Find the basis so that the matrix will be diagonal

[Hall]

Homework Statement

Let $V$ be the Linear space of all real polynomials of degree $\leq3$. Let D denote the differentiation operator and let $T : V \to V$ be the Linear transformation which maps $p(x)$ onto $x p'(x)$.

Let $W$ be the image of $V$ under $TD$. Find the bases for V and W relative to which the matrix of $TD$ is in diagonal form.

Relevant Equations

See the main body, please

First of all, it is clear that we can find such a bases (the theorem is given in almost all of the books, but if you want to share some insight I shall be highly grateful.)

We can show that $W$ will be the set of all real polynomials with degree $\leq 2$. So, let's have $\{1,x,x^2\}$ as the basis for $W$.

We want this matrix (having this idea that V is four dimensional):
$$
\begin{bmatrix}
1 &0&0&0\\
0&1&0&0\\
0&0&1&0
\end{bmatrix}
$$
Let the basis for $V$ be $\{e_1,e_2,e_3,e_4\}$. Then, $e_1 = c_1 + c_2x +c_3x^2 + c_4x^3$, we have
$$
TD (e_1)= 1 $$
Which implies
$$
6c_4 x^2 + 2c_3x -1=0$$

But I cannot solve c's. Now, I seek for your gentle guidance.

 

[PeroK]

Homework Statement:: Let $V$ be the Linear space of all real polynomials of degree $\leq3$. Let D denote the differentiation operator and let $T : V \to V$ be the Linear transformation which maps $p(x)$ onto $x p'(x)$.

Let $W$ be the image of $V$ under $TD$. Find the bases for V and W relative to which the matrix of $TD$ is in diagonal form.
Relevant Equations:: See the main body, please

We can show that $W$ will be the set of all real polynomials with degree $\leq 2$. So, let's have $\{1,x,x^2\}$ as the basis for $W$.

What polynomial maps to the polynomial $1$?

We want this matrix (having this idea that V is four dimensional):
$$
\begin{bmatrix}
1 &0&0&0\\
0&1&0&0\\
0&0&1&0
\end{bmatrix}
$$

I would say that only a square matric can be diagonal.

Let the basis for $V$ be $\{e_1,e_2,e_3,e_4\}$. Then, $e_1 = c_1 + c_2x +c_3x^2 + c_4x^3$, we have
$$
TD (e_1)= 1 $$
Which implies
$$
6c_4 x^2 + 2c_3x -1=0$$

But I cannot solve c's. Now, I seek for your gentle guidance.

Perhaps rethink this approach.

It's a strange question. Are you sure it's not supposed to be $DT$?

 

[Hall]

What polynomial maps to the polynomial 1

Well, I don't think (I mean I cannot find) there is a polynomial $p(x)$ such that $x p''(x)=1$.

I would say that only a square matric can be diagonal.

Sometimes, I have heard the book to say "the main diagonal".

a strange question. Are you sure it's not supposed to be DT?

Yes, it is TD.

Question number 20. (Apostle's Calculus Vol. 2)

 

[PeroK]

If you stick with the basis $1, x, x^2, x^3$ for $V$, can you find a suitable basis for $W$?

 

[Hall]

If you stick with the basis $1, x, x^2, x^3$ for $V$, can you find a suitable basis for $W$?

$TD(1)=0$, and that may make the whole first column zero, as zero element cannot fall in the basis for W.

 

[PeroK]

$TD(1)=0$, and that may make the whole first column zero, as zero element cannot fall in the basis for W.

Well, okay. The diagonal entries don't have to be non-zero.

 

[Office_Shredder]

$TD(1)=0$, and that may make the whole first column zero, as zero element cannot fall in the basis for W.

This is going to be unavoidable, since the domain is higher dimension than the codomain

 

[Hall]

So, should we consider "the main diagonal" as starting from the first element of the second column?

Should we seek for
$$
\begin{bmatrix}
0 &1& 0& 0\\
0 &0 &1&0\\
0 & 0 &0 &1
\end{bmatrix}
$$
?

 

[PeroK]

So, should we consider "the main diagonal" as starting from the first element of the second column?

Should we seek for
$$
begin{bmatrix}
0 &1& 0& 0\\
0 &0 &1&0\\
0 & 0 &0 &1
\end{bmatrix}
$$
?

I don't think the question makes much sense unless we consider $TD: V \to V$.

Asking for the dimension of the range of $TD$ would be a better question to get you started.

 

[Hall]

I don't think the question makes much sense unless we consider $TD: V \to V$.

Asking for the dimension of the range of $TD$ would be a better question to get you started.

Yes, of course the dimension of W is 3 and that of V is 4. But that's a very nice point to consider TD as a mapping from V to V, not to TD(V).

But still, if we consider the basis for the codomain V as $\{1, x, x^2, x^3\}$, we would still get the issue of finding a polynomial such that $x p''(x)=1$.

I'm getting this feeling that TD(V) shall not contain 1.

 

[PeroK]

Yes, of course the dimension of W is 3

Is it?

But still, if we consider the basis for the codomain V as $\{1, x, x^2, x^3\}$, we would still get the issue of finding a polynomial such that $x p''(x)=1$.

Why consider the codomain first? I'd look first at the action of $TD$ on the usual basis for $V$.

I'm getting this feeling that TD(V) shall not contain 1.

It's more than a feeling!

 

[Hall]

How about the following one?

Basis for W $\{x,x^2\}$. Let the basis for $V$ be $\{e_1, e_2, e_3, e_4\}$. And we want the matrix to look like
$$
\begin{bmatrix}
1 &0&0&0\\
0&1&0&0
\end{bmatrix}
$$
$TD(e_1) = x$, by inspection I can say $e_1= x^2/2$.

$TD(e_2)= x^2$ by inspection we have $e_2= x^3/6$

$TD (e_3) =0$, we can easily get $e_3 =1$

$TD(e_4)=0$, I think we have $e_4=x$

 

[pasmith]

I think that noting $$TD : x^k \mapsto k(k-1)x^{k-1}$$ gets you there immediately: $TD(\{1,x,x^2,x^3\}) = \{0, 2x, 6x^2\}$ must span $W$, so a basis for $W$ is $\{TD(x^2) = 2x, TD(x^3) = 6x^2\}$.

 

[Hall]

I think that noting $$TD : x^k \mapsto k(k-1)x^{k-1}$$ gets you there immediately: $TD(\{1,x,x^2,x^3\}) = \{0, 2x, 6x^2\}$ must span $W$, so a basis for $W$ is $\{TD(x^2) = 2x, TD(x^3) = 6x^2\}$.

I just wanted to hear from you how did you ensure
$$TD (\{1, x, x^2, x^3\}) = \{0, 2x, 6x^2\}$$
span W? I mean do we have theorem or a something which says that Linear transformation when applied to the basis of its domain results in values which we would span the range?

 

[PeroK]

I mean do we have theorem or a something which says that Linear transformation when applied to the basis of its domain results in values which we would span the range?

It follows from the linearity of $T$:
$$T(v) = T(v_1e_1 + v_2e_2 \dots + v_ne_n) = v_1T(e_1) + \dots + v_nT(e_n)$$