# Homework Help: Linear algebra - transformations

1. Jan 11, 2008

### Niles

[SOLVED] Linear algebra - transformations

1. The problem statement, all variables and given/known data
I actually have two questions:

1) I have a linear transformation L and it is represented by a matrix A. I also have a vector w, and I want to find out if w gets "hit" by L - see "answer-part" for my approach, and please comment.

2) Does the vectors that span R^n have to be orthogonal and linearly independant or only linearly independant? And is this the same for a vector-space in R^n? Please see comments in "answer-part" as well.

3. The attempt at a solution

1) Can I just solve the system Ax=w? If it is consistent, w gets "hit" by L?

2) The reason why I ask is that e.g. in R^3, the three unit vectors are orthogonal and linearly independant. And I have worked with vector-spaces in R^n where the vectors that span the space are not orthogonal. So I am a little confused here.

Last edited: Jan 11, 2008
2. Jan 11, 2008

### e(ho0n3

Concerning 1): I don't understand what you mean by 'w gets "hit" be L'.

Concerning 2): Which vectors are you talking about? If S spans R^n, the vectors in S need not be linearly independent. If they are though, then the vectors in S can be used as a basis for R^n.

3. Jan 11, 2008

### Dick

1) Yes, if the system of equations is 'consistent' that means you can find a solution. So w is in the range. It is hit.

2) You've asked this before and I've answered it before. A basis doesn't have to be orthogonal. Are you going to ask again?

4. Jan 12, 2008

### Niles

Thanks to both of you.

I know I asked it before, and it's not nice of me to ask again - but as I wrote, I got a little confused, but I've bookmarked this topic.

Thanks again.

5. Jan 12, 2008

### HallsofIvy

More correctly, w is in the image of the linear transformation. If L: U->V, the V is the "range" and the image is a subspace of V.

6. Jan 12, 2008

### HallsofIvy

Strictly speaking the vectors that span a vector space don't even have to be independent! A basis for a vector space must both span the space and be linearly independent. A basis still doesn't have to be orthogonal. That's just often used because orthonormal bases are particularly simple. In fact, a general vector space does not necessairily have an "inner product" defined and so the concepts of "orthogonal" and "normalized" may not even be defined.