Creating a Unit Vector after Gram Schmidt Process?

[bugatti79]

Folks,

Where am I going wrong?

$$v_1=(3,0,0)$$, $$v_2=(0,1,2)$$,$$v_3=(0,2,5)$$

$$u_1=(3,0,0)$$

$$u_2=(0,1,2)-\frac{(0,1,2)(3,0,0)}{||(3,0,0)||^2}(3,0,0)$$

The inner product (0,1,2)(3,0,0) yields 0...?

 

[Evgeny.Makarov]

The inner product (0,1,2)(3,0,0) yields 0...?

So what?

 

[bugatti79]

So what?

Hi,

Inner product of those 2 equal to 0 would suggest they are perpendicular but according to wolfram,

gram schmidt {{3,0,0},{0,1,2},{0,2,5}} - Wolfram|Alpha Results

 

[Evgeny.Makarov]

Inner product of those 2 equal to 0 would suggest they are perpendicular

Yes.

but according to wolfram

Again, so what? (Smile) The second vector in W|A results is collinear with the original second vector; it is just multiplied by $1/\sqrt{5}$ so that its length becomes 1. The first vector is also normalized.

Addition: If, on the other hand, one of the vectors produced by the algorithm is 0, it means that the corresponding original vector belongs to the span of the previous vectors. Such vector can be simply omitted.

 

[bugatti79]

Yes.

Again, so what? (Smile) The second vector in W|A results is collinear with the original second vector; it is just multiplied by $1/\sqrt{5}$ so that its length becomes 1. The first vector is also normalized.

Addition: If, on the other hand, one of the vectors produced by the algorithm is 0, it means that the corresponding original vector belongs to the span of the previous vectors. Such vector can be simply omitted.

ok, I calculate

$$u_3=v_3-\frac{(v_3,u_1)}{||u_1||^2} u_1-\frac{(v_3,u_2)}{||u_2||^2}u_2$$

$$=(0,2,5)-0-\frac{12}{5}(0,1,2)=(0,-2/5,1/5)$$

I don't know where the sqrt is coming from? ie, $$||u_2||^2=(0^2+1^2+2^2)=5$$...?

 

[Evgeny.Makarov]

I don't know where the sqrt is coming from? ie, $$||u_2||^2=(0^2+1^2+2^2)=5$$...?

Could you formulate your questions more precisely? I suppose you are asking about the square root in the W|A results; you should say so. And how exactly does the fact that $$\|u_2\|^2=5$$ contradict that?

The norm of $u_3$ is
\[
\left\|\left(0,-\frac25,\frac15\right)\right\|=\sqrt{\left(-\frac25\right)^2+\left(\frac15\right)^2}=\sqrt{\frac{4}{25}+\frac{1}{25}}=\sqrt{\frac{5}{25}}=\sqrt{\frac15}.
\]
The unit vector is
\[
\frac{u_3}{\|u_3\|}=\frac{u_3}{\sqrt{1/5}}=\sqrt{5}u_3=\left(0,-\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}\right).
\]

 

[bugatti79]

Could you formulate your questions more precisely? I suppose you are asking about the square root in the W|A results; you should say so. And how exactly does the fact that $$\|u_2\|^2=5$$ contradict that?

The norm of $u_3$ is
\[
\left\|\left(0,-\frac25,\frac15\right)\right\|=\sqrt{\left(-\frac25\right)^2+\left(\frac15\right)^2}=\sqrt{\frac{4}{25}+\frac{1}{25}}=\sqrt{\frac{5}{25}}=\sqrt{\frac15}.
\]
The unit vector is
\[
\frac{u_3}{\|u_3\|}=\frac{u_3}{\sqrt{1/5}}=\sqrt{5}u_3=\left(0,-\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}\right).
\]

Ah yes, one still has to create the unit vector after the G-S process is performed.
Ok, I see it now. Thanks very much for your time.