Can i have a column filled with zeroes when computer $T$?

Question

Let $T: P2 \rightarrow P2$ be the liner transformation defined by $T(p(x)) = xp'(x)$. Find the matrix $A= [T]$ of $T$ relative to the standard basis of $P2$ : B=$(1,x,x^2) $

My answer basically was:

.$[T(1)]$ relative to $B$ : $(0,0,0) $

.$[T(x^2)]$ relative to $B$ : $(0,0,2)$

.$[T(x)]$ relative to $B$ : $(0,1,0)$

In $[T(1)]$ I have a whole column of zero, did I make a mistake somewhere or this is normal?

I also obtained the same in the second question: Find the matrix $A= [T]$ of $T$ relative to the standard basis of $B' = \{1+x^2, 2x, 1\}$:

.$[T(1)]$ relative to $B'$ : $(0,0,0) $

.$[T(2x)]$ relative to $B'$ : $(0,1,0)$

.$[T(1+x^2)]$ relative to $B'$ : $(2,0,0)$

Thank you.

Bonus question: How to I compute the transition matrix from $B'$ to $B$?

Answer 1

To explain why things work the way they do,

Let $T$ be our linear transformation in question with respect to basis $B=\{e_1,e_2,e_3\}$ where $e_1=\begin{bmatrix}1\\0\\0\end{bmatrix},e_2=\begin{bmatrix}0\\1\\0\end{bmatrix},e_3=\begin{bmatrix}0\\0\\1\end{bmatrix}$ are our standard basis vectors.

An arbitrary vector then, $v := v_1\cdot e_1 + v_2\cdot e_2 + v_3\cdot e_3=\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}$

We wish to come up with a matrix, $A_T$, such that following through the matrix multiplication yields the same result as applying the transformation directly. That is to say, $A_T\cdot v = T(v)$. To do this, we garner information from the basis vectors $e_1,e_2,e_3$.

Since we know that applying this transformation takes us from a space to the same space, we know that $A_T$ must be square (this is not always the case).

If $T(e_1) = \alpha e_1 + \beta e_2 + \gamma e_3=\begin{bmatrix}\alpha\\\beta\\\gamma\end{bmatrix}$, that implies the following:

$$A_T \begin{bmatrix}1\\0\\0\end{bmatrix} = \begin{bmatrix}\color{red}{?}&?&?\\\color{blue}{?}&?&?\\\color{green}{?}&?&?\end{bmatrix}\begin{bmatrix}1\\0\\0\end{bmatrix} = T(e_1) = \begin{bmatrix}\alpha\\\beta\\\gamma\end{bmatrix}$$

By carrying out the matrix multiplication in the center, you get $\begin{bmatrix}\color{red}{?}\\\color{blue}{?}\\\color{green}{?}\end{bmatrix}=\begin{bmatrix}\alpha\\\beta\\\gamma\end{bmatrix}$ and so we have learned what the first column of $A$ looks like.

$A_T = \begin{bmatrix}\alpha & ? & ?\\\beta &?&?\\\gamma&?&?\end{bmatrix}$

You can figure out the other columns of $A_T$ as well by using the rest of the standard basis vectors.

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For your specific example, $T(p(x)) = xp'(x)$ going from $P_2\to P_2$ using the standard basis $\{1,x,x^2\}$, we calculate:

• $T(e_1) = T(1) = x\cdot\frac{d}{dx}[1] = x\cdot 0 = 0 = 0\cdot 1 + 0\cdot x + 0\cdot x^2=\color{red}{\begin{bmatrix}0\\0\\0\end{bmatrix}}$
• $T(e_2) = T(x) = x\cdot\frac{d}{dx}[x] = x\cdot 1 = x = 0\cdot 1 + 1\cdot x + 0\cdot x^2=\color{blue}{\begin{bmatrix}0\\1\\0\end{bmatrix}}$
• $T(e_3) = T(x^2) = x\cdot \frac{d}{dx}[x^2]=x\cdot 2x = 2x^2 = 0\cdot 1 + 0\cdot x + 2\cdot x^2 = \color{green}{\begin{bmatrix}0\\0\\2\end{bmatrix}}$

(note: it is much more organized to do these calculations in order as they appear in the definition for the basis)

Thus, $A_T = \begin{bmatrix}\color{red}{0} & \color{blue}{0} & \color{green}{0}\\ \color{red}{0} & \color{blue}{1} & \color{green}{0} \\ \color{red}{0} & \color{blue}{0} & \color{green}{2}\end{bmatrix}$ with respect to $B$.

(note: it is perfectly acceptable for an entire column or row to equal zeroes. That does not contradict the definition of a transformation being linear)

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Let us see what happens when we change the basis we are working with to something else. I will not use the same basis as your problem asks to allow you to work it on your own, but instead a similar example. Suppose our basis is instead $B''=\{2x^2-1, x, 2\}$. We will work with $T$ as defined above: $T=x\cdot p'(x)$ from $P_2\to P_2$.

We do the same process:

• $T(e_1) = T(2x^2-1) = 4x^2 =4x^2 - 2 + 2 = 2\cdot (2x^2-1)+0\cdot (x) +1\cdot (2) = \begin{bmatrix}2\\0\\1\end{bmatrix}$
• $T(e_2) = T(x) = x = \begin{bmatrix}0\\1\\0\end{bmatrix}$
• $T(e_3) = T(2) = 0 = \begin{bmatrix}0\\0\\0\end{bmatrix}$

So $A_T = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0\\ 1 & 0 & 0\end{bmatrix}$ with respect to $B''$.

The only way that the problem changed is that we need to think in terms of our new basis, not our old.

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For computing a change of basis matrix, try to apply the same ideas. We wish to use a matrix such that by inputting a standard basis vector with respect to the one basis, it outputs the vector that it equals with respect to the other basis.

Using my modified example, letting $K_{B''\to B}$ represent our change of basis matrix (and through an abuse of notation operation as well):

• $K_{B''\to B} (e_1) = K_{B''\to B}(2x^2-1) = 2x^2-1 = -1\cdot (1) + 0\cdot (x) + 2\cdot (x^2) = \begin{bmatrix}-1\\0\\2\end{bmatrix}$
• $K_{B''\to B} (e_2) = K_{B''\to B}(x)=x=0\cdot (1)+1\cdot (x) + 0\cdot (x^2) = \begin{bmatrix}0\\1\\0\end{bmatrix}$
• $K_{B''\to B}(e_3) = K_{B''\to B}(2) = 2 = 2\cdot (1) + 0\cdot (x) + 0\cdot (x^2) = \begin{bmatrix}2\\0\\0\end{bmatrix}$

Thus the transition matrix for $K_{B''\to B}$ is:

$$\begin{bmatrix}-1 & 0 & 2\\ 0 & 1 & 0\\ 2 & 0 & 0\end{bmatrix}$$

Answer 2

The standard basis for P2 is {$1$, $x$, $x^2$}.

And $T: P_{2} \rightarrow P_{2}$ is defined by: $T(f)(x) = xf'(x)$

So $T(1)=0, T(x)=x, T(x^2)=2x^2$

So the matrix $A_T$ of T for $T\left[ \begin{array}{} 1 \\ x \\ x^2 \\ \end{array} \right]$ is

$ A_T= \left[ \begin{array}{} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \\ \end{array} \right] $