# Linear transformation and change of basis

1. May 23, 2014

### negation

1. The problem statement, all variables and given/known data

Let B = {(1, -2),(2, -3)} and S be the standard basis of R2
and [-8,-4;9,4]
be a linear transformation expressed in terms of the standard basis?

3. The attempt at a solution

1) What is the change of basis matrix PSB ?
1,2
-2,-3

2)What is the change of basis matrix PBS ?
-3,-2
2,1

What is the linear transform expressed in terms of the basis B?
i spent 4 hours going through the proof for this part but unable to bridge a connection. Something is lacking in my understanding.
Give me a step by step guidance on this one.

2. May 23, 2014

### pondzo

3. May 23, 2014

### negation

Khan is good but his proof and subscript can at times be sloppy. I enjoy Mathispower4u (discovered it before Physicforum did) but it seems there are no videos on change on basis on Mathispower4u.

But anyway, I have a much rigorous proof on hand that I wish to understand fully.

The transformation machine that maps the vector v with respect to the standard basis, s, is given as [f(v)]s.
[f(v)]s = Ass [v]s
Ass = [ [f(e1)]s [f(e2)]s...[f(en)]s ] (I don't get this part)

There's more to come but I hope to understand this part first before proceeding...

Last edited: May 23, 2014
4. May 23, 2014

### HallsofIvy

Staff Emeritus
The basic point is this: if we have basis {u, v} then every vector can be written in the form au+ bv and represented as "$\begin{bmatrix}a \\ b\end{bmatrix}$". In particular, u= 1u+ 0v and is represented as $\begin{bmatrix}1 \\ 0\end{bmatrix}$ while v= 0u+ 1v and is represented as $\begin{bmatrix}0 \\ 1 \end{bmatrix}$.

Now suppose linear transformation, A, is represented by the matrix
$$\begin{bmatrix}a & b \\ c & d\end{bmatrix}$$
Then Au is
$$\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}a \\ c \end{bmatrix}$$
and Av is
$$\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix}= \begin{bmatrix}b \\ d\end{bmatrix}$$

That is, applying the linear transformation to the basis vectors, and then writing the result as a linear combination of the basis vectors, the coefficents give the columns.

In this particular case,
$$\begin{bmatrix}-8 & -4\\ 9 & 4\end{bmatrix}\begin{bmatrix}1 \\ -2\end{bmatrix}= \begin{bmatrix}0 \\ 1\end{bmatrix}$$
and we can write $\begin{bmatrix}0 \\ 1\end{bmatrix}= -2\begin{bmatrix}1 \\ -2\end{bmatrix}+ 1\begin{bmatrix}2 \\ -3\end{bmatrix}$

so the first column of the matrix representation is $\begin{bmatrix}-2 \\ 1\end{bmatrix}$

Similarly,
$$\begin{bmatrix}-8 & -4\\ 9 & 4\end{bmatrix}\begin{bmatrix}2 \\ -3\end{bmatrix}= \begin{bmatrix}-4 \\ 6\end{bmatrix}$$
and $\begin{bmatrix}-4 \\ 6\end{bmatrix}= 0\begin{bmatrix}1 \\-2\end{bmatrix}- 2\begin{bmatrix}2 \\ -3\end{bmatrix}$
so the second column is $\begin{bmatrix}0 \\ -2\end{bmatrix}$

The matrix representation in this (ordered) basis is
$$\begin{bmatrix}-2 & 0 \\1 & -2\end{bmatrix}$$

5. May 23, 2014

### negation

Would the outcome be any different if I first went down the route of

[-8;-9] = α1[1;0] + α2[0;1]
[-4;4] = β1[1;0] + β1[0;1]

that is, I express each column of the matrix A as a linear combination of the vectors in S.

6. May 23, 2014

### negation

Ok I got this. Basically, all I had to find was [T(x)s]B.