# Linear Transformations,Nullity and Rank. Need to know if my work is right

Question $1$. $$T\begin{pmatrix}a & b\\\ c & d\end{pmatrix}=(d+b)+(b+2c+d)x+(3c+3d)x^2+4dx^3$$

What is the kernel and nullity of $T$? Is it $1-1$? What is the image and rank of $T$? Is it onto.

Answer for $Q1$: $b=c=d=0$. Nullity$=0$ and its $1-1$. The image is $\{1,x,x^2,x^3\}$ and rank$=4$. Its onto since $T$ is $1-1$.

Question $2. T(a_1,a_2,a_3)= (a_2,a_3,a_1)$. Is it $1-1$ and onto?

Answer for $Q2$: $(a_1, a_2, a_3)=0$ so nullity$=0$ and its $1-1$. $R(T)= \mathbb{R}^3$ and its onto since $T$ is $1-1$.

For question $1$, Note that $$T\left(\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}\right) = 0$$
$$(d+b)+(b+2c+d)x+(3c+3d)x^2+4dx^3=b(1+x)+c(2x+3x^2)+d(x+3x^2)$$
You can use a systematic approach via matrices to the problem. Consider the standard basis $$\left\{ E_1=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, E_2=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, E_3=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, E_4=\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right\}$$ of the domain and the standard basis $\{1,x,x^2,x^3\}$ of the codomain. Since $$T(E_1)=0,\quad T(E_2)=1+x,\quad T(E_3)=2x+3x^2\quad T(E_4)=1+x+3x^2+4x^3$$ we conclude that the matrix of $T$ with respect to these bases is $$\begin{bmatrix} 0 & 1 & 0 & 1 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 3 & 3 \\ 0 & 0 & 0 & 4 \end{bmatrix}$$ Performing Gaussian elimination we have $$\to \begin{bmatrix} 0 & 1 & 0 & 1 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 3 & 3 \\ 0 & 0 & 0 & 4 \end{bmatrix} \to \begin{bmatrix} 0 & 1 & 0 & 1 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 4 \end{bmatrix} \to \begin{bmatrix} 0 & 1 & 0 & 1 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ This tells us that the rank of the matrix is $3$, which is the same as the rank of $T$. Therefore the nullity is $4-3=1$.
Since clearly $E_1$ belongs to the kernel of $T$, a basis for the kernel is $\{E_1\}$. A basis for the image is $\{T(E_2),T(E_3),T(E_4)\}$.