hi people, just started following a linear algebra course and i've run into some trouble trying to solve some problems in the book.

 

I've been asked to calculate the projection of (1,2,3,4) unto [(1,0,0,0),(3,4,0,0)]

 

Obviously they are not orthagonal as their in product is 3 and not 0.

So you project the in sum of (3,4,0,0) and (1,0,0,0) = (3,0,0,0)

 

(3,0,0,0) - (3,4,0,0) = (0,4,0,0)

 

Now it is orthagonal, I project vector x on (1,0,0,0) = (1,0,0,0)

Projection of vector x onto (0,4,0,0) I think is (0,1,0,0)

 

This leads to V = (1,1,0,0)

 

I was just wondering if anyone can do the calculations as well as I'm not able to confirm this.

 

Thanx in advance

I think I know how to project a vector onto another vector or onto any subspace of the parent space, but I'm not sure I understand what your question is asking of you. For instance, what does [(1,0,0,0),(3,4,0,0)] represent - a pair of vectors, or something else ?

No, with [(1,0,0,0),(3,4,0,0)] i mean the the space it spans. I hope this helps

I've been asked to calculate the projection of (1,2,3,4) unto [(1,0,0,0),(3,4,0,0)]

 

Look at it like this. The space spanned by your set of vectors is really the xy plane. your space is an xyzw four space. Okay but lets pretend you space was actually and xyz three space, the projection of (1,2,3,0) onto the xy plane is (1,2,0,0) and has length SQRT(5). Can you visualize this? You can actually make a nice diagram of this. Unless something really spooky happens when we move to four dimensions the projection of (1,2,3,4) onto the xy plane is still (1,2,0,0). You can convince yourself of this more by arguing that the projection of (1,2,0,4) onto the xy plane is also (1,2,0,0).

 

The tricky part is putting it back into the basis you have. So you have to form a linear combination of (1,0,0,0) and (3,4,0,0) that is equal to (1,2,0,0).

 

x*(1,0,0,0)+y*(3,4,0,0)=(1,2,0,0)

 

x+3y = 1

4y=2

 

y=1/2

x=-1/2

 

(-1/2,1/2,0,0) in the basis you were given. You can draw this too and verify why the first element is now negative.

 

A bit rusty on linear algebra but I think that is correct.

 

EJ

 

Note: my method here was to assume that (1,2,3,4) was taken over the standard basis. That the point was to find the projection to the xy plane then transform it into the basis {(1,0,0,0),(3,4,0,0)}. The question really depends on what basis you have.