Show that the span of two vectors is T-invariant Show that the span of \begin{bmatrix}1\\1\\0\\1\end{bmatrix} and \begin{bmatrix}1\\0\\2\\0\end{bmatrix} is a $T$-invariant subspace of the linear map given by 
\begin{bmatrix}4&-2&-1&-1\\ 3&-1&-1&-1\\-2&2&2&0\\1&-1&0&1\end{bmatrix}

I tried to take some general vector in the span and multiply it by the matrix in the hope of getting something that was clearly a linear combination of my two original spanning vectors, but this did not work, that is, the vector was clearly not in the span.
So how am I meant to show $T$-invariance?
Note: Apologies for the formatting, my first time

As the map given by $x\mapsto Tx$ is linear, it suffices to prove that the image of the spanning vectors is again in the span of the two vectors. This is done by the following computation:
\begin{align*}\begin{bmatrix}4&-2&-1&-1\\ 3&-1&-1&-1\\-2&2&2&0\\1&-1&0&1\end{bmatrix}\begin{bmatrix}1\\1\\0\\1\end{bmatrix}&=\begin{bmatrix}4\cdot 1+(-2)\cdot 1+(-1)\cdot 0+(-1)\cdot 1\\3\cdot 1+(-1)\cdot 1+(-1)\cdot 0+(-1)\cdot 1\\(-2)\cdot 1+2\cdot 1+2\cdot 0+0\cdot 1\\1\cdot 1+(-1)\cdot 1+0\cdot 0+1\cdot 1\end{bmatrix}=\begin{bmatrix}1\\1\\0\\1\end{bmatrix}\\
&=1\cdot \begin{bmatrix}1\\1\\0\\1\end{bmatrix}+0\cdot\begin{bmatrix}1\\0\\2\\0\end{bmatrix}\end{align*}
\begin{align*}\begin{bmatrix}4&-2&-1&-1\\ 3&-1&-1&-1\\-2&2&2&0\\1&-1&0&1\end{bmatrix}\begin{bmatrix}1\\0\\2\\0\end{bmatrix}&=\begin{bmatrix}4\cdot 1+(-2)\cdot 0+(-1)\cdot 2+(-1)\cdot 0\\3\cdot 1+(-1)\cdot 0+(-1)\cdot 2+(-1)\cdot 0\\(-2)\cdot 1+2\cdot 0+2\cdot 2+0\cdot 0\\1\cdot 1+(-1)\cdot 0+0\cdot 2+1\cdot 0\end{bmatrix}=\begin{bmatrix}2\\1\\2\\1\end{bmatrix}\\
&=1\cdot \begin{bmatrix}1\\1\\0\\1\end{bmatrix}+1\cdot \begin{bmatrix}1\\0\\2\\0\end{bmatrix}
\end{align*}
Another way to see this (but unnecessary complicated is to see that the spanned set is the kernel of $(T-I)^2$ and as such is $T$-invariant. (This is suggested by the theory of Jordan normal form.
