Question

Linear algebra problem @ganeshie8

Answer 1

a) Plugging in the values I get -20

for b) is this correct \[|<p(x),q(x)>| \le ||p(x)||~||q(x)||\] so I have \[|-20| \le \sqrt{(p(-1))^2+(p(0))^2+(p(1))^2} \sqrt{(q(-1))^2+(q(0))^2+(q(1))^2}\] it looks weird to me but if I keep going i get \[20 \le 7\sqrt{20}\]

Answer 2

Then for c) I would do \[||p(x)+q(x)||\le ||p(x)||+||q(x)||\]

Answer 3

And then d) I think we do same thing as a sort of part a) but it has to equal 0 but I'm not exactly sure what \(P^2\) here is, was that meant to be \(P_2\)

Answer 4

\(||p(x)|| = \sqrt{20}\)

\(||q(x)||=7\)

Then for part c \[||p(x)+q(x)|| = \sqrt{(p(x)+q(x)) \cdot (p(x)+q(x))}\] oh wait is this just the square root of the inner product

Answer 5

Squared then square rooted so 20

Answer 6

Sort of like the length haha but rather it's a sum

Answer 7

.

Answer 8

You did part a completely right I checked. What you wrote for part b was correct too, although I didn't go through the computation, the form of it is correct. Part c also looks good, you are doing the right thing, I am just a bit hesitant to say you're doing it right because you've written the dot product so as long as you mean:

\[(p(x)+q(x)) \cdot (p(x)+q(x)) = \langle (p(x)+q(x)),(p(x)+q(x))\rangle\]

then you're good.

for part d I think you're on the right track but unless you really have a strategy to make two polynomials orthogonal it won't exactly be easy, although you can probably guess one. I'll just give you a hint, you can take the projection of \(p(x)\) onto \(q(x)\) out of \(p(x)\) itself to make it orthogonal. In symbols, let's say:

The projection of \(p(x)\) on \(q(x)\) is a multiple of \(q(x)\) so let's call the projection \(k*q(x)\), then when we take this inner product we expect it to be zero, since there's no longer anything in common:

\[\langle p(x) - k*q(x) , q(x) \rangle = 0\]

You can use the linearity of the inner product to get:

\[\langle p(x), q(x) \rangle - k \langle q(x) , q(x) \rangle = 0\]

But you've already calculated both of these inner products, so all you have to do is solve for k and then you have a set of two orthogonal polynomials \(q(x)\) and \([p(x)-k q(x)]\).

Answer 9

Great thanks, I'll work on d in just a minute for part c would this be right because I'm sort of confusing the concept I think haha...so something like \[||p(x)+q(x)|| = \sqrt{[p(-1)+q(-1)]^2+[p(0)+q(0)]^2+[p(1)+q(1)]^2}\] if I got the components right

Answer 10

Yeah that is right

Answer 11

Sweet xD

Answer 12

It might help next time you come to a problem like this to make some kind of subsitution, like \(f(x) = p(x)+q(x)\) and then calculate \(\langle f(x),f(x) \rangle\) that way and then you don't have to worry as much about mixnig stuff up. I make substiutions for large stuff I don't want to write all the time cause I'm lazy lol.

Answer 13

Yeah you're right, but first I usually like to do it the long way then I realize later I can make a sub xD, I will keep that in mind.

Also for part d) I'm going to read up on some of this a bit more than do it, because I don't know the linear algebra theory well so yeah I need to read it lol

Answer 14

In pictures, what I am talking about in part D is this:

Answer 15

Yeah I think I see it now, I'm just comparing what you said to the drawing haha

Answer 16

This is vector addition, and these are the components of the vector p(x), the proof is just add head to tail (seems kind of silly to write out, but I think it makes it clear that these are really vectors): \[kq(x) + [ p(x) - k q(x)] = p(x)\]