1. ## [SOLVED] Linear Transformations

Could anyone walk me through these two? I need some help getting going.

1) Prove that the map $T:P_n\rightarrow P_n$ defined by $T\left[p\left(t\right)\right]=\frac{dp}{dt}$ is a linear transformation and determine if $T$ is one-to-one.

2) Let $T_A:\mathbb{R}^n\rightarrow \mathbb{R}^m$. Prove that the range of $T_A$ is a subspace of $\mathbb{R}^m$.

2. Originally Posted by leverin4
Could anyone walk me through these two? I need some help getting going.

1) Prove that the map $T:P_n\rightarrow P_n$ defined by $T\left[p\left(t\right)\right]=\frac{dp}{dt}$ is a linear transformation and determine if $T$ is one-to-one.
Take two polynomials $p,q\in P_n$. Therefore,

\begin{aligned}T\left(ap+bq\right)&=\frac{\,d}{\,d t}\left[ap+bq\right]\\ &= \frac{\,d}{\,dt}\left[ap\right]+\frac{\,d}{\,dt}\left[bq\right]\\&=\ldots\end{aligned}

Can you finish this?

Note that $T$ is not one to one. Can you think of an easy counterexample?

2) Let $T_A:\mathbb{R}^n\rightarrow \mathbb{R}^m$. Prove that the range of $T_A$ is a subspace of $\mathbb{R}^m$.
Let $w_1,w_2\in\text{range}\!\left(T_A\right)$. Thus, $w_1=T_A\!\left(v_1\right)$ and $w_2=T_A\!\left(v_2\right)$ for some $v_1,v_2\in\mathbb{R}^n$.

Now note that

\begin{aligned}aw_1+bw_2&=aT_A\!\left(v_1\right)+b T_A\!\left(v_2\right)\\ &= T_A\!\left(av_1\right)+T_A\!\left(bv_2\right)\\ &= T_A\!\left(av_1+bv_2\right)\end{aligned}

Since $T_A\!\left(av_1+bv_2\right)\in\mathbb{R}^m$, we can now say that $\text{range}\!\left(T_A\right)$ is a subspace of $\mathbb{R}^m$.

Does this make sense?

3. Originally Posted by Chris L T521
Take two polynomials $p,q\in P_n$. Therefore,

\begin{aligned}T\left(ap+bq\right)&=\frac{\,d}{\,d t}\left[ap+bq\right]\\ &= \frac{\,d}{\,dt}\left[ap\right]+\frac{\,d}{\,dt}\left[bq\right]\\&=\ldots\end{aligned}

Can you finish this?

Note that $T$ is not one to one. Can you think of an easy counterexample?
I'm a little confused as to where $a$ and $b$ came from.

If I'm interpreting the problem right, a counter example would be $x^3$ because then the transformation would lead to $3x^2$

EDIT: Technically, I guess it would be $p=t^3$ as a counter example. Gotta keep the variables straight.

Let $w_1,w_2\in\text{range}\!\left(T_A\right)$. Thus, $w_1=T_A\!\left(v_1\right)$ and $w_2=T_A\!\left(v_2\right)$ for some $v_1,v_2\in\mathbb{R}^n$.

Now note that

\begin{aligned}aw_1+bw_2&=aT_A\!\left(v_1\right)+b T_A\!\left(v_2\right)\\ &= T_A\!\left(av_1\right)+T_A\!\left(bv_2\right)\\ &= T_A\!\left(av_1+bv_2\right)\end{aligned}

Since $T_A\!\left(av_1+bv_2\right)\in\mathbb{R}^m$, we can now say that $\text{range}\!\left(T_A\right)$ is a subspace of $\mathbb{R}^m$.

Does this make sense?
Again, I'm confused about $a$ and $b$. Are they just arbitrary constants? If so, the math makes sense to me, but I don't see how it proves that it's a subspace.

4. Originally Posted by leverin4
I'm a little confused as to where $a$ and $b$ came from.

If I'm interpreting the problem right, a counter example would be $x^3$ because then the transformation would lead to $3x^2$
To clarify, a and b are arbitrary constants. Now, to show that injectivity fails, you want to pick polynomials $p,q\in P_n:\frac{\,dp}{\,dt}=\frac{\,dq}{\,dt}$. Don't make it too hard.

Again, I'm confused about $a$ and $b$. Are they just arbitrary constants? If so, the math makes sense to me, but I don't see how it proves that it's a subspace.
Yes, a and b are arbitrary constants. Recall that if $W\subset V$ and when $u,v\in W$ we have $au+bv\in W$, then $W$ is a subspace of $V$.

In this case we let $W=\text{range}\!\left(T_A\right)$ and $V=\mathbb{R}^m$.

Does this clarify things?

5. Originally Posted by Chris L T521
To clarify, a and b are arbitrary constants. Now, to show that injectivity fails, you want to pick polynomials $p,q\in P_n:\frac{\,dp}{\,dt}=\frac{\,dq}{\,dt}$. Don't make it too hard.
woops...forgot we were working in $n$ dimensions. So something simple like, in your example in the quote box above, setting

$p: p=t^n$
$q: p=t^n+4$

To finish the proof, I just add:

\begin{aligned}T\left(ap+bq\right)&=\frac{\,d}{\,d t}\left[ap+bq\right]\\ &= \frac{\,d}{\,dt}\left[ap\right]+\frac{\,d}{\,dt}\left[bq\right]\\&=T(ap)+T(bq)\\&=aT(p)+bT(q)\end{aligned}

Then I do the same thing with an arbitrary scalar multiple, right?

Yes, a and b are arbitrary constants. Recall that if $W\subset V$ and when $u,v\in W$ we have $au+bv\in W$, then $W$ is a subspace of $V$.

In this case we let $W=\text{range}\!\left(T_A\right)$ and $V=\mathbb{R}^m$.

Does this clarify things?
This part makes sense now. Thank you.

6. Originally Posted by leverin4
woops...forgot we were working in $n$ dimensions. So something simple like, in your example in the quote box above, setting

$p: p=t^n$
$q: p=t^n+4$

To finish the proof, I just add:

\begin{aligned}T\left(ap+bq\right)&=\frac{\,d}{\,d t}\left[ap+bq\right]\\ &= \frac{\,d}{\,dt}\left[ap\right]+\frac{\,d}{\,dt}\left[bq\right]\\&=T(ap)+T(bq)\\&=aT(p)+bT(q)\end{aligned}
The second to last line isn't correct. We can't assume its linear! However, note that $\frac{\,d}{\,dt}\left[ap\right]+\frac{\,d}{\,dt}\left[bq\right]=a\frac{\,dp}{\,dt}+b\frac{\,dq}{\,dt}=a T\!\left(p\right)+bT\!\left(q\right)$.

Taking $T\!\left(ap+bq\right)$ takes care of the constant muliple part as well. So you only need to check this linear combination, instead of the sum and constant multiple parts separately (in other words, its easier and quicker to do both steps in one step).

Does this make sense?

7. Originally Posted by Chris L T521
The second to last line isn't correct. We can't assume its linear! However, note that $\frac{\,d}{\,dt}\left[ap\right]+\frac{\,d}{\,dt}\left[bq\right]=a\frac{\,dp}{\,dt}+b\frac{\,dq}{\,dt}=a T\!\left(p\right)+bT\!\left(q\right)$.

Taking $T\!\left(ap+bq\right)$ takes care of the constant muliple part as well. So you only need to check this linear combination, instead of the sum and constant multiple parts separately (in other words, its easier and quicker to do both steps in one step).

Does this make sense?
ugh, I see my mistake. I'm not a fan of these. Thanks for your help. It's been great.