Find $\mathrm{Ker}(T)$ and $\mathrm{Im}(T)$ of the following linear transformation with bases

Consider the three standard unit vectors,

$$\begin{align*} T(1,0,0) &= \pmatrix{1&5\\0&0}\\ T(0,1,0) &= \pmatrix{0&0\\0&0}\\ T(0,0,1) &= \pmatrix{0&0\\1&3}\\ T(a,b,c) &= a\pmatrix{1&5\\0&0} + c\pmatrix{0&0\\1&3} \end{align*}$$

Kernal: Solving the equation $T(a,b,c) = \pmatrix{0&0\\0&0}$ gives $a=c=0$ and $b$ is a free variable, so a basis of the kernal is $\{(0,1,0)\}$.

Image: The two matrices $\pmatrix{1&5\\0&0}$ and $\pmatrix{0&0\\1&3}$ are already linearly independent, so they form the basis of the image of $T$.

Note that $T$ maps to the space of $2\times 2$ matrices, which, as a vector space is just a $4$ dimensional space: arranging the $4$ coordinates in $2\times 2$ matrix or column or row has no significance in such a context.

The image of $T$ just consists of all matrices of the form $\pmatrix{a&5a\\c&3c}$, so it is spanned by $$B_1:=\pmatrix{1&5\\0&0}\quad\quad B_2:=\pmatrix{0&0\\1&3}$$ because $T(a,b,c)=a\cdot B_1+c\cdot B_2$.
Clearly, they are not scalar multiples of each other, so $\dim{\rm im}T=2$.

By the dimension theorem, then we'll have $\dim\ker T=1$, and as $b$ is not used in the formula for $T$, any $(0,b,0)$ is in the kernel (a basis of it is the one vector $(0,1,0)$).

The "kernel" of a linear transformation is the subspace of the domain space of all vectors that are transformed into the 0 vector. Here that would give $T(a, b, c)= \begin{bmatrix}a & 5a \\ c & 3c\end{bmatrix}= \begin{bmatrix}0 & 0 \\ 0 & 0 \end{bmatrix}$ so that we must have $ a= 0, 5a= 0, c= 0, 3c= 0$. That is, $a= c= 0.$ Because there is no $'b'$ in that, $b$ can be anything. So the kernel consists of all vectors of the form $(0, b, 0)$, the one dimensional subspace spanned by $(0, 1, 0).$

The "image" of a linear transformation is the subspace of the range space of all vectors that something in the domain is mapped to. Here, any $(a, b, c)$ is mapped into $\begin{bmatrix}u & v \\ w & x\end{bmatrix}= \begin{bmatrix} a & 5a \\ c & 3c\end{bmatrix}$. That means that $v= 5u$ and $x= 3c$. So the image consists of matrices $\begin{bmatrix}u & 5u \\ c & 3c\end{bmatrix}= u\begin{bmatrix}1 & 5 \\ 0 & 0 \end{bmatrix}+ v\begin{bmatrix}0 & 0 \\ 1 & 3 \end{bmatrix}$.