Let $T:V \rightarrow W$ be a linear transformation. The kernel of $T$, denoted by $\ker(T)$, is the set of all vectors in $V$ that are mapped by $T$ to $0\in W$.
That is, $\ker(T) = \{v \in V \mid T(v) = 0\}$.
It can be shown that $\ker(T)$ is a vector subspace of $V$, but you don’t have to verify that.
Now, let $W$ be the vector space of all symmetric $2$ by $2$ matrices.
Moreover, let $T$ be the linear transformation $T: W \rightarrow \textbf{P}_{2}(\textbf{R})$ defined by
\begin{align*}
T\begin{bmatrix}
a & b\\
b & c
\end{bmatrix} = (a – b) + (b – c)x + (c – a)x^{2}
\end{align*}
Find the dimension of $\ker(T)$ by first identifying a basis for the vector space $\ker(T)$.
Best Answer
Let $A=\begin{pmatrix} a & b\\ b & c \end{pmatrix} \in \ker(T)$. Then, we must have $a-b=b-c=c-a=0$ (all coefficients of the polynomial must vanish). This means $a=b=c$. We have shown that the elements of the kernel are exactly the matrices with equal coefficients.
A basis for $\ker(T)$ is, for example, $\left\{ \begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix} \right\}$, because every matrix with equal coefficients may be obtained by multiplying this particular one by some real number