Question

Help Problem Below....

Let .........

Answer 1

\[A=\left[\begin{matrix}1 & 1 \\ 2 & -3\end{matrix}\right]\]
Find \[[A]_{v}\]

Answer 2

What is \[[A]_{v}\]

Answer 3

???? what does that mean???

Answer 4

@ivanmlerner do you know what this means?

Answer 5

No I don't, is that the hole problem?

Answer 6

yes... is just says give that matrix find \[[A]_{v}\]
and if you cant tell that subscript is a v.

Answer 7

and the closest thing in the book that resembles this it states...
The matrix associated with L in the basis V is calculated in terms of dot product... the matrix is \[[L]_{v}=(L(v _{j})(v _{i}))\]

Answer 8

but i am not sure what they mean by that either.

Answer 9

Sorry, can't figure it out either

Answer 10

thank you for looking.

Answer 11

@DanielxAK do you know what this problem is asking me?

Answer 12

I don't recall ever seeing that notation. Give me a minute to look.

Answer 13

okay. thank you.

Answer 14

Is there more than just that line in your book?

Answer 15

as far as the notation that looks like that, there are only two parts in the book that mention that I will attach a picture from the book

Answer 16

this page is the only part that has notation like that.

Answer 17

What's definition 9.3.4?

Answer 18

Let W={w1......wn} be a basis for the vector space V. The n x n matrix [L]w associated with the linear map L:V->V and the basis W is defined as follows: The jth column of [L]w is [L(wj)]w --the coordinates of L(wj) relative to the basis W.

Answer 19

I didnt think part (a) was part of the problem, but part (a) says find the coordinates of the vector v=(1,4) in the orthonormal basis V.
v1=1/sr5(1,2) and v2=1/sr5(2,-1)

now part a. i already solved. but now part b. is exactly what is listed.

(b) Let A = (matrix above). Find [A]v.

And that is all the problem states. But I didnt think (a) had anything to do with it. but maybe it does.

Answer 20

Okay. I think I get it now. First, the notation they use is terrible. (L(vj)vi) is very poor as the way it's written the sizes don't match up. It'd be better written as: <<vj,L>,vi>.

I'm not sure how in depth I can be about this as I haven't really worked with this problem, but your linear map is taking all possible vector combinations of your basis and building a matrix that way. It's an orthonormal basis, so it should map back to itself if you repeat the process. (This is due to the fact that the magnitude is one, and that the inner product of any two vectors in that basis will equal 0 as long as it isn't the same vector.)

So, to calculate it, just do what the book does very similarly. Define your A as L, v1 and v2 from part (a), and then calculate each entry in your new matrix. Notice that Av1*v1 is still awful notation. It actually means v1^T*A*v1.

Answer 21

\[\left(\begin{matrix}1/\sqrt{5} \\ 2/\sqrt{5}\end{matrix}\right) \left[\begin{matrix}1 & 1 \\ 2 & -3\end{matrix}\right](1/\sqrt{5},2/\sqrt{5})\]

Answer 22

is that what that is?

Answer 23

as the first entry of the matrix?

Answer 24

Yes

Answer 25

but how do you multiply that b/c doesnt the first vector have to be a row vector to multiply by the matrix A

Answer 26

Actually, you have the backwards. Your first vector is 2x1, not 1x2. (Vectors are normally written in columns.) Your book is writing them in rows, which is an odd habit and not very useful in linear algebra.

Answer 27

so the vector v=(1,4) has nothing to do with this problem... only the orthogonal vectors v1 and v2

Answer 28

Right

Answer 29

and what exactly is this matrix? how is it different than the matrix v1 and v2 make?

Answer 30

Not sure what you mean. v1^t*A*v1 will give you something very different than the first entry in V.

Answer 31

well if v1 and v2 are orthogonal to the v=(1,4), and they give a matrix A... the [A]v, is a matrix that is what? orthogonal to A, or orthogonal to v1 and v2? or how is [A]v related to the matrix A, and why is it that you use a random set of orthogonal vectors whose original vector v you dont even use.

Answer 32

Not sure I can fully explain it right now, sorry. Again, this is something I haven't had much time to look at fully.

But, your matrix here is a function (specifically, a linear function). So, in two dimensions, you're applying your basis to your function to shift or rotate your function (you can try it with the standard basis in R^2, which is (1,0) and (0,1), you'll end up getting the same matrix).

Anytime you think basis, think coordinates, because that's all they really are.

Answer 33

so I computed [A]v... I got \[\left[\begin{matrix}-1 & 3 \\ 2 & -1\end{matrix}\right]\]

and so then that is a reasonable solution, assuming I did calculations right.

Answer 34

it just looks like a transformation to the original A right?

Answer 35

Think that's right. I closed the program I used to calc it already.

Also, I'm fairly sure you can regard it as the inverse function of A. You can play around with it, but consider it like you're solving Ax = b. You'll have two lines (or equations), for each row of the matrix. Plot them like you would a line. Then, take the [A]v you found, and do the same thing, using the same b (you can pick anything for b, it's just your y intercept). Your lines should be inverses of each other, I think.

I'm tired, so I'm going to bed now. But, I think you have a pretty good idea of it.

Answer 36

thank you very much. I need to as well. The mornings come to quick.