# Homework Help: Linear Algebra problem

1. Jun 13, 2010

### degs2k4

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

3. The attempt at a solution
on P1), they only ask for the coefficient matrix, which I think is the following one:
1 -6 2
0 1 4
0 0 15/8

on P2), they ask for the orthogonal matrix P used for the transformation. I suppose I have to apply Gram-Schmidt of the coefficient matrix from P1) to get P, but I am not sure of it...
Could someone guide me to solve this problem ?

Thanks in advance

2. Jun 13, 2010

### lanedance

that matrix gives the quadratic form, but the normal way to do this is to consider the symmetric matrix that gives your equation, so if we call your matrix M, we can find the symmetric part by:
$$A = \frac{1}{2}(M + M^T) = \frac{1}{2}(\begin{pmatrix} 1 & -6 & 2 \\0 & 1 & 4 \\0 & 0 & \frac{15}{8} \end {pmatrix} + \begin{pmatrix} 1 & 0 & 0 \\-6 & 1 & 0 \\2 & 4 & \frac{15}{8} \end {pmatrix}) = \begin{pmatrix} 1 & -3 & 1 \\-3 & 1 & 2 \\1 & 2 & \frac{15}{8} \end {pmatrix}$$

you can check this gives you the same quadratic form, and the properties of the symmetric matrix will be useful later on

3. Jun 13, 2010

### lanedance

for the next parts, i would start thinking about eignevectors...

note that you can prove symmetric matricies with real entries have real eigenvalues & are diagonalisable

4. Jun 13, 2010

### degs2k4

Why? I mean, that is something you thought of by looking at the rest of the problem? I didn't know that was the normal way to do that...

Thanks for the advice. I'm going to try that and post back the results here again.

5. Jun 13, 2010

### lanedance

i had a look at my text book.... ;)

the explanation is as follows, first any matrix can be written in terms of its symmetric (A=AT) and anti symmetric (B=-BT) parts, say M = A + B, where they are given by:
$$A = \frac{1}{2}(M+ M^T), \ \ B = \frac{1}{2}(M- M^T)$$

now consider the quadratic form
$$Q = x^T A x$$

now consider the quadratic form
$$Q = x^T M x = x^T A +x ^T B x$$

Q is scalar, so clearly QT = Q, then
$$Q = x^T M x = x^T A +x ^T B x = Q^T = x^TA x - x ^T B x$$

which gives
$$x ^T B x = 0$$

so the anti symmetric part doesnt really do anything anyway..

6. Jun 13, 2010

### lanedance

then when you do it, you get all of the nice properties that come with symmetric matricies

7. Jun 27, 2010

### degs2k4

Hello again,

Sorry about the late reply. Have been quite busy with other work.

After learning the properties of symmetric matrices and eigenvalues/vectors, I think I could solve most of the problem. But have doubts in the last parts of the problem. My main doubt now is how could I express the translation of section (3) as a matrix to be used in (4) for the composite transformation...

I uploaded 4 scanned pages of what I have done so far:

http://i50.tinypic.com/zsvlmg.jpg"
http://i49.tinypic.com/1zyydkh.jpg"
http://i45.tinypic.com/2mynbqb.jpg"
http://i47.tinypic.com/bjipmc.jpg"

I would be very grateful if someone could check it...
Thanks in advance...

Last edited by a moderator: Apr 25, 2017
8. Jun 30, 2010

### degs2k4

Could some please check if this is ok, specially parts 3-5?

Thanks

Page 1 : sections (1) and (2) of the problem

Page 2 : section (2) (cont)

Page 3 : section (3)

Page 4 : section (4) and (5)

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