Let $T: M_{2×2} \rightarrow P_3$ be the linear transformation
$$T( \begin{bmatrix}
a & b \\
c & d
\end{bmatrix})=(a+c)x^3+(a+b+c)x^2+(d-a-c)x+(a+b+c+d)$$
a) Find bases for the kernel and range of $T$ and compute their dimensions. For the kernel, do I make all of the coeffients=0? Like $(a+c)=0, (a+b+c)=0$… and solve? If so, I just get that $a=-c, b=0,$ $c:$ free,$d=0,$ so would the basis for kernel be the matrix
\begin{bmatrix}
-1 & 0 \\
1 & 0
\end{bmatrix}
with dimension 1? For the range, I reduced the same matrix that I did for the kernel, found the pivot columns (columns 1,2, and 4) and wrote the basis as the set of those original columns: {(x^3+x^2-x+1),(x^2+1),(x+1)}. Is this right? I'm just skeptical since I can't find a way to get the polynomial "x" from that set.
b) Determine whether or not $T$ is one-to-one and/or onto.
If I found the right kernel then this would not be one-to-one since the kernel isn't 0, but for the range, how do I know if it is onto? Since there wasn't a pivot in every row of the matrix I reduced in part (a), does that mean that it is not onto?
Best Answer
For the kernel, you are right and just have to explicitly write down the set.
For the range, a hint is to write the polynomial as $$(a+c)x^3+(a+b+c)x^2+(d−a−c)x+(a+b+c+d)$$ $$=a(x^3+x^2-x+1)+b(x^2+1)+c(x^3+x^2-x+1)+d(x+1)$$ and try to see what the image could be.
In summary, by your calculation, we can have $$\text{Ker}~T=\{\begin{bmatrix} -t & 0 \\ t & 0 \end{bmatrix}|t\in\mathbb{R}\}$$ which is a one-dimensional subspace.
Besides, the image of $T$ is spanned by the basis $\{x^3+x^2-x+1,x^2+1,x+1\},$ so the image is of dimension $3$ while the dimension of $P_3$ is 4, and hence $T$ is not onto.