More linear algebra - Perpindicular Line spanned by a vector

[succubus]

I'm back, I'm just brushing up on some linear algebra for my test tomorrow and am doing some recommended questions. I have a bit of confusion on some of them, and some of them I just want to make sure my thinking is correct.

So on to the problem...

Consider the line spanned by

1
2
3

in $$\Re^{3}$$ Find a basis of $$\bot$$L


So the basis would be the set of all independent column? vectors... Since it's $$\Re^{3}$$ I would assume that it is at least n x 3? So would I find a vector that when dotted against this would equal 0? :/

 

[Mark44]

Do you know what the phrase "line spanned by [1 2 3]^T in R^3" means?

So the basis would be the set of all independent column? vectors

No, I don't think so. I'm not sure what you mean by independent column vectors. I think you might be getting ahead of yourself by thinking in terms of a matrix before you fully comprehend the geometry here.

... Since it's $\Re^{3}$
I would assume that it is at least n x 3?

What is? Again, I think you're getting ahead of yourself by thinking about matrices. At least that's what it seems like.

So would I find a vector that when dotted against this would equal 0?

That's an excellent idea! The perpendicular subspace consists of vectors that are perpendicular to the line, so the dot product of anyone of them with the vector that generates the line should be zero.

Write an equation that expresses this idea and you'll be on your way to solving this problem.

 

[succubus]

Well, I've come up with $$\overline{x_{1}}$$ + 2$$\overline{v_{2}}$$ + 3$$\overline{v_{3}}$$ = 0

And since the first column vector has a 1 coefficient (or can be made that way in any other case), then I solve for

$$\overline{v_{1}}$$ = -2$$\overline{v_{2}}$$ - 3$$\overline{v_{3}}$$

Then where do I go?

"Do you know what the phrase "line spanned by [1 2 3]^T in R^3" means?"
Could you explain a little in more detail. I somewhat understand the concept of a span and basis. We just started getting into these concepts in class. I'm shaky in my confidence. The concept is testing me spatially :(

 

[Mark44]

Well, I've come up with $$\overline{x_{1}}$$ + 2$$\overline{v_{2}}$$ + 3$$\overline{v_{3}}$$ = 0

Nope, that's not it at all. You shouldn't have an equation with vectors in it after you have taken a dot product. Also, you're mixing up x's and v's, which makes things only slightly worse.

And since the first column vector has a 1 coefficient (or can be made that way in any other case), then I solve for

$$\overline{v_{1}}$$ = -2$$\overline{v_{2}}$$ - 3$$\overline{v_{3}}$$

Then where do I go?

"Do you know what the phrase "line spanned by [1 2 3]^T in R^3" means?"
Could you explain a little in more detail. I somewhat understand the concept of a span and basis. We just started getting into these concepts in class. I'm shaky in my confidence. The concept is testing me spatially :(

When the text talks about a space spanned by a collection of vectors, it means the set of all linear combinations of those vectors. For example, the span of {v1, v2, v3} is the set of all possible sums c1 * v1 + c2 * v2 + c3 * v3, where the c's are scalars and the v's are the vectors. If we're talking about the space spanned by one vector, the set of all linear combinations of one thing is c1 * v1. In other words, scalar multiples of the vector v1. This is a line through the origin. The line spanned by [1 2 3] is the line through the origin that goes through (1, 2, 3), and (2, 4, 6), and (-3, -6, -9), and (sqrt(7), 2sqrt(7), 3sqrt(7), and ...

Back to the problem at hand. You have a vector, [1 2 3]. You want to find the subspace of R^3 consisting of vectors that are perpendicular (hint: dot product) to [1 2 3]. What does your arbitrary, garden-variety vector in R^3 look like?