Find $\mathrm{Ker}(T)$ and $\mathrm{Im}(T)$ of the following linear transformation with bases Question: Let T: $\mathbb R^{3} → M_{2\times 2}(\mathbb R)$ be the linear transformation defined by $$T((a,b,c)) = \begin{bmatrix}
         a & 5a\\
         c & 3c\\
        \end{bmatrix} $$ 
Consider the bases $\alpha = \{(0, 1, 0),(0, 1, 1),(1, 1, 0)\}$ of $\mathbb R^{3}$ and
, 
$\beta = \{{\begin{bmatrix}\
         1 &−1\\
         0 & 0\\
        \end{bmatrix}} ,{\begin{bmatrix}
         0 & 1\\
         -1 & 0\\
        \end{bmatrix}} ,{\begin{bmatrix}
         0 & 0\\
         1 & 1\\
        \end{bmatrix}}, {\begin{bmatrix}
         0 &1\\
         0 & -1\\
        \end{bmatrix}}\}$ of $M_{2\times 2}(\mathbb R)$. Find $\mathrm{Ker}(T)$ and $\mathrm{Im}(T)$.
Context: This question is a practice problem given to us by our introductory linear algebra professor for solving and understanding concepts (Note: this is not an assignment or graded homework question) 
Attempt: 
Finding $\mathrm{Ker}(T)$: Row reducing the transformation matrix to RREF would give the matrix \begin{bmatrix}
         1 &0\\
         0 & 1\\
        \end{bmatrix} leaving no free variables thus $\mathrm{Ker}(T)=\mathrm{span}\{0\} $
Finding $\mathrm{Im}(T)$: Since there are no free variables, we use the original columns from $T((a,b,c))$ to get our Image i.e.
$\mathrm{Col}(A) =\mathrm{span}\{\begin{bmatrix}
         a \\
         c \\
        \end{bmatrix}, \begin{bmatrix}
         5a \\
         3c \\
        \end{bmatrix}\}$. I know we use this with the basis $\beta$ to obtain our image. How do we proceed from here?
Do we perform $a\begin{bmatrix}
         1&-1 \\
         0&0 \\
        \end{bmatrix} + c\begin{bmatrix}
         0&0 \\
         1&1 \\
        \end{bmatrix}$ and $5a\begin{bmatrix}
         1&-1 \\
         0&0 \\
        \end{bmatrix} + 3c\begin{bmatrix}
         0&0 \\
         1&1 \\
        \end{bmatrix}$? 
This gives us $\mathrm{Im}(T) = \mathrm{span} \{\begin{bmatrix}
         a&-a \\
         c&c \\
        \end{bmatrix}, \begin{bmatrix}
         a&-5a \\
         3c&3c \\
        \end{bmatrix}\}$. 
Doubt: Is this correct or am I going about this the wrong way? I have a feeling I made a mistake because according to the R-N theorem, my rank and nullity must add up to $3$ but at this point, it adds up to $2$. 
 A: 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 $T$.
A: 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 $(0,1,0)$).
A: 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 $\begin{bmatrix}u & 5u \\ c & 3c\end{bmatrix}= u\begin{bmatrix}1 & 5 \\ 0 & 0 \end{bmatrix}+ v\begin{bmatrix}0 & 0 \\ 1 & 3 \end{bmatrix}$.
A: You say that the kernel $Ker(T) = \{0\}$ 
What happends if you calculate $T(0,b,0)$?
Remember that $ImT = \{ M \in M_{2\times 2} : M = T(a,b,c) \ (a,b,c) \in \mathbb{R^3}\}$
