Finding Unknown Linear Transformations Using Given Cases

[Lancelot59]

I'm given the following linear transformations:
$$T \left[ \begin{array}{cccc} 2 \\ 2 \end{array} \right] = \left[ \begin{array}{cccc} 1 \\ 2 \end{array} \right]$$
$$T \left[ \begin{array}{cccc} -4 \\ 0 \end{array} \right] = \left[ \begin{array}{cccc} -5 \\ 1 \end{array} \right]$$

I need to find:
$$T \left[ \begin{array}{cccc} -2 \\ 3 \end{array} \right] = ?$$

I solved prior problems by just reverse engineering it. That method seemed to fail here.
I came up with the following

$$T \left[ \begin{array}{cccc} x_{1} \\ x_{2} \end{array} \right] = \left[ \begin{array}{cccc} x_{1}-1 \\ \frac{x_{2}}{3}+1 \end{array} \right]$$

And it worked for both of the given cases, but it didn't for the unknown.

I got $$T \left[ \begin{array}{cccc} -3 \\ 1 \end{array} \right]$$, and the answer is: $$T \left[ \begin{array}{cccc} -4 \\ 3 \end{array} \right]$$

So how do I go about solving this problem?

 

[Mark44]

I'm given the following linear transformations:
$$T \left[ \begin{array}{cccc} 2 \\ 2 \end{array} \right] = \left[ \begin{array}{cccc} 1 \\ 2 \end{array} \right]$$
$$T \left[ \begin{array}{cccc} -4 \\ 0 \end{array} \right] = \left[ \begin{array}{cccc} -5 \\ 1 \end{array} \right]$$

I need to find:
$$T \left[ \begin{array}{cccc} -2 \\ 3 \end{array} \right] = ?$$

I solved prior problems by just reverse engineering it. That method seemed to fail here.
I came up with the following

$$T \left[ \begin{array}{cccc} x_{1} \\ x_{2} \end{array} \right] = \left[ \begin{array}{cccc} x_{1}-1 \\ \frac{x_{2}}{3}+1 \end{array} \right]$$

And it worked for both of the given cases, but it didn't for the unknown.

I got $$T \left[ \begin{array}{cccc} -3 \\ 1 \end{array} \right]$$, and the answer is: $$T \left[ \begin{array}{cccc} -4 \\ 3 \end{array} \right]$$

So how do I go about solving this problem?

Use the linearity of the transformation to find out what T does to the standard basis, {<1, 0>^T, <0, 1>^T}.

$$T \left( \begin{array}{c} 2 \\ 2 \end{array} \right) = T\left( 2\left[\begin{array}{c} 1 \\ 0\end{array} \right] + 2\left[\begin{array}{c} 0 \\ 1\end{array}\right]\right) = \left[ \begin{array}{c} 1 \\ 2 \end{array} \right]$$

Break up the other bit of known information the same way to solve for T(<0, 1>^T), and then solve for T(<1, 0>^T). Once you know what T does to the standard basis vectors, you know what it does to any vectors.

 

[jbunniii]

I came up with the following

$$T \left[ \begin{array}{cccc} x_{1} \\ x_{2} \end{array} \right] = \left[ \begin{array}{cccc} x_{1}-1 \\ \frac{x_{2}}{3}+1 \end{array} \right]$$

A quick sanity check will tell you that your answer can't be right, because it does not represent a linear map. The elements on the right hand side have to be linear combinations of $x_1$ and $x_2$. The "-1" and "+1" standing on their own are not allowed.

To see why, consider what happens to the output vector when you multiply the input vector by 2.

The quickest way to solve this is to recognize that you need to find two numbers, $a$ and $b$, such that

$$2a - 4b = -2$$

and

$$2a + 0b = 3$$

 

[Lancelot59]

A quick sanity check will tell you that your answer can't be right, because it does not represent a linear map. The elements on the right hand side have to be linear combinations of $x_1$ and $x_2$. The "-1" and "+1" standing on their own are not allowed.

To see why, consider what happens to the output vector when you multiply the input vector by 2.

The quickest way to solve this is to recognize that you need to find two numbers, $a$ and $b$, such that

$$2a - 4b = -2$$

and

$$2a + 0b = 3$$

Okay, so I tried solving that system and got:
$$T \left[ \begin{array}{cccc} 1 0 1 \\ 0 1 1 \end{array} \right]$$

I see now! That other one is a linear combination of the first two! Yay!

 

[Mark44]

Are you saying that a = -5 and b = 1 are solutions to the system
2a - 4b = -2
2a + 0b = 3
?

Obviously those aren't solutions.

 

[Lancelot59]

Are you saying that a = -5 and b = 1 are solutions to the system
2a - 4b = -2
2a + 0b = 3
?

Obviously those aren't solutions.

No, I messed up when entering in the LaTEX. The solutions are a=1 and b=1.

 

[Mark44]

They satisfy the first equation, but not the second...
2(1)$\neq$ 3

 

[Lancelot59]

Hang on, shouldn't the system be:

2a-4b=-2
3a-0b=3

I'll go with that being a typo, in which case it does work.

 

[Mark44]

I don't know if that was a typo or not. You were solving a pair of equations that jbunniii had in post 3. He's using a different method than I suggested.

In any case, the goal is not to find a and b, but to find T(<-2, 3>). (All vectors are really the transposes of the shown vectors.)

You reported that the answer was T(<-2, 3>) = <-4, 3>.

That's not what I get, so maybe you have some information about the problem written down incorrectly.

From
$$T \left[ \begin{array}{c} -4 \\ 0 \end{array} \right] = \left[ \begin{array}{c} -5 \\ 1 \end{array} \right]$$
it follows pretty quickly that
$$T \left[ \begin{array}{c} 1 \\ 0 \end{array} \right] = \left[ \begin{array}{c} 5/4 \\ -1/4 \end{array} \right]$$

Use that fact in your other equation to solve for T(<0, 1>), like I suggested in post #2.

Once you know T(<1, 0>) and T(<0, 1>), it's pretty simple to find T(<-2, 3>).

 

[Lancelot59]

I don't know if that was a typo or not. You were solving a pair of equations that jbunniii had in post 3. He's using a different method than I suggested.

In any case, the goal is not to find a and b, but to find T(<-2, 3>). (All vectors are really the transposes of the shown vectors.)

You reported that the answer was T(<-2, 3>) = <-4, 3>.

That's not what I get, so maybe you have some information about the problem written down incorrectly.

From
$$T \left[ \begin{array}{c} -4 \\ 0 \end{array} \right] = \left[ \begin{array}{c} -5 \\ 1 \end{array} \right]$$
it follows pretty quickly that
$$T \left[ \begin{array}{c} 1 \\ 0 \end{array} \right] = \left[ \begin{array}{c} 5/4 \\ -1/4 \end{array} \right]$$

Use that fact in your other equation to solve for T(<0, 1>), like I suggested in post #2.

Once you know T(<1, 0>) and T(<0, 1>), it's pretty simple to find T(<-2, 3>).

I mean it was a typo in the original post.
It should be:

$$T \left[ \begin{array}{cccc} 2 \\ 3 \end{array} \right] = \left[ \begin{array}{cccc} 1 \\ 2 \end{array} \right]$$
$$T \left[ \begin{array}{cccc} -4 \\ 0 \end{array} \right] = \left[ \begin{array}{cccc} -5 \\ 1 \end{array} \right]$$

Find $$T \left[ \begin{array}{cccc} -2 \\ 3 \end{array} \right]$$

My logic for solving the problem was this:

$$a*T (\left[ \begin{array}{cccc} 1 \\ 2 \end{array} \right]) + b*T (\left[ \begin{array}{cccc} -4 \\ 0 \end{array} \right]) = T(\left[ \begin{array}{cccc} -2 \\ 3 \end{array} \right])$$

Solving the system gave me a=1 and b=1.

$$\left[ \begin{array}{cccc} 1 \\ 2 \end{array} \right] + \left[ \begin{array}{cccc} -5 \\ 1 \end{array}] = \left[ \begin{array}{cccc} -4 \\ 3 \end{array} \right]\right$$

So I posted the wrong problem.

In any case, does that logic make sense? I'll go over your posts a few times and see if I can make sense of it.

 

[Mark44]

I think your logic is flawed, plus you have another typo in what you wrote.
It should be
$$a*T (\left[ \begin{array}{c} 2 \\ 3 \end{array} \right]) + b*T (\left[ \begin{array}{c} -4 \\ 0 \end{array} \right]) = T(\left[ \begin{array}{c} -2 \\ 3 \end{array} \right])$$

The problem is that you don't know what T(<-2, 3>) is, so how can you find a and b? If you are assuming that the answer is as given in your book, and then reverse engineering, that's not a good strategy.

The basic idea about a linear transformation is that T(au + bv) = aT(u) + bT(v)

1. Figure out what T(<1, 0>) and T(<0, 1>) are. I've already shown you T(<1, 0>).
2. Rewrite T(<-2, 3>) so that the vector argument is a linear combination of <1, 0> and <0, 1>. IOW, find constants a and b so that T(<-2, 3>) = T(a<1, 0> + b<0, 1>) = aT(<1, 0>) + bT(<0, 1>).

 

[jbunniii]

Hang on, shouldn't the system be:

2a-4b=-2
3a-0b=3

I'll go with that being a typo, in which case it does work.

Well, it might be a typo, but it doesn't necessarily have to be. The problem has a solution as originally stated.

My argument is as follows. If you can calculate what linear combination $a, b$ makes the following true:

$$a \begin{bmatrix} 2 \\ 2 \end{bmatrix} + b \begin{bmatrix} -4 \\ 0 \end{bmatrix} = \begin{bmatrix} -2 \\ 3 \end{bmatrix}$$

then linearity of $T$ ensures that the same $a,b$ will make the following true:

$$T\begin{bmatrix} -2 \\ 3 \end{bmatrix} = a \begin{bmatrix} 1 \\ 2 \end{bmatrix} + b \begin{bmatrix} -5 \\ 1 \end{bmatrix}$$

Expressing the first equation elementwise, you need to solve

$$2a - 4b = -2$$

and

$$2a + 0b = 3$$

If this led you to conclude that the original problem must have had a typo, I'm not sure why. This pair of equations has a perfectly legitimate solution. (Hint: it's REALLY easy to solve for $a$.)

P.S. This is a quick and dirty solution if you only need to find $Tx$ for one specific $x$. Mark44's method will allow you to do it for ANY $x$.

 

[Lancelot59]

Okay, well I'm having an issue with that.

$$T \left( \begin{array}{c} 2 \\ 3 \end{array} \right) = \left[ \begin{array}{c} 1 \\ 2 \end{array} \right]$$

What I currently have is:

$$T \left( \begin{array}{c} 2 \\ 3 \end{array} \right) = T\left( 1\left[\begin{array}{c} 1 \\ 0\end{array} \right] + 2\left[\begin{array}{c} 0 \\ 1\end{array}\right]\right) = \left[ \begin{array}{c} 1 \\ 2 \end{array} \right]$$
and

$$T \left( \begin{array}{c} -4 \\ 0 \end{array} \right) = T\left( -5\left[\begin{array}{c} 1 \\ 0\end{array} \right] + 1\left[\begin{array}{c} 0 \\ 1\end{array}\right]\right) = \left[ \begin{array}{c} -5 \\ 1 \end{array} \right]$$

Did I do this correctly?

 

[Mark44]

No, neither one.
In the first one,
$$T \left( \begin{array}{c} 2 \\ 3 \end{array} \right) \neq T\left( 1\left[\begin{array}{c} 1 \\ 0\end{array} \right] + 2\left[\begin{array}{c} 0 \\ 1\end{array}\right]\right)$$

because
$$\left( \begin{array}{c} 2 \\ 3 \end{array} \right) \neq \left( 1\left[\begin{array}{c} 1 \\ 0\end{array} \right] + 2\left[\begin{array}{c} 0 \\ 1\end{array}\right]\right)$$

In short, you're saying that <2,3> = <1, 0> + <0, 2>, which isn't true.

It should be obvious that
$$T \left( \begin{array}{c} 2 \\ 3 \end{array} \right) =T\left( 2\left[\begin{array}{c} 1 \\ 0\end{array} \right] + 3\left[\begin{array}{c} 0 \\ 1\end{array}\right]\right)$$

You have exactly the same problem with the other equation you wrote.

 

[Lancelot59]

No, neither one.
In the first one,
$$T \left( \begin{array}{c} 2 \\ 3 \end{array} \right) \neq T\left( 1\left[\begin{array}{c} 1 \\ 0\end{array} \right] + 2\left[\begin{array}{c} 0 \\ 1\end{array}\right]\right)$$

because
$$\left( \begin{array}{c} 2 \\ 3 \end{array} \right) \neq \left( 1\left[\begin{array}{c} 1 \\ 0\end{array} \right] + 2\left[\begin{array}{c} 0 \\ 1\end{array}\right]\right)$$

In short, you're saying that <2,3> = <1, 0> + <0, 2>, which isn't true.

It should be obvious that
$$T \left( \begin{array}{c} 2 \\ 3 \end{array} \right) =T\left( 2\left[\begin{array}{c} 1 \\ 0\end{array} \right] + 3\left[\begin{array}{c} 0 \\ 1\end{array}\right]\right)$$

You have exactly the same problem with the other equation you wrote.

Yes, then how does that equal (1,2)?

 

[Mark44]

You are given that
$$T \left( \left[\begin{array}{c} 2 \\ 3 \end{array}\right] \right) =\left[\begin{array}{c} 1 \\ 2\end{array} \right]$$

but the work you showed in the middle was incorrect.

You seem to be going around in circles. Take a good look at the last part in post #11. I laid out one approach that you can use. Pay more attention to where you're trying to go, and a little less on making it look good in LaTeX.