Question

Help!

Answer 1

I need help with (d) if I can figure out d, i can figure out c too

Answer 2

I had to be afk, I'll look at it now.

Answer 3

ok thank you...I think I got first two, and also c and d, not sure if they are correct

Answer 4

Alright, this is pretty straight forward actually you're lucky! You want to express v in terms of the basis B, then that means there is a linear combination of the 3 vectors in B that can get you to v, correct? So in an equation, what would that look like? \[\LARGE x \bar b_1 + y \bar b_2 + z \bar b_3 = \bar v\]

here I just said x, y, and z are the scalars that we need to get there. Remember, there's only one possible combination of these three to get to v, so this has a unique solution right? Ok, so can you do some math on this and turn it into a matrix equation?

Answer 5

I got\[v=3(1,0,0)^T+3(0,1,0)^T+0(0,0,1)^T\]

Answer 6

for (a)? this is correct right?

Answer 7

No, you can't just solve it by inspection, that's cheating!

Answer 8

I mean how? I am expressing it

Answer 9

Wait, no this is actually incorrect

Answer 10

Sorry, this isn't the same equation I wrote earlier. Look at what I wrote and write that equation with the basis vectors given not the ones you're assuming.

Answer 11

Sure, what you're saying is a true statement, I don't know if that's the answer for (a) or not, I didn't read that, I am only looking at part (d).

Answer 12

oh please look at part B and check if this is correct. The above answer was for part a
\[v=3(1,1,0)^T+3(1,-1,0)^T+0(0,0,1)^T\]

Answer 13

What? I thought you asked about part (d)? Can we just do one thing at a time?

Answer 14

haha yes. Sorry. Sure here is what I got \[[v]_{B}=(3,3,0)^T\]

Answer 15

Why do you think that is the representation of v in the basis B?

Answer 16

because if V = a1b1+a2b2+....+anbn \[If v=a_1b_1+...+a_nb_n, then -->[v]_B = (a_1,...a_n)^T\]

Answer 17

@Kainui

Answer 18

Go to the post where I wrote:

"Alright, this is pretty straight forward actually you're lucky!" etc etc

Take the equation there!

Look at your problem and plug in the values of your basis B. So the scalar x is multiplied by: \[\LARGE \bar b_1 =\left(\begin{matrix}1 \\ 1 \\ 0\end{matrix}\right)\]

Now plug in the other two basis vectors and v.

Answer 19

oh so is it \[v=[(3,3,0)^T,(3,-3,0)^T,(0,0,0)^T]\]

Answer 20

whoops that \[[v]_B \] not \[v\]

Answer 21

No. Where did the 3's come from? They are an invention you have come up with that don't exist. The only coefficients on these basis vectors are x, y, and z! Adding up these vectors you have shown gives you the vector (6,0,0) which is clearly wrong!

Answer 22

\[v=(3,3,0)^T\]
multiply \[v*b \] correct?

Answer 23

You are trying to do more than I am telling you. Simply do as I say. Replace in my given equation simply b1,b2,b3, and v with what you see simply written. Do not replace x,y, and z, as they are just coefficients. Do just this. No more, you are needlessly complicating this and when you see, you will be relieved I hope!

Remember, we're trying to find out what combination of these new basis vectors bring us to v. We must set up our matrix equation to solve for this, and that is what we will do next. For now all you are writing is that v is a linear combination of these basis vectors with three unknown weights on them.

Answer 24

\[v={x(1,1,0)^T,y(1,-1,0)^T,z(0,0,1)^T}\]

Answer 25

Awesome. =)

Answer 26

i haven't expressed them as \[[v]_B \] have i?

Answer 27

Don't forget, v is really just (3,3,0)^T but that's ok I think you could do that. Now here's where we solve for x, y, and z.

Answer 28

\[\LARGE \left(\begin{matrix}3 \\ 3 \\ 0\end{matrix}\right) = x\left(\begin{matrix}1 \\ 1 \\ 0\end{matrix}\right)+y\left(\begin{matrix}1 \\ -1 \\ 0\end{matrix}\right)+z\left(\begin{matrix}0 \\ 0 \\ 1\end{matrix}\right)\]

So this is literally what you have at the moment. This IS v expressed in the basis B. The problem is, how much of each vector do we need?

Answer 29

x=3, y=z=0?

Answer 30

An alternate way to express this is to simply take the x,y, and z out as a vector! Now perhaps this will make it clearer where to go. Any ideas of how we can solve for the unknown weights this way? \[\LARGE \LARGE \left(\begin{matrix}3 \\ 3 \\ 0\end{matrix}\right) = \left(\begin{matrix}1 &1&0 \\1&-1& 0 \\ 0&0&1\end{matrix}\right) \left(\begin{matrix}x \\ y \\ z\end{matrix}\right)\]

Answer 31

Sure, we can do it that way, but that way you've described really won't help you in general. It's just a lucky guess.

Answer 32

oh ya, I really don't wanna guess, let's go your way please

Answer 33

So is there any way to solve for the vector with our unknowns in it?

Answer 34

x+y=3
x-y=3
z=0?

Answer 35

Think in terms of matrices. Can we divide both sides by that 3x3 matrix?

Answer 36

like is there an inverse?

Answer 37

@Kainui also i think there might be

Answer 38

If there wasn't an inverse, it wouldn't be a basis, now would it?

Answer 39

nope

Answer 40

but we still haven't figured out what \[[v]_B\] is

Answer 41

What do you think we are doing?

Answer 42

solving for x, y, z

Answer 43

So what's x, y, and z? Why do we care about those letters?

Answer 44

x=3
y=z=0?
I don't know why we care about them. Sorry

Answer 45

What are they coefficients on?

Answer 46

b1 b2 and b3

Answer 47

Good, so if I give you 6 of the vector b1, then what does that mean? It's just the vector (6,6,0) represented in this basis right? I'd have to give you 6(1,0,0)+6(0,1,0) to get that same vector in the regular old basis set yeah?

Answer 48

yeah right

Answer 49

So now remember xb1+yb2+zb3, what does this mean? This could be any vector, after all, every vector is just a linear combination of basis vectors. However we wanted to find one vector in particular right? Are you on board or not? It's ok if you're not, just try to explain what you don't understand to me so I can help.

Answer 50

i get it what you trying to say...the only thing i am not getting is that the final answer will just be some numbers. but this has x,y,z components in it

Answer 51

Yeah, v represented in the basis B will just be three numbers. Of course, it's just (3,0,0) because we figured it out easily by just looking. But if you left multiply the equation by the inverse matrix, we get the vector (x,y,z)^T multiplied by the identity matrix and we have a simply matrix equation on the left to multiply, solving for our vector.

Answer 52

u know i WAS THINKING UMMM YOU CAN MAKE THEM A UNIT BASIS AND
THEN
\[\LARGE x \bar b_1 + y \bar b_2 + z \bar b_3 = \bar v\]
the inner product
(b1,v)=x, if b1 is a unit basis