Understanding Linear Transformations: How to Write the Matrix with Symmetry

[DWill]

Homework Statement

Hi all, we just started doing linear transformations in class and I still don't fully understand them. Here's one question I've been stuck on:

Let P₂(x,y) be the vector space of polynomials in the variables x and y of degree at most 2. Recall the monomial basis for this space is {1, x, y, x², xy, y²}

Let S: P₂(x,y) -> P₂(x,y) be the linear transformation which swaps the role of x and y in a polynomial. Write the matrix of S with respect to the monomial basis. Note that S is symmetric.

Homework Equations

The Attempt at a Solution

What are the basic steps I need to take to solve problems of this kind? The thing is I don't know how to start and what to do, so if somebody can teach me I will try to do this on my own. Thanks

 

[DWill]

Ok I think I figured out how to find the matrix of S on my own. It's just (0, 1, 1, 0) right? Now I just need to figure out how to write it with respect to the monomial basis?

 

[e(ho0n3]

First of all, (0, 1, 1, 0) is not a matrix. Find out what S(1), S(x), S(y), etc. is written in terms of the monomial basis. S will be the matrix whose columns are, respectively, the coefficients of S(1), S(x), S(y), etc.

 

[Mark44]

Your matrix will need to be 6 x 6. You can represent your monomial basis {1, x, y, x^2, xy, y^2} as column vectors, (1, 0, 0, 0, 0, 0), (0, 1, 0, 0, 0, 0), and so on. Now look at what S does to each of these vectors. The output vectors will be the columns of your transformation matrix.

 

[DWill]

e(hoon3, what is S(1), S(x), S(y)? I think that's my problem, I'm not sure what you mean by write them in terms of the monomial basis.

Mark, how do I write the basis as column vectors? Factor out the x and y's respectively? I'm not sure what you mean there.

Thanks for the help though!

 

[e(ho0n3]

S(1) is the image of 1 under S, i.e. S applied to 1. And similarly for the rest. "In terms of the monimial basis" means writing S(1) as a monomial.

 

[DWill]

Hm, so a 6x6 matrix that looks like this then?

1 0 0 0 0 0
0 0 1 0 0 0
0 1 0 0 0 0
0 0 0 0 0 1
0 0 0 0 1 0
0 0 0 1 0 0

 

[Office_Shredder]

For example, let's say I have a vector space with two basis vectors, a and b. I pick a linear transformation T where T(a) = a+b and T(b) = -b. Then I want to construct a basis for T by representing a as (1,0) and b as (0,1). Notice the space is 2 dimensional, so the matrix will be 2x2. If the matrix is

[w x]
[y z] = A

say, then
T(a) = a+b means A(1,0) = (1,1) (the coefficients of the basis vectors match up with the coordinates) But A(1,0) = (w,y) so w=1, y=1

T(b) = -b means A(0,1) = (0,-1). But A(0,1) = (x,z) so x=0, z=-1

And the matrix would be

[1 0]
[1 -1]

Notice how each column is the image of one of the basis vectors... this isn't a coincidence, if you look at how the matrix was calculated you can see clearly how this happened. So instead, when given T(a) = a+b and T(b) = -b, instead of doing all those equations above (well, 2 here, but for a 6x6 matrix you'd be screwed) you can just read off the columns of the matrix by looking at the image of each basis vector.

Now do the same for the map that you have

 

[Tedjn]

It is important to recognize that you can turn this question into a question about a linear transformation $S : \mathbb{R}^6 \rightarrow \mathbb{R}^6$ because you can identify each element in P₂(x,y) uniquely with a vector in R⁶. You do this by considering each element of the monomial basis {1, x, y, x², xy, y²} as if it were a different direction in R⁶. Therefore, what must the size of the matrix S?

How would you construct this matrix? Do what Mark44 and e(ho0n3 are suggesting. It is a theorem that the columns of the matrix of a linear transformation T : Rⁿ to R^m can be found by applying each of the standard basis vectors in Rⁿ to T.

EDIT: DWill, that's correct.

 

[DWill]

That's the correct matrix? whew...thanks

Now if I want to find the eigenvalues of this matrix I need to take the det(A-lambda*I) right? Since this is a 6x6 matrix I would need to do cofactor expansion? Is there some kind of trick of pattern I'm not seeing here, because doing that might take a while...

 

[Office_Shredder]

You can expand along the top row... which is convenient. Notice all but one of the entries are zero. So you really have (1-lambda)*(5x5 determinant). And only two entries in the 5x5 matrix's top row are non-zero, so expand into two 4x4 determinants. A lot of stuff will start out being 0 and it shouldn't take that long

 

[Tedjn]

In fact, you may be able to write down six linearly independent eigenvectors just by considering how S changes polynomials.

 

[DWill]

Ok similar question, need a bit of help:

Let C = (1 2 ; 3 4). Let L be the linear transformation L(A) = CA. Write the matrix representation of L with respect to the standard basis of M₂.

Here's what I got so far:

The standard basis looks like S = { (1 0 ; 0 0), (0 1 ; 0 0), (0 0 ; 1 0), (0 0 ; 0 1) }. If I call each element of the basis v_i, e.g. first element v₁ = (1 0 ; 0 0), etc., I found L(v_i) for each element: L(v₁) = (1 0 ; 3 0) and so on

So now I have 4 2x2 matrices. ehoon said somethingn earlier that the matrix L will be the coefficients of the 4 matrices I now have. So I take each column of these matrices and join them together to get L? Wouldn't that be a 2x8 matrix?

 

[Tedjn]

No. The theorem that let's you construct the matrix of a linear transformation from its values on the standard basis vectors works only for a map from Rⁿ to Rⁿ. What you have is a map from M₂ to M₂.

What you must do in this question and in most questions like this one is uniquely identify each element in M₂ with an element in Rⁿ for some appropriate n. For a 2 by 2 matrix, what would n have to be? How could you identify each 2 by 2 matrix with a vector in Rⁿ?

After you have done this, you will see that L is really a map from Rⁿ to Rⁿ. Thus, a matrix for L would be 4 by 4, its columns given by the values of the transformation at the standard basis vectors.

 

[Defennder]

You have to express each L(v_i) in terms of your ordered basis. So for example, (1,0 ; 3,0) would be written as a coordinate vector of the standard basis as: (1 0 3 0)^T. Doing that 4 times gives you a 4x4 matrix.

 

[DWill]

The matrix would be this then?

1 0 2 0
0 1 0 2
3 0 4 0
0 3 0 4

 

[Defennder]

That looks ok.