# How do you get a matrix from this basis?

1. Apr 29, 2015

### bartersnarter

1. The problem statement, all variables and given/known data
Here's my problem. I only need help with the bottom part, but if you could explain the problem more vividly that would help too.

2. Relevant equations
A = S-1BS (?)
There aren't really any relevant equations. This part of linear algebra is getting really abstract, at least I think so. If there are any equations, I'm not aware of them.
3. The attempt at a solution
The problem I'm assigned for homework is very similar to this. I'm almost sure I got the answer right, but I don't completely understand the question. Why does the basis give that specific matrix B? What do you do with the basis to get B?

2. Apr 30, 2015

### BvU

Hello BS, welcome to PF !

Your hitting an extremely important subject in physics, math, engineering and what have you here !

Unfortunately, the explanation given in your text is already quite clear, so I find it difficult to add to it meaningfully.

Do you have any problem with the interpretation of matrix B on the basis of $\mathfrak {B}$ ?

How do you write vectors that are written as $(\lambda, \mu)$ on the basis of $\mathfrak {B}$ if you have to write them on the usual basis on $\Re^2$ ?

And if you apply the transformation B to $(\lambda, \mu)$ what do you get? And how does it look in $\Re^2$ ?

The other way around: what do you do to find the $(\lambda, \mu)$ corresponding to a vector $(x, y)$ in $\Re^2$ ?

---
Sorry about the $\Re$, looks awful. Someone will know which fonts are available here, but I have trouble finding out

 ah, found \mathbb : $\mathbb R$
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Last edited: Apr 30, 2015
3. Apr 30, 2015

### BvU

Perhaps I can save a few words with a picture:

T is reflection about the line spanned by (2,3).
What is T of vector $\vec v = (5,2)$ ?

We found that T is diagonal on a basis $\mathfrak B$ consisting of $\vec e_1 = (2,3)$ and $\vec e_2 = (-3,2)$
with $T(\vec e_1) = (1) \vec e_1$ and $T(\vec e_2) = (-1) \vec e_1$
T is linear, which means that
the projection $\lambda \vec e_1$ of $\vec v$ on $\vec e_1$
transforms into $\lambda T(\vec e_1) = \lambda \vec e_1$

and the projection $\mu \vec e_2$ of $\vec v$ on $\vec e_2$
transforms into $\mu T(\vec e_2) = - \mu \vec e_2$

4. May 1, 2015

### Fredrik

Staff Emeritus
The formula for the $\mathfrak B$-matrix of $T$ (for arbitrary $\mathfrak B$) is certainly relevant.