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I'm having issues with understanding one of the exercises I'm making.

I have to find an orthogonal basis for the column space of $A$, where:

$$A = \begin{bmatrix} 0 & 2 & 3 & -4 & 1\\ 0 & 0 & 2 & 3 & 4 \\ 2 & 2 & -5 & 2 & 4\\ 2 & 0 & -6 & 9 & 7 \end{bmatrix}.$$

The first question was to find a basis of the column space of $A$, clearly this is simply the first $3$ column vectors (by reducing it to row echelon form, and finding the leading $1$'s).

However, then I had to find an orthogonal basis out of the column space of $A$, and here is where I get lost. I started off with finding the first vector:

$$u_1 = \begin{bmatrix}0\\0\\2\\2\\\end{bmatrix}.$$

Then I thought I would find the second vector like this:

$$u_2 = \begin{bmatrix}2\\0\\2\\0\\\end{bmatrix}-\left(\begin{bmatrix}2\\0\\2\\0\\\end{bmatrix}\cdot\begin{bmatrix}0\\0\\2\\2\\\end{bmatrix}\right)*\begin{bmatrix}0\\0\\2\\2\\\end{bmatrix} = \begin{bmatrix}2\\0\\2\\0\\\end{bmatrix}-4*\begin{bmatrix}0\\0\\2\\2\\\end{bmatrix} = \begin{bmatrix}2\\0\\-6\\-8\\\end{bmatrix}.$$

However, according to the result sheet we were given, instead of having a $4$, I should have $\frac{4}{8}$. I somehow can not figure out what I am missing, since the dot product of the two vectors clearly is $4$.

Also, as a second question: if I had to find a orthonormal basis I would only have to take the orthogonal vectors found here, and multiply them by their $1$/length, correct?

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1 Answer 1

up vote 3 down vote accepted

Your basic idea is right. However you can easly verify that the vectors $u_1, u_2$ you found are not orthogonal, by calculating $$<u_1,u_2> = (0,0,2,2)\cdot \left( \begin{matrix} 2 \\ 0 \\ -6 \\ -8 \end{matrix} \right) = -12-16 = -28 \neq 0$$ So something is going wrong in you process.

I suppose you want to use the Gram-Schmidt Algorithm to find the orthogonal basis. I think you skiped the normalisation part of the algorithm because you only want an orthogonal not and orthonormal basis. However even if you dont want to have an ortho*normal* basis you have to take care about the normalisation of your projections. If you only do $u_i<u_i,u_j>$ it will go wrong. Instead you need to normalize and take $u_i\frac{<u_i,u_j>}{<u_i,u_i>}$. If you do the normalisation step of the Gram-Schmidt Algorithm, of curse $<u_i,u_i>=1$ so it's usually is left out. The wikipedia article should clear it up quite well.

Update

Ok, you say that $v_1 = \left( \begin{matrix} 0 \\ 0 \\ 2 \\ 2 \end{matrix} \right), v_2 = \left( \begin{matrix} 2 \\ 0 \\ 2 \\ 0 \end{matrix} \right), v_3 = \left( \begin{matrix} 3 \\ 2 \\ -5 \\ -6 \end{matrix} \right)$ is the basis you start from. As you did you can take the first vector $v_1$ as it is. So you first basis vector ist $u_1 = v_2$ Now you want to calculate a vector $u_2$ that is orthogonal to this $u_1$. Gram Schmidt tells you, that you recive such a vector by

$$u_2 = v_2 - \text{proj}_{u_1}(v_2)$$

And then a third vector $u_3$ orthogonal to both of them by $$u_3 = v_3 - \text{proj}_{u_1}(v_3) - \text{proj}_{u_2}(v_3)$$

You did do this approach. What went wrong, is your projection. You calculated it as $$ \text{proj}_{u_1}(v_2) = v_2<u_1,v_2>$$ but this is incorrect. The true projection is $$ \text{proj}_{u_1}(v_2) = v_2\frac{<u_1,v_2>}{<u_1,u_1>}$$ As I tried to point out some textbooks will skip the devision by $<u_1,u_1>$ in the explanation of Gram-Schmidt. But this is because in most cases you want to construct an orthonormal basis. In that case you normalize every $u_i$ before proceding to the next step. Therefore $<u_i,u_i> = 1$ can be skipped.

So what you need to change is to divide by $<u_2,u_2> = 8$ in your projection.

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Noted should be, I know that the the second vector I received is incorrect. The correct vector should be: (2, 0, 1, -1). However, I want to know WHAT I did wrong. Namely, the 4 should be replaced by a 4/8. I believe that I somehow do not understand the calculation of the projection. –  Ruddie Jun 28 '12 at 11:19
    
@Ruddie : I extended my answer a bit. Hope this helps –  Haatschii Jun 28 '12 at 11:57
    
ah, I see! So we take the dot product of u1 with v2, and divide this with the dot product of u1 with itself. And then we get the orthogonal basis. I indeed was never told that we divide by the dot product of u1 with itself. Thank you for the explaination! –  Ruddie Jun 28 '12 at 12:11

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