Determine the dimension of a subspace of $\operatorname{Hom}(\mathbb{R}^3, \mathbb{R}^3)$

Question

The exercise is the following. I have to find the dimension of the subspace $$ V\subset\operatorname{Hom}(\mathbb{R}^3, \mathbb{R}^3),\qquad V=\{L\in\operatorname{Hom}(\mathbb{R}^3, \mathbb{R}^3):\operatorname{Im}L\subset Z\} $$ where $$ Z=\{(x_1, x_2, x_3)\in\mathbb{R}^3:x_1+x_2+x_3\leq1,\ x_1+x_2+x_3\geq-1\}. $$ My attempt. I know that $\operatorname{dim}\operatorname{Im}L\leq3$ and it cannot be $3$, because otherwise it means that all the space $\operatorname{Im}L$ is between two planes. So its dimension can be only $2$ or $1$. Now, the book excludes the case $\operatorname{dim}\operatorname{Im}L=1$ and it it says that $\operatorname{dim}\operatorname{Im}L=2$. I cannot understand why. Someone can help me?

Thank You

Answer 1

Let $\nu=(1,1,1)$, check that a subspace $U$ is contained in $Z$ if and only if $\nu\perp U$. Complete $\nu$ to an orthogonal basis $\{\nu,u_1,u_2\}$. Conclude that $\mathrm{Im}~L\subset Z$ if and only if $$ L = \begin{bmatrix} 0 & 0& 0 \\ * & *&* \\ * & *&* \\ \end{bmatrix} $$ in the basis $\{\nu,u_1,u_2\}$. Hence $\dim V= 6$.

Answer 2

It is possible for $\dim \text{im }L$ to be $2,1$ or even $0$, but the dimension of the image of any particular $L\in V$ is irrelevant.

The key idea is that $\text{im }L$ must be a subspace of $W\{(x_1,x_2,x_3):x_1+x_2+x_3=0\}$, which is a subspace of dimension $2$. Furthermore, $L$ can be any map from $\mathbb R^3$ to $W$, so $$ \dim V=\dim \operatorname{Hom}(\mathbb R^3,W)=\dim\operatorname{Hom}(\mathbb R^3,\mathbb R^2), $$ since $W\cong \mathbb R^2$. Can you figure out $\dim\operatorname{Hom}(\mathbb R^3,\mathbb R^2)$?