20}}\ \frac1{\sqrt{20}}\ \frac1{\sqrt{20}}\ -\frac4{\sqrt{20}}\right)$.

Now, take the linear map $T\colon\Bbb R^5\longrightarrow\Bbb R^5$ such that

• $Tw_1=w_1$;
• $Tw_2=\cos\left(\frac{2\pi}5\right)w_2+\sin\left(\frac{2\pi}5\right)w_3$;
• $Tw_3=-\sin\left(\frac{2\pi}5\right)w_2+\cos\left(\frac{2\pi}5\right)w_3$;
• $Tw_4=\cos\left(\frac{4\pi}5\right)w_4+\sin\left(\frac{4\pi}5\right)w_5$;
• $Tw_5=-\sin\left(\frac{4\pi}5\right)w_4+\cos\left(\frac{4\pi}5\right)w_5$.

It is clear that $T^5=\operatorname{Id}$ and that the eigenvalues of $T$ are $1$, $\zeta$, $\zeta^2$, $\zeta^3$ and $\zeta^4$. The matrix of $T$ with respect to the basis $\Omega$ is, of course,$$T^\ast=\begin{bmatrix}1&0&0&0&0\\0&\cos\left(\frac{2\pi}5\right)&-\sin\left(\frac{2\pi}5\right)&0&0\\0&\sin\left(\frac{2\pi}5\right)&\cos\left(\frac{2\pi}5\right)&0&0\\0&0&0&\cos\left(\frac{4\pi}5\right)&-\sin\left(\frac{4\pi}5\right)\\0&0&0&\sin\left(\frac{4\pi}5\right)&\cos\left(\frac{4\pi}5\right)\end{bmatrix},$$which is an orthogonal matrix. So, take $A$ as the matrix of $T$ with respect to the canonical basis, which is $M.T^\ast.M^{-1}$, with$$M=\begin{bmatrix}\frac1{\sqrt5}&\frac1{\sqrt2}&\frac12&0&\frac1{\sqrt{20}}\\\frac1{\sqrt5}&-\frac1{\sqrt2}&\frac12&0&\frac1{\sqrt{20}}\\\frac1{\sqrt5}& 0 &-\frac12&\frac1{\sqrt2}&\frac1{\sqrt{20}}\\\frac1{\sqrt5}&0&-\frac12&-\frac1{\sqrt2}&\frac1{\sqrt{20}}\\\frac1{\sqrt5}&0&0&0& -\frac4{\sqrt20}\end{bmatrix}$$(the columns of $M$ are the $w_i$'s). Since both $M$ and $T^\ast$ are orthogonal matrices, then so is $A$. And the eigenvalues of $A$ are the same as those of $T$; in particular, it has distinct eigenvalues.