Orthogonal projection, orthonormal basis, coordinate vector of the polynomial?

[belleamie]

Hey there I'm working on questions for a sample review for finals I'm stuck on these three I think I'm starting to confuse all the different theorem, I'm so lost please help

1) Find the coordinate vector of the polynomial
p(x)=1+x+x^2

relative to the following basis of P2:
p1=1+x, p2=1-x, p3=1+2x+3x^2

?
I wasnt sure how to work this problem out:
Does it start out as?
b1=1,t,t^2
b2=t,1,t^2
b3= 1+t, 1-t, t-t^2

2) Let X be the linear span of the vectors
(1,1,1,1) (1,1,1,0) (1,1,0,0)
in R^4. Find the orthonormal basis for X?

It is:
[[u1]]^2
[[u2]]^2
[[u3]]^2

u1=1/2(1,1,1,1)
u2=1/6(1,1,1,0)
u3=1/4(1,1,0,0)

3) Let X be the linear span of the vectors
(1,2,1,2) (1,2,1,0) (1,1,0,0)
in R^4. Find the orthogonal projection of the vector (1,1,1,1) on th esubspace X?
It is solved like this:
c1=(v,u1)/(u1/u2)=(1+2+1+2)/(1+4+1+4)
c2=(v,u2)/(u2/u2)=(1+2+1+0)/(1+4+1+0)
c3=(v,u3)/(u3/u3)=(1+1+0+0)/(1+1+0+0)

there for x=proj(v,x) = c1u1+c2u2+c3u3

 

[shmoe]

1) Find the coordinate vector of the polynomial
p(x)=1+x+x^2

relative to the following basis of P2:
p1=1+x, p2=1-x, p3=1+2x+3x^2

You want to find the vector [a,b,c] where p(x)=a*p1(x)+b*p2(x)+c*p3(x). This is the coordinate vector of p(x) with respect to your basis. I don't understand what followed, with the b's.

2) Let X be the linear span of the vectors
(1,1,1,1) (1,1,1,0) (1,1,0,0)
in R^4. Find the orthonormal basis for X?

It is:
[[u1]]^2
[[u2]]^2
[[u3]]^2

u1=1/2(1,1,1,1)
u2=1/6(1,1,1,0)
u3=1/4(1,1,0,0)

An orthonormal basis is made up of orthogonal unit vectors.
Do you know the Gram-Schmidt orthogonalization algorithm? Use it to find an orthogonal basis, then make them unit vectors by dividing by their norms. Alternatively, you might be able to find an orthogonal basis by staring long enough, but Gram-Schmidt will work.

3) Let X be the linear span of the vectors
(1,2,1,2) (1,2,1,0) (1,1,0,0)
in R^4. Find the orthogonal projection of the vector (1,1,1,1) on th esubspace X?
It is solved like this:
c1=(v,u1)/(u1/u2)=(1+2+1+2)/(1+4+1+4)
c2=(v,u2)/(u2/u2)=(1+2+1+0)/(1+4+1+0)
c3=(v,u3)/(u3/u3)=(1+1+0+0)/(1+1+0+0)

there for x=proj(v,x) = c1u1+c2u2+c3u3

In order to use that formula for the projection, you must use an orthogonal basis. Use Gram-Schmidt to get one.

 

[M98Ranger]

= (1/2,1/2,1/2,1/2)

1) The coordinate vector of a polynomial relative to a basis is a representation of the polynomial in terms of the basis vectors. In this case, the basis is given as p1=1+x, p2=1-x, p3=1+2x+3x^2. To find the coordinate vector of p(x)=1+x+x^2, we can express the polynomial as a linear combination of the basis vectors:

p(x)=1(1+x)+1(1-x)+0(1+2x+3x^2)

Therefore, the coordinate vector of p(x) relative to the given basis is [1, 1, 0].

2) To find an orthonormal basis for X, we can use the Gram-Schmidt process. This process takes a set of vectors and produces an orthonormal set of vectors with the same span.

In this case, we have the vectors (1,1,1,1), (1,1,1,0), and (1,1,0,0). First, we normalize the first vector by dividing it by its length:

u1=(1,1,1,1)/√4=1/2(1,1,1,1)

Next, we subtract the projection of the second vector onto u1 from the second vector:

v2=(1,1,1,0)-proj((1,1,1,0),u1)=(1,1,1,0)-1/2(1,1,1,1)=(1/2,1/2,1/2,-1/2)

Then, we normalize v2 to obtain u2:

u2=(1/2,1/2,1/2,-1/2)/√3=1/6(1,1,1,-1)

Finally, we repeat the process for the third vector:

v3=(1,1,0,0)-proj((1,1,0,0),u1)-proj((1,1,0,0),u2)=(1,1,0,0)-1/2(1,1,1,1)-1/6(1,1,1,-1)=(1/3,1/3,-1/6,-1/6)

Normalizing v