Finding base and dimension of the intersection of Ker(T) and Im(T) Given the Linear Transformation $T:\mathbb{R}^3\rightarrow\mathbb{R}^3 $ that satisfies:
$$
T\left(
\begin{array}{c}
1\\
1\\
1\\
\end{array}
\right)=\left(
\begin{array}{c}
2\\
3\\
5
\end{array}
\right),T\left(
\begin{array}{c}
1\\1\\0
\end{array}
\right)=\left(
\begin{array}{c}
2\\2\\4
\end{array}
\right),T\left(
\begin{array}{c}
1\\0\\0
\end{array}
\right)=\left(
\begin{array}{c}
1\\2\\3
\end{array}
\right)$$
I need to find the base & dimension of $ker(T)\cap Im(T)$
I've found the bases for $ker(T)$ and $Im(T)$ and after doing the old wishy washy way my lecturer taught me to find the intersection, I've gotten this result:
$ker(T)\cap Im(T)=${$\left(
\begin{array}{c}
0\\0\\0
\end{array}
\right)$}
thus resulting in $dim(ker(T)\cap Im(T))=0$.
I am uncertain that I did right, I am also not sure about how to (and if I'm able to) find a base for group I got.
(My guess is that the group represented by the zero-vector has no base, as it's dimension is $0$).
 A: The set $\beta=\{(1,1,1),(1,1,0),(1,0,0)\}\subseteq \mathbb{R}^{3}$ is a basis for vector space $\mathbb{R}^{3}$ over $\mathbb{R}$.
Then $$\begin{bmatrix} a\\b\\ c\end{bmatrix}=\alpha_{1}\begin{bmatrix} 1\\ 1\\ 1\end{bmatrix}+\alpha_{2}\begin{bmatrix} 1\\ 1 \\ 0\end{bmatrix}+\alpha_{3}\begin{bmatrix} 1\\ 0 \\ 0\end{bmatrix} \implies \begin{cases} &\alpha_{1}= c,\\ \quad &\alpha_{2}= b-c,\\ \quad &\alpha_{3}=a-b\\ \end{cases}$$
Hence since $T$ is a linear trasnformation, we have
$$\color{red}{T}\begin{bmatrix} a\\b\\ c\end{bmatrix}=c \color{red}{T}\begin{bmatrix} 1\\ 1\\ 1\end{bmatrix}+(b-c)\color{red}{T}\begin{bmatrix} 1\\ 1 \\ 0\end{bmatrix}+(a-b)\color{red}{T}\begin{bmatrix} 1\\ 0 \\ 0\end{bmatrix}$$
$$T\begin{bmatrix} a\\b\\ c\end{bmatrix}=\begin{bmatrix} a+b\\ 2a+c\\ 3a+b+c\end{bmatrix}$$
By definition,
\begin{align*}
{\rm ker}(T)&=\{(a,b,c)\in \mathbb{R}^{3}: T(a,b,c)=(0,0,0)\}\\
&=\left\{(a,b,c)\in \mathbb{R}^{3}: a=-\frac{1}{2}c, b=\frac{1}{2}c, c\in \mathbb{R}\right\}\\
&={\rm span}\left\{\begin{bmatrix}-\frac{1}{2}\\ \frac{1}{2}\\ 1\end{bmatrix} \right\} \implies \dim {\rm ker}(T)=1
\end{align*}
By definition,
\begin{align*}
{\rm im}(T)&=\left\{(x,y,z)\in \mathbb{R}^{3}\exists (a,b,c)\in \mathbb{R}^{3}: T(a,b,c)=(x,y,z)\right\}\\
&=\left\{(x,y,z)\in \mathbb{R}^{3}: x+y-z=0\right\}\\
&=\left\{(x,y,z)\in \mathbb{R}^{3}: x=-y+z,y,z\in \mathbb{R}\right\}\\
&={\rm span}\left\{\begin{bmatrix} -1\\ 1\\ 0\end{bmatrix}, \begin{bmatrix} 1\\ 0 \\ 1\end{bmatrix} \right\} \implies \dim {\rm im}(T)=2
\end{align*}
Therefore,
\begin{align*}
{\rm ker}(T)+{\rm im}(T)&={\rm span}\left\{ \beta_{{\rm ker}(T)}\cup \beta_{\rm im}(T) \right\}\\
&={\rm span}\left\{\color{red}{\begin{bmatrix}-\frac{1}{2}\\ \frac{1}{2}\\ 1\end{bmatrix}}, \color{green}{\begin{bmatrix} -1\\ 1\\ 0\end{bmatrix}, \begin{bmatrix} 1\\ 0 \\ 1\end{bmatrix}} \right\}
\end{align*}
Since $$\det \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} & 1\\ -1 & 1 & 0\\ 1 & 0 & 1\end{bmatrix} =-1\not=0$$
so  ${\rm ker}(T)\oplus {\rm im}(T)= \mathbb{R}^{3}$ hence a basis for ${\rm ker}(T)+ {\rm im}(T)$ is given by
$$\beta_{{\rm ker}(T)+ {\rm im}(T)}=\left\{\begin{bmatrix}-\frac{1}{2}\\ \frac{1}{2}\\ 1\end{bmatrix}, \begin{bmatrix} -1\\ 1\\ 0\end{bmatrix}, \begin{bmatrix} 1\\ 0 \\ 1\end{bmatrix} \right\}$$
Hence $$\dim {\rm ker}(T)+ {\rm im}(T)=3.$$
Therefore,
$${\rm ker}(T)\cap {\rm im}(T)={\rm span} \left\{ \begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix} \right\} \implies \dim {\rm ker}(T)\cap {\rm im}(T)=0.$$
Alternatively, direct by definition
\begin{align*}
{\rm ker}(T)\cap {\rm im}(T)&=\left\{(x,y,z)\in \mathbb{R}^{3}: \begin{cases} x+y-z=0\\ x+\frac{1}{2}z=0,\\ y-\frac{1}{2}z=0\end{cases} \right\}\\
&=\left\{(x,y,z)\in \mathbb{R}^{3}: x=y=z=0 \right\}\\
&={\rm span}\left\{ \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}\right\}
\end{align*}
Therefore a basis for ${\rm ker}(T)\cap {\rm im}(T)$ is given by $\emptyset$ and then the dimension is $0$.
A: Let $\beta_1=(1,1,1)$, $\beta_2=(1,1,0)$ and let $\beta_3=(1,0,0)$. Let $e_1,e_2,e_3$ denote the standard basis on $\mathbb{R}^3$. Note that $e_1=\beta_3$, $e_2=\beta_2-\beta_3$ and that $e_3=\beta_1-\beta_2$. Let $v_i=T\beta_i, \ i=1,2,3.$ Then $Te_1=T\beta_3=v_1=(1,2,3)$, $Te_2=T(\beta_2-\beta_3)=v_2-v_3=(1,0,1)$, $Te_3=T(\beta_1-\beta_2)=v_1-v_2=(0,1,1)$. This gives you the matrix corresponding to $T$ with respect to the standard basis, and is sufficient for what you need and perhaps the more familiar.
You can also ponder the question by drawing a square split in two between $\ker T$ and some other set $U$ - call it the complement of $\ker T$ -  and another square split between Image $T$ and some  other set $V$, the latter you may call the complement of Image $T$. You could ask yourself where elements of $U$ go, and where elements of $\ker T$ end up, draw some arrows and so on.
