Finding a $3 \times 3$ matrix that only has 1 non-trivial invariant subspace of $\mathbb{R}^3$.

Question

Does such a matrix exist, and if so how can I find it? My intuition was the matrix $$M = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$$ as the only eigenspace is $\text{span} \{\mathbf{e}_1\}$, but I'd also need to prove that there's no other 2-dimensional invariant subspace.

If we let $\mathbf{u}$ and $\mathbf{w}$ be arbitrary linearly independent nonzero vectors in $\mathbb{R}$, we want $M\mathbf{u} = \alpha \mathbf{u} + \beta \mathbf{w}$ and $M\mathbf{w} = \gamma \mathbf{u} + \delta \mathbf{w}$.

We then get the system of equations $$\begin{align*} u_1 + u_2 &= \alpha u_1 + \beta w_1 \\ u_2 + u_3 &= \alpha u_2 + \beta w_3 \\ u_3 &= \alpha u_3 + \beta w_3 \\ w_1 + w_2 &= \gamma u_1 + \delta w_1 \\ w_2 + w_3 &= \gamma u_2 + \delta w_3 \\ w_3 &= \gamma u_3 + \delta w_3 \end{align*}$$

I'm sure we could solve this by turning it into a matrix and RREF and finding values of the constants to make it nonzero, but I don't feel like this is the way.

With some research I found the theorem:

Let $T: V \to V$ be a linear operator on a finite dimensional vector space. If $T$ is diagonalizable, and $W$ is a $T$-invariant subspace of $W$, then the restriction of $T$ to $W$, $T_W$, is also diagonalizable.

Does this mean, because the only eigenspace is the above and is one-dimensional, any 2-dimensional subspace cannot be invariant because the eigenspaces of $T_W$ must be eigenspaces of $T$? Also, does this imply that any $n \times n$ Jordan matrix only has a single non-trivial invariant subspace? And does this all generalise to $\mathbb{C}$ as well?

Thanks for any help!

Answer 1

If the Frobenius normal form of the matrix has more than one block then there is more than one invariant subspace.

Hence there must be at most one block, since the matrix has odd degree the block must be a $\lambda$ block.

Notice that for the $\lambda$ block the space of vectors $(x,0,0)$ and the space of vectors $(x,y,0)$ are both invariant.

So there is no such matrix.

Answer 2

Does this mean, because the only eigenspace is the above and is one-dimensional, any 2-dimensional subspace cannot be invariant because the eigenspaces of TW must be eigenspaces of T?

Invariant subspaces need not be eigenspaces! 1-dimensional invariant subspaces do need to consist of eigenvectors (along with $0$), but that's it. So I don't think this works -- for example, consider the linear transformation which rotates the $xy$-plane by $\pi/2$ and fixes the $z$-axis. This has a unique eigenspace, and also has an invariant $2$-dimensional subspace.

Indeed, your matrix has another nontrivial invariant subspace, namely $\operatorname{span}\{e_1,e_2\}$.

Note that we'd have this same problem for any upper-triangular matrix, and thus also for any matrix similar to an upper-triangular matrix! This happens if and only if the minimal polynomial of the matrix has no non-real roots.

So, if we want this to work, we need the matrix $A$ to have complex eigenvalues $\{\lambda, \bar{\lambda}, r\}$ where $\lambda \in \mathbb{C} \setminus \mathbb{R}$ and $r \in \mathbb{R}$. The $r$-Eigenspace of $A$ (which is $1$-dimensional) will need to be our unique nontrivial invariant subspace. Pick a (complex) $\lambda$-eigenvector $v \in \mathbb{C}^3$ of $A$, and let $w = v + \bar{v}$. Note that $\bar{v}$ is a $\bar{\lambda}$-eigenvector of $A$ (hence linearly independent to $v$), and that that $w \in \mathbb{R}^3 \setminus \{0\}$ as a result. I claim that $w$ and $Aw$ are linearly independent in $\mathbb{R}^3$. To see this, suppose there exist $\alpha, \beta \in \mathbb{R}$ such that $\alpha w + \beta Aw = 0$. In other words, we have $(\alpha + \lambda \beta) v + (\alpha + \bar{\lambda} \beta) \bar{v} = 0$, so by linear independence of $v$ and $\bar{v}$ we get $\alpha + \lambda \beta = 0 = \alpha + \bar{\lambda} \beta$. By comparing imaginary parts, we find that $\beta = 0$. Then our original assumption becomes $\alpha w = 0$, which forces $\alpha = 0$ since $w \neq 0$. We conclude that $\operatorname{span}\{w, Aw\}$ is a $2$-dimensional subspace of $\mathbb{R}^3$. Moreover, I claim that this is an invariant subspace. This reduces to showing that $A^2w \in \operatorname{span}\{w,Aw\}$. To see this, note that $\lambda^2 = (\lambda + \bar{\lambda})\lambda - \lambda \bar{\lambda}$ and likewise $\bar{\lambda}^2 = (\lambda + \bar{\lambda})\bar{\lambda} - \lambda\bar{\lambda}$. Then $$A^2 w = \lambda^2 v + \bar{\lambda}^2 \bar{v} = ((\lambda + \bar{\lambda})\lambda - \lambda \bar{\lambda})v + ((\lambda + \bar{\lambda})\bar{\lambda} - \lambda\bar{\lambda})\bar{v} = (\lambda + \bar{\lambda})Aw - \lambda\bar{\lambda}w \in \operatorname{span}\{w,Aw\}.$$

So, we're busted: there is no example of a linear transformation $T : \mathbb{R}^3 \to \mathbb{R}^3$ which has a unique nontrivial (i.e. not of dimension $0$ or $3$) invariant subspace.