Gram Schmidt Method (calculating orthogonal vector set)

[mess1n]

Hey, I'm going over the Gram Schmidt method, and need some help understanding it. I understand that you're intending to create an orthogonal vector (i'll call them v) set based on a set of vectors you already have (i'll call them u). Then:

Let v₁ = u₁

Now, construct the second orthogonal vector v₂ using a linear combination of u₂ and v₁ i.e:

v₂ = u₂ + c₁v₁

where the condition [ v₂.v₁ = 0 ] is satisfied.

However, what I wonder is why at this point I can't use the following equation instead:

v₂ = v₁ + c₁u₂

Surely I'm still creating a linear combination for v₂ which will still satisfy the condition? However whenever I use this second form, my answer comes out wrong! Why is this?

Andrew

 

[FieldDuck]

Excuse me if I'm wrong, but I don't think the only condition for forming v_2 is that v_2 is a linear combination of u_2 and v_1. I believe v_2 needs to be a vector formed by subtraction from u_2 its projection onto the subspace. If my reasoning is correct I think that doing this is motivated from the orthogonal decomposition theorem.

 

[Studiot]

Field Duck is essentially correct.

Just using c₁ disguises the true nature of your constant. How do you determine c₁?

 

[lurflurf]

^obviously
c1=-(v1.u2) /(v1.v1)

For the original question one can infact use
v2 = a1v1 + b1u2 often

 

[mess1n]

Thanks for the responses. So I can use the second form of the equation? How comes I get wrong answers when I use it! Is it just a coincidence and I'm making a mistake with the numbers? It makes sense that the vectors would be different... but when I check if they're all orthogonal afterwards it turns out that they aren't. This is the problem when I use the second form of the equation.

 

[Studiot]

Perhaps if you answered the question about c₁ someone may be able to help?

You have two constants c₁ which must be different.

How did you obtain each?

 

[mess1n]

Studiot:

I only have one constant c₁.

It is obtained as lurflurf explained.

Perhaps you're misunderstanding my question. I'm wondering why I can't use:

"v2 = v1 + c1u2"

INSTEAD OF:

"v2 = u2 + c1v1"

 

[Studiot]

How have you only got one constant?

Compare

\begin{array}{l}<br /> {v_2} = {v_1} + {c_1}{u_2} \\ <br /> {v_2}.{v_1} = \left( {{v_1} + {c_1}{u_2}} \right).{v_1} \\ <br /> 0 = {v_1}.{v_1} + {c_1}{u_2}.{v_1} \\ <br /> {c_1} = - \frac{{{v_1}.{v_1}}}{{{u_2}.{v_1}}} \\ <br /> \end{array}

With

\begin{array}{l}<br /> {v_2} = {u_2} + {c_1}{v_1} \\ <br /> {v_2}.{v_1} = \left( {{u_2} + {c_1}{v_1}} \right).{v_1} \\ <br /> 0 = {u_2}.{v_1} + {c_1}{v_1}.{v_1} \\ <br /> {c_1} = - \frac{{{u_2}.{v_1}}}{{{v_1}.{v_1}}} \\ <br /> \end{array}

 

[mess1n]

How have you only got one constant?

Compare

\begin{array}{l}<br /> {v_2} = {v_1} + {c_1}{u_2} \\ <br /> {v_2}.{v_1} = \left( {{v_1} + {c_1}{u_2}} \right).{v_1} \\ <br /> 0 = {v_1}.{v_1} + {c_1}{u_2}.{v_1} \\ <br /> {c_1} = - \frac{{{v_1}.{v_1}}}{{{u_2}.{v_1}}} \\ <br /> \end{array}

With

\begin{array}{l}<br /> {v_2} = {u_2} + {c_1}{v_1} \\ <br /> {v_2}.{v_1} = \left( {{u_2} + {c_1}{v_1}} \right).{v_1} \\ <br /> 0 = {u_2}.{v_1} + {c_1}{v_1}.{v_1} \\ <br /> {c_1} = - \frac{{{u_2}.{v_1}}}{{{v_1}.{v_1}}} \\ <br /> \end{array}

So why can't I use the first constant instead of the second? Surely I've still constructed an orthogonal vector using the first method?

Cheers,
Andrew