Question

Suppose that T:R^3 --> R^3 is a linear transformation and we know that T(1,1,0)=(2,1,-1), T(0,1,2)=(1,0,2), T(2,1,3)=(4,0,1). Find T(3,2,-1)

Answer 1

How do I even start this question? This was in the section basis and all of a sudden it's about transformations..

Answer 2

Remember that linear transformations have the property
T(av) = aT(v)

Answer 3

What you want to do is make (3,2-1) a linear combination of the basis vectors.

Answer 4

T(av) = aT(v)
T(u + v) = T(u) + T(v)

Answer 5

Okay I don't really get like say T(1,1,0)=(2,1,-1), what does the numbers 1,1,0 represent and what does (2,1,-1) mean

Answer 6

is T(1,1,0) just like a vector V=(a,b,c)?

Answer 7

(1,1,0) is a vector in R^3

Answer 8

then what does (2,1,-1) mean?

Answer 9

T(1,1,0)=(2,1,-1) indicates that the linear transformation maps (1,1,0) to (2,1,-1)

Answer 10

Okay so would I do addition and subtraction with T(1,1,0) T(0,1,2) and T(2,1,3) until I get T(3,2,-1)?

Answer 11

Start by finding the linear combination of (1,1,0), (0,1,2) and (2,1,3) that yields (3,2,-1)

Answer 12

okay so to find linear combination I would rref and find the pivots and then get the equations?

Answer 13

You need to solve a system of equations. To do this you need to row reduce a matrix consisting of columns of those vectors.

Answer 14

okay so I put this in a matrix sorry just want to make sure it's correct, [1 0 2 3; 1 1 1 2; 0 2 3 -1]?

Answer 15

Row reduce this:
\[\left[ \begin {array}{cccc} 1&0&2&3\\ 1&1&1&2
\\ 0&2&3&-1\end {array} \right] \]

Answer 16

\[\left[ \begin {array}{cccc} 1&0&0&{\frac {13}{5}}
\\ 0&1&0&-\frac{4}{5}\\ 0&0&1&\frac{1}{5}
\end {array} \right] \]

Answer 17

when you put it in matrix form in generally with v=(a,b,c) v_2=(d,e,f) etc. are the vectors always put in columns first? sometimes I do it rows..

Answer 18

yeah I got that too

Answer 19

This should be correct, check yourself if you like
\[\frac{13}{5}\cdot (1, 1, 0) + -\frac{4}{5}\cdot(0, 1, 2) + \frac{1}{5}\cdot (2, 1, 3) = (3, 2, -1)\]

Answer 20

but why is the answer (26/5,13/5,-4)?

Answer 21

That's not the end of the problem. But I will explain in a moment

Answer 22

So we now have a linear combination that yields the vector we want to take the image of.
The vector we want to transform.

Answer 23

\[T(3, 2, -1) = T\left(\frac{13}{5}\cdot (1, 1, 0) + -\frac{4}{5}\cdot(0, 1, 2) + \frac{1}{5}\cdot (2, 1, 3)\right)\]

Answer 24

Are you following this so far?

Answer 25

Yes I am except for the part where you got 13/5(1,1,0) + -4/5(0,1,2)..etc.

Answer 26

I see how it's from the matrix but I thought the last column would be like x_4 if you know what I mean

Answer 27

I will explain that at the end, lets continue

Answer 28

Okay

Answer 29

By linearity we know the following holds:
\[\begin{eqnarray*}
T(3, 2, -1) &=& T\left(\frac{13}{5}\cdot (1, 1, 0) + -\frac{4}{5}\cdot(0, 1, 2) + \frac{1}{5}\cdot (2, 1, 3)\right) \\
&=& T\left(\frac{13}{5}\cdot (1, 1, 0)\right) + T\left(-\frac{4}{5}\cdot(0, 1, 2)\right) + T\left(\frac{1}{5}\cdot (2, 1, 3)\right) \\
&=& \frac{13}{5}\cdot T\left( 1, 1, 0\right) +\left(-\frac{4}{5}\right) \cdot T\left(0, 1, 2\right) + \frac{1}{5}\cdot T\left(2, 1, 3\right)
\end{eqnarray*}\]

Answer 30

Can you finish the problem now?

Answer 31

Yeah I think so you just take those scalars and multiply them to the corresponding vector? of the linear transformation

Answer 32

and then add them up. Just one last thing would this method work for T:R^3 and R^4?

Answer 33

they are in different dimensions nowT:R^3 --> R^4?

Answer 34

Its the same process. And now I will explain the origin of the "magic numbers".
You are solving a system of equations.

You've setup a matrix A consisting of a basis.
You wish to find out a linear combination of that basis yielding a certain vector so you end up with.

Ax = (3, 2, -1)
And you are solving for x
So what you were looking at before was an augmented matrix setup to solve the system of equations.

Answer 35

What! That make so much more sense so really it was Ax=(3,2,-1)?

Answer 36

Where A is
\[\left[ \begin {array}{cccc} 1&0&2\\ 1&1&1
\\ 0&2&3\end {array} \right]\]

Answer 37

haha whoa thanks for your help it makes more sense now