# How to find the basis for $\mathrm{Ker}(T)$ of the following linear transformation?

$$\newcommand{\Ker}{\mathrm{Ker}}$$Find a basis for the vector space $$V$$ so that the first $$\dim(\Ker(T))$$ vectors are a basis for $$\Ker(T)$$, $$T:V\rightarrow W$$ a linear transformation.

1. $$T:R^4 \rightarrow P_2(R)$$ defined by $$T(a_1, ..., a_4)=(a_1+a_2)+(a_2+a_3)x+(a_3+a_4)x^2$$

2. $$T:P_n(R) \rightarrow P_n(R)$$, which is given by differentiation.

In terms of 1), does the matrix look something like $$\begin{bmatrix}a_1+a_2&0&0\\0&a_2+a_3&0\\0&0&a_3+a_4\end{bmatrix}$$? The answer says that $$\{(1,-1,1,-1), e_1, e_2, e_3\}$$ is the basis, but why?

For 2), the answer says $$\{1, x, ..., x^n\}$$ is the basis, but I still cannot write out the matrix.

Hints:

• Do you see how to get each of $1$, $x$, and $x^2$ in the image (using $e_1$, $e_2$, and $e_3$)?
• Do you see that this implies that the image is all of $P_2(R)$ so the kernel is one dimensional?
• Are you able to solve $a_1 + a_2 = 0, a_2 + a_3 = 0, a_3 + a_4 = 0$? Do you see that the first given vector solves this system?
• Are you able to take the derivative of $x$? What is it? Where does this tell you is a $1$ in your matrix?
• Are you able to take the derivative of $x^2$? What is it? Where does this tell you is a $2$ in your matrix?
• Are you able to take the derivative of $x^3$? What is it? Where does this tell you is a $3$ in your matrix?
• $\vdots$
for (1), choose the standard basis $\{(1,0,0,0)^\top, (0,1,0,0)^\top, \cdots\}$ in $R^4$ and $\{1, x, x^2\}$ for $P_2(R).$ with respect to these basis $T$ has the matrix representation $\pmatrix{1&1&0&0\\0&1&1&0\\0&0&1&1}.$