Determine ON-base of the column space of matrix

• November 1st 2010, 07:56 AM
maria88
Determine ON-base of the column space of matrix
1 3 1 2 2
-2 1 3 3 -2
1 2 1 1 1
1 2 2 1 0
3 1 2 -2 -1

Gauss elimination gives me the pivot positions in colonn number 1,2 and 3.

I think i have to use a normalised Gram-Smith method but i dont understand how to do it. plz help.
/Maria
• November 1st 2010, 08:06 AM
Ackbeet
Your (correct) work so far enables you to say that the first three columns span the column space of the matrix. So take the original three columns and apply Gram-Schmidt to them. Do you understand the Gram-Schmidt procedure?
• November 2nd 2010, 01:53 AM
maria88
I tried looking in my textbook but i didn't really get it, I actually missed the class when we went over the Gram-Schmidt procedure.
If anyone can explain the steps I would really appreciate it.

/Maria
• November 2nd 2010, 02:27 AM
Ackbeet
To produce an ON basis $\{e_{1},e_{2},\dots,e_{n}\}$ by using the Gram-Schmidt procedure on a linearly independent set $\{x_{1},x_{2},\dots,x_{n}\},$ follow these steps:

1. Let $\displaystyle e_{1}=\frac{x_{n}}{\|x_{n}\|}.$

2. For the remaining vectors in the ON basis, follow a two-step cumulative procedure: first, take the next vector $x_{2},$ and subtract off the amount of that vector that is in the direction of all previously computed vectors in the ON basis. In this case, set $f_{2}=x_{2}-\langle x_{2},e_{1}\rangle e_{1}.$ Here $\langle x,y\rangle$ is the inner product, or the dot product.

3. Normalize to get the next ON vector: $e_{2}=\dfrac{f_{2}}{\|f_{2}\|}.$

4. Set $f_{3}=x_{3}-\langle x_{3},e_{1}\rangle e_{1}-\langle x_{3},e_{2}\rangle e_{2}.$ (You need to subtract off the amount in the directions of both the previous two ON vectors.)

5. Then $e_{3}=\dfrac{f_{3}}{\|f_{3}\|}.$

Continue. Make sense?
• November 3rd 2010, 03:09 AM
maria88
I'm not sure i understand but ill try.
The first thing I do is divide my first vector [1,2,1,1,3] with the length of the vector (sqrt(16)). This gives me [1/4,2/4,1/4,1/4,3/4], and this is the first vector in my On base right?

After this step is done (if its right) I am supposed to subtract the inner product from the next vector in my original matrix.
This is what i don't understand how to do.

Also I think I am mixing up the terms, in my textbook it says I should use the following formula to calculate a Orthogonal basis. Can i use this and then normalize the answers that i get?

v1=x1
v2=x2-((x2v1)/(v1^2))*v1
v3=x3-((x3v1)/(v1^2))*v1-((x3v2)/(v2^2))*v1

/Maria
• November 3rd 2010, 05:19 AM
Ackbeet
The formulas you wrote down are the orthogonal Gram-Schmidt, not the orthonormal Gram-Schmidt. You can normalize all the vectors afterwards if you like, as you suggested. Either way should work. The amount of work will be about the same in either case. Let's say you continue down the path I laid out. (Incidentally, you have sign error in your first vector. It should be a negative 2/4 in the second position.) Your next calculation would be

$f_{2}=\begin{bmatrix}3\\1\\2\\2\\1\end{bmatrix}-(3/4+1(-1/2)+2/4+2/4+3/4)\begin{bmatrix}1/4\\-1/2\\1/4\\1/4\\3/4\end{bmatrix}
=\begin{bmatrix}3\\1\\2\\2\\1\end{bmatrix}-\begin{bmatrix}1/2\\-1\\1/2\\1/2\\3/2\end{bmatrix}=\begin{bmatrix}5/2\\2\\3/2\\3/2\\-1/2\end{bmatrix}.$

To get the normalized vector, divide by its length, which in this case is $\sqrt{15}.$ Therefore, we have

$e_{2}=\dfrac{1}{\sqrt{15}}\begin{bmatrix}5/2\\2\\3/2\\3/2\\-1/2\end{bmatrix}.$

See how this works?
• November 3rd 2010, 07:56 AM
maria88
I think i got it now,

$
e_{3}=\dfrac{1}{\sqrt{5/42}}\begin{bmatrix}-5/4\\19/10\\-9/20\\11/20\\33/20\end{bmatrix}.$

Does this look right to you?

I´m pretty sure I got $f_3$ right since the inner products of any two bases always amount to 0.
• November 3rd 2010, 08:14 AM
Ackbeet
Your vector's direction is correct, but I think you're getting too clever with your multiplication factor. You've obviously got the right idea, but you should multiply by the denominator, or reciprocate that factor one way or another.
• November 4th 2010, 03:41 AM
maria88
Yes I noticed that later. Thank you very much for your help, it´s much appreciated!
• November 4th 2010, 04:52 AM
Ackbeet
You're welcome! Have a good one.