Let $T: \mathbb{R}^5 \rightarrow P_2(\mathbb{R})$ be a linear transformation that has the matrix $$A=\begin{pmatrix} 1 & 3 & 2 & 0 & -1 \\ 2 & 6 & 4 & 6&4 \\ 1& 3 & 2 & 2&1 \end{pmatrix}$$ relative to the bases ${(1, 1, 1, 1, 1),(1, 1, 1, 1, 0),(1, 1, 0, 0, 0),(1, 0, 0, 0, 0),(0, 0, 0, 1, 0)}$ of $\mathbb{R}^5$ and ${(1 + x + x^2 , x, 1)}$ of $P_2(R)$
Find $Ker(T)$ and find $Im(T)$
I know the steps to find kernel and image, but i'm not sure how to first make the matrix with respect to the two bases, please help!!!
Let $$\alpha = \left\lbrace \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 0\\ 0\\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} \right\}.$$ Let $\beta = \{1 + x + x^2,x,1 \} \subset P_2(\Bbb R)$. Then $$[T]_\alpha^\beta = A = \begin{bmatrix} 1&3&2&0&-1\\ 2&6&4&6&4\\ 1&3&2&2&1 \end{bmatrix}.$$ Row reducing, you can find that $$\text{null} A = \text{span} \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ -1 \\ 1 \end{bmatrix} , \begin{bmatrix} -2\\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -3\\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \right\};$$ $$\text{col}(A) = \text{span} \left\{ \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 6 \\ 2 \end{bmatrix} \right\}. $$ So the image of $T$ is spanned by the vectors $$\beta_1 + 2\beta_2 + \beta_3 = x^2+3x+2 \ \ \text{ and }\ \ 6\beta_2 + 2\beta_3 = 6x + 2. $$ There is a trick to convert your $\text{null}(A)$ into your desired $\ker(T)$. (We could have applied a similar trick to find the image of $T$.) Take the column space of the following matrix. $$[I]_\alpha \begin{bmatrix} 1 & -2 & -3 \\ 0 & 0 & 1 \\ 0 & 1 & 0\\ -1 & 0 & 0\\ 1& 0 & 0 \end{bmatrix}= \begin{bmatrix} 1 & 1 & 1 & 1 & 0\\ 1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} 1 & -2 & -3 \\ 0 & 0 & 1 \\ 0 & 1 & 0\\ -1 & 0 & 0\\ 1& 0 & 0 \end{bmatrix} = \begin{bmatrix} 0&-1&-2\\ 1&-1&-2\\ 1&-2&-2\\ 2&-2&-2\\ 1&-2&-3 \end{bmatrix}.$$ See if you can work out why this is correct.