Conceptual questions on linear algebra

[Niles]

Homework Statement

Hi all. I have two questions (and two attempts) which I hope you can answer.

1) When I have a linear transformation L : V -> W, and I am asked to find the range (image) of this transformation, what is it exactly I am required to do? (I am not given the matrix A that corresponds to this transformation). Is the range simply the set of functions we get from our transformation L(x)? Sadly, my book doesn't give a proper answer.

2) I have four matrices that span out a vector space W. The four matrices are 2x2 matrices, and they are:

A_1 = (1 0 , 0 0) - (that is 1 0 in top, 0 0 in bottom).
A_2 = (0 1 , 0 0)
A_3 = (0 0 , 1 0)
A_4 = (0 0 , 0 1).

We have another matrix A = (a b , c d) and a linear transformation F : W -> W given by:

F(X) = AX-XA, X is a matrix in W.

I have to find the matrix for F with respect to the basis W spanned by A_1 .. A_4. What I did was to find F(A_1) up to F(A_4) and then express this result as a linear combination of A_1 to A_4, e.g.:

F(A_1) = 0*A_1 - b*A_2 + c*A_3 - 0*A_4. Then (0,-b,c,0)^T is the first column in my matrix. This gives me a 4x4 matrix - is this approach correct?

Thanks in advance,
sincerely Niles.

 

[Dick]

Part 2) is just fine. The first question is a little unclear. If you are not given a matrix then you need to be told 'something' about the linear transformation. Then you need to use that 'something' to deduce which elements of W can be expressed in the form L(v) for v in V. It's hard to be more specific.

 

[Niles]

Part 2: Great :-)

Part 1: I can take an example from my book. A linear operator on P_3 (the vector space spanned by polynomials with degree less than three) is given by:

L(p(x)) = p(x) - p'(x).

To determine the image/range of L, I would do this:

L(p(x)) = ax^2+bx+c - 2ax+b = ax^2 + (2ax+b)x + (b+c). Can I say that this is the range?

 

[HallsofIvy]

Homework Statement

Hi all. I have two questions (and two attempts) which I hope you can answer.

1) When I have a linear transformation L : V -> W, and I am asked to find the range (image) of this transformation, what is it exactly I am required to do? (I am not given the matrix A that corresponds to this transformation). Is the range simply the set of functions we get from our transformation L(x)? Sadly, my book doesn't give a proper answer.

The range of any function is the set of all possible "values" of the function. In particular, if a L is a linear transformation (linear function) from V to W, then the Image of L is defined as the set of all w in W such that L(v)= w for some v in V. How you would determine that range depends on exactly how you are "given" V. One example, that does not require a "matrix", is the differentiation operator, D, from the vector space of polynomials of degree 2 or less to itself: any "v" is of the form $ax^2+ bx+ c$ and D($ax^2+ bx+ c$)= ax+ b. The range is the set of all polynomials of degree 1 or less.

2) I have four matrices that span out a vector space W. The four matrices are 2x2 matrices, and they are:

A_1 = (1 0 , 0 0) - (that is 1 0 in top, 0 0 in bottom).
A_2 = (0 1 , 0 0)
A_3 = (0 0 , 1 0)
A_4 = (0 0 , 0 1).

We have another matrix A = (a b , c d) and a linear transformation F : W -> W given by:

F(X) = AX-XA, X is a matrix in W.

I have to find the matrix for F with respect to the basis W spanned by A_1 .. A_4. What I did was to find F(A_1) up to F(A_4) and then express this result as a linear combination of A_1 to A_4, e.g.:

F(A_1) = 0*A_1 - b*A_2 + c*A_3 - 0*A_4. Then (0,-b,c,0)^T is the first column in my matrix. This gives me a 4x4 matrix - is this approach correct?

Thanks in advance,
sincerely Niles.

A standard way of writing a linear transformation as a matrix in a specific basis[/itex] is to apply the linear transformation to each of the basis vectors, writing the result in terms of the basis. The coefficients in each of those form a column of the matrix.
F(A_1)= AA1- A1A=
$$\left[\begin{array}{cc}a & 0 \\ c & 0\end{array}\right]-\left[\begin{array}{cc}a & b \\ 0 & 0\end{array}\right]= \left[\begin{array}{cc} 0 & -b \\ c & 0\end{array}\right]$$
That can be written as -bA_2 + cA_3. The first column is [0 -b c 0]. Yes, that's exactly what you did and exactly what you got!


Apparently you got this in while I was typing my response:

Part 1: I can take an example from my book. A linear operator on P_3 (the vector space spanned by polynomials with degree less than three) is given by:

L(p(x)) = p(x) - p'(x).

To determine the image/range of L, I would do this:

L(p(x)) = ax^2+bx+c - 2ax+b = ax^2 + (2ax+b)x + (b+c). Can I say that this is the range?

A little more calculation: $L(p(x))= (3a)x^2+ bx+ (b+c)$. Since a, b, c can be any real numbers, 3a, b, and b+ c can be any real numbers. The range is just P_3 itself.

Notice, by the way, that the kernel of L is all polynomials in P_3 such that $L(p)= 3ax^2+ bx+ b+c= 0$. That requires that 3a= 0, b= 0, b+c= 0 so a= b= c= 0. The kernel is just {0}. Notice that the sum of the dimensions of the Kernel and Image is 3+ 0= 3, the dimension of P_3. It is, in general, true that if L:V->W, then the dimension of the kernel of L plus the dimension of the image of L is equal to the dimension of V.

 

[Niles]

First I want to thank you for your very thorough reply, which was very helpful.

I have one last issue: I want to try and find the kernel of L(p(x)) = p(0)*x+p(1). What I did was:

ker L is L(p(x)) = p(0)*x+p(1) = c*x + (a+b+c) = 0. So all the polynomials that makes c*x + (a+b+c) = 0 is the kernel of L. Since c = 0, I get that a = -b - is this correct?

Again, thank you both for taking the time to help.

 

[Dick]

First I want to thank you for your very thorough reply, which was very helpful.

I have one last issue: I want to try and find the kernel of L(p(x)) = p(0)*x+p(1). What I did was:

ker L is L(p(x)) = p(0)*x+p(1) = c*x + (a+b+c) = 0. So all the polynomials that makes c*x + (a+b+c) = 0 is the kernel of L. Since c = 0, I get that a = -b - is this correct?

Again, thank you both for taking the time to help.

Exactly right.