Linear Algebra Stuff. Find a Basis, I Need Your Help!

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January 29th 2010, 11:34 AM
mmmboh

Linear Algebra Stuff. Find a Basis, I Need Your Help!

Hi, I have an assignment due and I have done most of the questions there is just one part of a question I have left, if someone can help that would be amazing :)

http://i49.tinypic.com/2dadgsj.jpg

Edit: I answered the second question, now all I need help with is 1B).

I already did the first question part A), but I can't figure out B :(

Thanks :)
January 30th 2010, 09:55 AM
mmmboh

OK did the second one, can someone help me with 1B) please!
January 31st 2010, 07:00 AM
mmmboh

For the first question part B, I know all the vectors in B are independent because B is a basis for W, so I know the inverse of all these vectors are also independent. And I know that A maps the vectors in U onto W, so the inverse of A will map the vectors in W, or B, onto U...but how should I write this out?
January 31st 2010, 08:36 AM
HallsofIvy

Something obviously happened to your last post!

The one you are having difficulty with is, I assume, 3b since there is no "problem 1" given. It says:

If A is an n by n invertible matrix from U to W and $\{v_1, v_2, \cdot\cdot\cdot, v_n\}$ is a basis for W, then $\{A^{-1}v_1, A^{-1}v_2, \cdot\cdot\cdot, A^{-1}v_n\}$ is a basis for U. Since A is n by n, both U and W must be of dimension n so we need only show that $\{A^{-1}v_1, A^{-1}v_2, \cdot\cdot\cdot, A^{-1}v_n\}$ is independent. Suppose $\alpha_1 A^{-1}v_1+ \alpha_2 A^{-1}v_2+ \cdot\cdot\cdot+ \alpha_n A^{-1}v_j= 0$ for some scalars, $\alpha_1$ , $\alpha_2$ , $\cdot\cdot\cdot$ , $\alpha_n$ . Apply to A to both sides of that equation. What does that, together with the fact that [tex]\{v_1, v_2, \cdot\cdot\cdot, v_n\}[tex] is a basis, tell you about the scalars?
January 31st 2010, 08:42 AM
mmmboh

Sorry I didn't really understand your last sentence, but what I think you want to know is that the scalars are all zero, I think?
January 31st 2010, 09:03 AM
mmmboh

I'm a little bit confused on how this proves that (A^-1)(V1+...Vk) is a basis for U though (Speechless)
January 31st 2010, 11:46 AM
hatsoff

Quote:

Originally Posted by mmmboh

I'm a little bit confused on how this proves that (A^-1)(V1+...Vk) is a basis for U though (Speechless)

EDIT: First of all, as HallsofIvy pointed out, your notation is off, here. You need to prove that $A^{-1}\{\vec v_1,\vec v_2,\cdots,\vec v_k\}$ is a basis for $U$ , not $A^{-1}(\vec v_1+\vec v_2+\cdots+\vec v_k)$ . That said....

There are a couple of ways to do this. Going by HallsofIvy's method, we need to show that $A^{-1}\mathcal{B}$ is linearly independent. To do this, let

$\sum_{i=1}^k\alpha_iA^{-1}\vec v_i=\textbf{0}$ for some scalars $\alpha_1,\alpha_2,\cdots,\alpha_k\in F$ .

Multiplying both sides by $A$ , we have

$\sum_{i=1}^k\alpha_i\vec v_i=\textbf{0}$ .

But since $\mathcal{B}$ is linearly independent, then $\alpha_i=0$ for each $i$ , which is to say $A^{-1}\mathcal{B}$ is linearly independent, and the conclusion follows.
January 31st 2010, 12:55 PM
HallsofIvy

Quote:

Originally Posted by mmmboh

I'm a little bit confused on how this proves that (A^-1)(V1+...Vk) is a basis for U though (Speechless)

It doesn't. That's not true and not what you were asked to show!
The point is that, since all the scalars are 0, the vectors $A^{-1}v_1$ , $A^{-1}v_2$ , ..., $A^{-1}v_n$ (NOT their sum) are independent and so form a basis for U.