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 $0$).
Best Answer
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 $0$.