Gram-Schmidt Orthogonalizationh

[happyg1]

Homework Statement

Use the Gram-Schmidt method to find an orthogonal matrix Q and an upper triangular matrix R so that A=QR, where
$$A=\begin{bmatrix}16 & -4 & 8\\-4 & 5 & -4\\8 & -4 & 14\end{bmatrix}$$

Homework Equations

$$A^{(1)}=Q^{(1)}r_{11}$$
$$A^{(2)}=Q^{(1)}r_{12}+Q^{(2)}r_{22}$$
$$A^{(3)}=Q^{(1)}r_{13}+Q^{(2)}r_{23}+Q^{(3)}r_{33}$$
$$r_{11}=||A^{(1)}||$$
$$r_{12}=(Q^{(1)})^H A^{(2)}$$
$$r_{22}=}||A^{(2)}-r_{12}Q^{(1)}||$$

Those are the formulas we derived form the information in our book.
The $$A^{(1)}$$ represents the first column of the matrix A, and the same goes for the Q's. The $$r_{11}$$ is the 1,1 entry in the R matrix and so forth. The H is Hermetian.

The Attempt at a Solution

OK.
In order to find the r11 entry in my R matrix, I took the norm of the first column of A:
$$r_{11}=\sqrt{16^2+-4^2+8^2}=4\sqrt{21}$$
Then I divide the first column of A by $$4\sqrt{21}$$ to gert my first column of Q.

That gives $$Q^{(1)}= \left( \begin{matrix}\frac{4}{\sqrt{21}}\\ \frac{-1}{\sqrt{21}}\\ \frac{2}{\sqrt{21}}\end{matrix}\right)$$Then I need my r12 entry of my R matrix. So,

$$r_{12}=(Q^{(1)})^HA^{(2)}$$
which is:
$$\left( \begin{matrix}\frac{4}{\sqrt{21}}&\frac{-1}{\sqrt{21}}& \frac{2}{\sqrt{21}}\end{matrix}\right) \left( \begin{matrix}-4\\5\\-4\end{matrix}\right) =\frac{-29}{\sqrt{21}}$$

Then I need my second column of my Q matrix, which is supposed to be given by
$$r^{22}Q^{(2)}=A^{(2)}-r_{12}Q^{(1)}$$
according to the equation. I just solved it for r22Q2.
Anyway, this is where my trouble starts.
I got:
$$\left(\begin{matrix}-4\\5\\-4\end{matrix}\right) - (-\frac{29}{\sqrt{21}}) \left(\begin{matrix}\frac{4}{\sqrt{21}}\\ \frac{-1}{\sqrt{21}}\\ \frac{2}{\sqrt{21}}\end{matrix}\right) = \left( \begin{matrix}\frac{-200}{24}\\ \frac{134}{21}\\ \frac{-142}{21}\end{matrix}\right)$$

Which is totally wrong. I know I'm missing something, but I can't see it.
The norm of that thing is SUPPOSED to be 1, but it's not.

I hoipe my latex came out OK. Bear with me while I try to edit it.
CC

 

[StatusX]

No, as you said, that's r22Q2, not Q2. You pick r22 so that Q2 has norm 1. Also, by "Hermetian", I'm pretty sure you mean to spell "Hermitian", and you mean to use the word "transpose".

 

[happyg1]

Ok, so I need to solve for the norm of$$\left( \begin{matrix}\frac{-200}{24}\\ \frac{134}{21}\\ \frac{-142}{21}\end{matrix}\right)$$
so that's
$$\sqrt{(\frac{-200}{24})^2 (\frac{134}{21})^2 (\frac{-142}{21})^2}=\frac{2\sqrt{2170}}{7}$$

so
$$r_{22}=\frac{2\sqrt{2170}}{7}$$
so then I multiply r22Q2 by 1 over r22 to get Q2:

$$Q^{(2)}=\frac{7}{2 \sqrt{2170} \left( \begin{matrix}\frac{-200}{24}\\ \frac{134}{21}\\ \frac{-142}{21}\end{matrix}\right)=$$

$$\left( \begin{matrix}\frac{(10)*\sqrt{2170}}{651}\\ \frac{(67)*\sqrt{2170}}{6510}\\ \frac{(71)*\sqrt{2170}}{6510}\end{matrix} \right)$$

and the norm of that IS 1. but it doesn't match what the TI89 coughed out. I don't know if I'm right or not here.
TI89 got
$$Q^{(2)}=\left[ \begin{matrix}\frac{16}{\sqrt{1869}}\\ \frac{38}{\sqrt{1869}}\\ \frac{-13}{\sqrt{1869}}\end{matrix} \right]$$

They're not the same at all. I just don't know what's wrong here.
HELP
CC