1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conceptual questions on linear algebra

  1. Dec 27, 2007 #1
    1. The problem statement, all variables and given/known data
    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.
  2. jcsd
  3. Dec 27, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    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.
  4. Dec 27, 2007 #3
    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?
  5. Dec 27, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    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 [itex]ax^2+ bx+ c[/itex] and D([itex]ax^2+ bx+ c[/itex])= ax+ b. The range is the set of all polynomials of degree 1 or less.

    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=
    [tex]\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][/tex]
    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:
    A little more calculation: [itex]L(p(x))= (3a)x^2+ bx+ (b+c)[/itex]. 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 [itex]L(p)= 3ax^2+ bx+ b+c= 0[/itex]. 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.
    Last edited: Dec 27, 2007
  6. Dec 27, 2007 #5
    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.
  7. Dec 27, 2007 #6


    User Avatar
    Science Advisor
    Homework Helper

    Exactly right.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Conceptual questions on linear algebra