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Let $V$ be the vector space $P_2$(x) of polynomials in $x$ of degree 2 or less, $W$ be the vector space $M_{2,2}$ of 2×2 real matrices and $T$ be the linear transformation

$T: V → W:$ $a + bx + cx^2$ $\mapsto$ \begin{bmatrix} a-b & b-c \\ 0 & c-a \end{bmatrix} Compute bases for the kernel and range of $T$ and for $V$ and hence verify the general rank-nullity theorem for $T : V → W$.

I don't understand how to get the kernel and range when the linear transformation maps to a vector space of matrices and so cannot compute bases. Any help/solutions would be appreciated.

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For the kernel of $T$ you want those polynomials in $V$ that map to the zero matrix in $W$. So \begin{align*} T(a+bx+cx^2) & = \begin{bmatrix}0&0\\0&0\end{bmatrix}\\ \begin{bmatrix} a-b & b-c \\ 0 & c-a \end{bmatrix} & = \begin{bmatrix}0&0\\0&0\end{bmatrix}. \end{align*} This gives $a=b=c$. So the polynomials which lie in the kernel are of the form $a(1+x+x^2)$, where $a \in \mathbb{R}$. So a basis for the kernel is $$\mathcal{B}_{\text{ker}}=\{1+x+x^2\} \implies \dim(\text{Ker} T)=1$$

Likewise we can go for a basis for the range. First we can get the range: assume that $\begin{bmatrix}p&q\\r&s\end{bmatrix} \in \text{Range }(T)$, then there exists some polynomial $a+bx+cx^2 \in V$ such that \begin{align*} T(a+bx+cx^2) & =\begin{bmatrix}p&q\\r&s\end{bmatrix}\\ \begin{bmatrix} a-b & b-c \\ 0 & c-a \end{bmatrix}& = \begin{bmatrix}p&q\\r&s\end{bmatrix} \end{align*} This gives the following system: $$ \begin{align*} a-b & = p\\ b-c & = q\\ 0 & = r\\ c-a&=s \end{align*} \Longrightarrow \begin{bmatrix} 1&-1&0&|&p\\ 0&1&-1&|&q\\ 0&0&0&|&r\\ -1&0&1&|&s \end{bmatrix} \Longrightarrow \begin{bmatrix} 1&-1&0&|&p\\ 0&1&-1&|&q\\ 0&0&0&|&r\\ 0&0&0&|&p+q+s \end{bmatrix} $$ From this it follows that the range only consists of matrices of the form $\begin{bmatrix}p&q\\0&-p-q\end{bmatrix}$. Now we can go for a basis for the range of $T$ as follows: $$\begin{bmatrix}p&q\\0&-p-q\end{bmatrix}=p\begin{bmatrix}1&0\\0&-1\end{bmatrix}+q\begin{bmatrix}0&1\\0&-1\end{bmatrix}$$ This shows that $$\mathcal{B}_{\text{range}}=\left\{\begin{bmatrix}1&0\\0&-1\end{bmatrix},\begin{bmatrix}0&1\\0&-1\end{bmatrix}\right\} \implies \dim(\text{Range } T)=2$$

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  • $\begingroup$ I'm struggling to show that the range only consists of matrices of that form i approached it by applying T to the standard basis vectors $e_1$,$e_2$ and $e_3$ which got me \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix} and \begin{bmatrix} 0 & 1 \\ 0 & -1 \end{bmatrix}. I'm not sure if this was the correct approach or where to go from here. $\endgroup$ – B.K97 Jan 12 '17 at 1:55
  • $\begingroup$ @B.K97 I have edited the solution to include more steps to find the range. $\endgroup$ – Anurag A Jan 12 '17 at 2:03

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