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 $2$.
Best Answer
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 $T$.