Linear map/transformation - linear alg

It maps from P4 to P4. What is the dimension of P4? Let's call it N. Then any matrix corresponding to this map should be NxN. You have written an 4x1 matrix, so that can't be right.

By the way, you don't need matrices to do this problem, and it might simply confuse things.

If we call the map T, and write a general input polynomial as

[tex]a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0[/tex]

then what does the output polynomial look like? (Just apply T to this input.) Then, for the kernel to have dimension exactly equal to one, what must be true?

First, let's check some basic things: can you explain why this map is linear? Can you explain why it maps P4 to P3? Hint: start by writing a general element of P4, similar to what I did above for P4. Then apply the map to it. Once you have done this, the next step is to identify exactly: what is its kernel? What is its image?

 

OK, let's start by understanding P4.

P4 is the set of all polynomials of degree <= 4. Therefore a general element (polynomial) of P4 looks like

[tex]p(x) = a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0[/tex]

Note that [tex]\{1, x, x^2, x^3, x^4\}[/tex] is a basis for P4, and it takes exactly five coefficients, [tex]a_0[/tex] through [tex]a_4[/tex] to specify a given element.

This means that P4 is a five-dimensional vector space.

Now, a given linear map from P4 to P4 (call it T) must because of its linearity obey the following:

[tex]T(p(x)) = a_4 T(x^4) + a_3 T(x^3) + a_2 T(x^2) + a_1 T(x) + a_0 T(1)[/tex]

Therefore T is completely defined by how it maps just five vectors, [tex]\{1, x, x^2, x^3, x^4\}[/tex]. (This amounts to choosing the columns of the matrix of T with respect to this basis.)

Now, what has to be true of the set

[tex]\{T(x^4), T(x^3), T(x^2), T(x^1), T(1)\}[/tex]

if the null space of T has dimension 0? What if it has dimension 1? What if it has dimension 2?

 

No, T maps from a five-dimensional vector space to a five-dimensional vector space. That determines the size of its matrix with respect to any choice of basis.

If you had a matrix that mapped R^5 to R^5, what would the size of that matrix be?

 

OK, now do you know how to write down an example of a 5 x 5 matrix that happens to have exactly 4 linearly independent columns?

If you had such a matrix M, what would the following dimensions be?

dim(null(M))
dim(range(M))

 

No, a specific concrete example is great - you can actually use this to answer the problem, but first you'll have to understand how:

Yes, those two terms mean the same thing.

No, I asked you to construct a 5x5 matrix that had exactly four linearly independent columns for a reason. The reason is that knowing this fact alone is enough to answer "what are dim(null(M)) and dim(range(M))".

Let's start with dim(range(M)). This is simply the dimension of the subspace spanned by the columns of M. If M has exactly four linearly independent columns, then what is dim(range(M))?

Then, the other important fact you need is this VERY FUNDAMENTAL rule:

dim(null(M)) + dim(range(M)) = D

where D is the dimension of the target space (the space that M maps to). If you don't know this rule, you need to scour back through your notes and find it! It's one of the key facts of linear algebra.

If the size of M is m x n, then what is D? (Hint: it's either m or n. Which one, and why?)

Once you have answered these questions, then you will know dim(range(M)) and D, which will allow you to solve for dim(null(M)). It will turn out in this case that dim(null(M)) is the the right number for one of the two problems you were asked to solve (i.e., either 1 or 2). (Which one?)

And you will be able to use the matrix you constructed to provide the example that the problem asked for.

 

dim(range(M)) = 4 because M spans a 4-dimensional SUBSPACE of R^5. (Not the same as spanning R^4.)

Correct.

D is the number of rows, i.e., if M maps from an m-dimensional space to an n-dimensional space, then dim(range(M)) + dim(null(M)) = m, it does not equal n. In your case both m and n are 5.

OK, so now you know dim(null(M)) = 1. This is good, because it the problem asked for an example where dim(null(M)).

So now how do you express your answer? You've already written down the matrix, so you can express the answer one of (at least) two ways:

(1) "Let T be the linear map from P4 to P4, whose matrix in terms of the basis [tex]\{x^4,x^3,x^2,x,1\}[/tex] is:" [insert your matrix here].

OR

(2) "[tex]\{x^4,x^3,x^2,x,1\}[/tex] is a basis for P4, and it suffices to specify T on this basis, so I choose to do so as follows:

[tex]T(x^4) = 1 x^4 + 2 x^3 + 1 x^2 + 1 x + 5[/tex] (I used the first column vector for the coefficients)

[tex]T(x^3) =[/tex] (linear combination using the second column vector)

etc.

Note that (1) and (2) are exactly equivalent. (You should convince yourself of this! This is exactly what a matrix representation MEANS.) Which way you choose to express it is really up to you.

In either case, explain as you did above why the null space of T has dimension 1.

 

Yes, exactly.

No, if it's mapping from a 5-dimensional space to a 4-dimensional space, that means the input vectors have 5 elements and the output vectors have 4 elements. That forces the matrix to be 5x4. Now your job is to find the elements of that matrix, assuming the standard basis for polynomials. The standard basis for P4 is [tex]x^4,x^3,x^2,x^1,1[/tex] and for P3 is [tex]x^3,x^2,x^1,1[/tex].

You need to use the given definition of the linear map in order to find the matrix. Hint: you can do this by "feeding" each element of the basis for P4, one at a time, to the map, and looking at the result. The result will be a linear combination of the elements of the basis for P3. The coefficients of the linear combination are precisely the number that go into the matrix columns.

 

Let's give the map a name - let's call it [tex]T[/tex], defined by:

[tex]T(f(x)) = 2f(x) - f(x-1) - f(x+1)[/tex]

where [tex]f(x)[/tex] is a polynomial in [tex]P^4[/tex].

For example, choose [tex]f(x) = x^4[/tex].

I'll construct [tex]T(x^4)[/tex] one piece at a time:

[tex]\begin{align*}
2f(x) &= 2x^4 \\
f(x-1) &= (x-1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1 \\
f(x+1) &= (x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1
\end{align*}[/tex]

Then

[tex]\begin{align*}T(x^4) &= 2f(x) - f(x-1) - f(x+1)\\
&= 2x^4 - (x^4 - 4x^3 + 6x^2 - 4x + 1) - (x^4 + 4x^3 + 6x^2 + 4x + 1) \\
&= 8x^3 -12x^2 + 8x - 2
\end{align*}[/tex]

Now you can see the coefficients which should go into the column corresponding to [tex]T(x^4)[/tex]. Repeat the process for the other basis elements and that gives you a matrix. You can use the matrix to answer the questions about the dimensions of the kernel and the image.

 

I think you made an algebra error when computing [tex]T(x^2)[/tex], but I don't think it affects the rest of the answer.

The dimension of the image is the number of linearly independent columns. The dimension of the kernel can be obtained from

dim(ker(A)) + dim(image(A)) = dim V

where V is the input vector space.

 

dim(image(A)) = 4? The matrix has four columns, one of which is all zeros. Can it really have four linearly independent columns? Or if you prefer to look at rows, are rows 3 and 4 linearly independent? What about any set of rows that includes row 5?