Finding the kernel and image of a linear transformation over the field $\Bbb Z_2$ Given the A vector space $V$ over the field $\Bbb Z_2$ and  a linear map $t:V \to V $ .following matrix $T =\begin{bmatrix}3& -1 &1\\-1 & 5 & -1\\ 1 & -1 & 3\end{bmatrix}$  where $T$ is a representation of $t$  Find the Kernel and Image. Finally say whether the following statement is true or false : $\operatorname{Ker} t \oplus \operatorname{Im} t = V $
Idea:
rewrite $T$ as such $T==\begin{bmatrix}1& 1 &1\\1 & 1 & 1\\ 1 & 1 & 1\end{bmatrix}$ however I don't know how to precede 
With @Jim help I found that kernel $=span(\lbrace\begin{bmatrix} -1\\1\\  0\end{bmatrix} \begin{bmatrix} -1\\0\\  1\end{bmatrix}  \rbrace )$
However I don't quite know how to find the kernel $\cap$ Image
 A: Rewriting $t$ as you have is the correct first step.  From there recall that as $\mathbb Z_2$ is a field everything you learned in linear algebra still holds.
You can see that the rank of $t$ is $1$ and the image is spanned by the vector
$$\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}.$$
You know that the kernel has dimension $2$ so you just need to find two linearly independent vectors that are killed by your rewritten $t$.  You can probably just guess them, so I'll let you do that.
As for kernel $\oplus$ image equaling $V$.  You now know that the dimensions add to be $3$ so if the sum is direct then it does indeed equal $V$.  To see that the sum is direct you need to show that the $0$ vector is the only vector in the intersection.  As you know exactly what the image is, just apply $t$ and directly check what's in the kernel.  You will find that the sum is indeed direct.
A: kernel ∩ Image is just the intersection of two subspaces , for  this
 find kernal of $\begin{bmatrix}-1& -1 &1\\1 & 0 & 1\\ 0 & 1 & 1\end{bmatrix}$
this matrix is formed by combining basis vectors of both kernal and img of T 
that will be kernel ∩ Image
