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Basis of the range of a Linear Transformation

  1. Mar 27, 2014 #1
    Mod note: fixed an exponent (% --> 5) on the transformation definition.
    1. The problem statement, all variables and given/known data
    A is a (4x5)-matrix over R, and L_A:R^5 --> R^4 is a linear transformation defined by L_a(x)=Ax. Find the basis for the range of L_A.

    2. Relevant equations


    3. The attempt at a solution
    ##A = \begin{bmatrix}1 & 2 & 3 & 4 & 5 \\2 & 3 & 4 & 5 & 6 \\3 & 4 & 5 & 6 & 7 \\4 & 5 & 6 & 7 & 8 \end{bmatrix}##

    ##A = \begin{bmatrix}1 & 2 & 3 & 4 & 5 & b_{1} \\2 & 3 & 4 & 5 & 6 & b_{2} \\3 & 4 & 5 & 6 & 7 & b_{3} \\4 & 5 & 6 & 7 & 8 & b_{4} \end{bmatrix} ##

    ##A = \begin{bmatrix}1 & 0 & -1 & -2 & -3 & -3b_{1}+2b_{2} \\0 & -1 & -2 & -3 & -4 & -2b_{1}+b_{2} \\0 & 0 & 0 & 0 & 0 & b_{1}-2b_{2}+b_{3} \\0 & 0 & 0 & 0 & 0 & 2b_{1}-3b_{2}+b_{4} \end{bmatrix}##

    Where do I go from there?
     
    Last edited by a moderator: Mar 27, 2014
  2. jcsd
  3. Mar 27, 2014 #2

    Mark44

    Staff: Mentor

    Let's put some words with the above so that you can understand what's going on. For some b ##\in## R4, the augmented matrix above represents this matrix equation:
    Ax = b
    A bit of row reduction yields the above. Note that I didn't check your work, so it's possible there were some arithmetic errors along the way. Also, you should multiply row 2 by -1 so that the matrix is in proper reduced form.
    The first two rows in the matrix tell us about x1 and the other coordinates of the input vector x. The first row tells us that x1 is a linear combination of x3, x4, and x5, plus a combination of some of the coordinates of b. The second row tells us that x2 is a different linear combination of x3, x4, and x5, plus some of the coordinates of b.

    The last two rows tell us about the restrictions on b itself, namely that b1 - 3b3 + 2b4 has to be zero, as does b2 - 2b3 + b4. Do you see why?

    From these restrictions you can solve for the coordinates of b, and thereby get a basis for the range of your transformation. I have glossed over things a bit here, but if what I've said isn't clear, ask for clarification.
     
  4. Mar 28, 2014 #3
    Yes, I see why b1 - 3b3 + 2b4 and b2 - 2b3 + b4 have to equal zero. Since they are limiting what sort of values b1,b2, b3, and b4 can take, I'm guessing they have something to do with range?
     
  5. Mar 28, 2014 #4

    Mark44

    Staff: Mentor

    In fact, they define the range.
     
  6. Mar 28, 2014 #5
    Ah Ok. So the answer to the problem is
    Rg(A) = {(b1,b2,b3): b1 - 3b3 + 2b4 = 0, b2 - 2b3 + b4=0}

    Is the notation for the solution correct?

    Would the basis in this case be two vectors (because there are two pivot columns?) that both fit the above rule for the range? For b1 - 3b3 + 2b4 = 0, b2 - 2b3 + b4=0, the free variables would be b3 and b4.

    If i choose b3 and b4 to equal 1 and 2 respectively, b1 and b2 are -1 and 0 respectively.
    If i choose b3 and b4 to equal 2 and 1 respectively, b1 and b2 are 4 and 3 respectively.

    the two vectors I get are (-1,0,1,2) and (4,3,2,1) assuming I did this correctly. Would these be the basis?
     
  7. Mar 28, 2014 #6

    Mark44

    Staff: Mentor

    Not quite. Each vector b in the range has four coordinates, not three as you show. And rather than showing the two equations, I would give the range as a set of two vectors in R4.
    What is usually done is to choose b3 = 1 and b4 = 0 to get one vector, and then choose b3 = 0 and b4 = 1 to get the other vector.
    If they satisfy the two equations you started with, they're fine. My choices would be <3, 2, 1, 0> and <-2, -1, 0, 1>, though. Your two vectors and my two vectors are linearly independent, since each vector in a pair is not a scalar multiple of the other one in the pair. If it turned out that there were three or more vectors in a basis, my technique would be better, since it would be obvious that the three (or more) vectors were linearly independent. This is because in certain coordinate positions, all but one vector would have 0's in that position, so it wouldn't be possible for any vector to be a linear combination of the others.
     
  8. Mar 28, 2014 #7
    So how about this?
    ##Rg(A)= \begin{bmatrix}1 \\0 \\-3 \\2 \end{bmatrix}, \begin{bmatrix}0 \\1 \\-2 \\1 \end{bmatrix} ##
     
  9. Mar 28, 2014 #8

    Mark44

    Staff: Mentor

    Nope. These are just the two rows of your reduced matrix. If you check, you'll see that they don't satisfy the two equations in the b coordinates.

    To get your basis vectors, note that the first row stands for b1 - 3b3 + 2b4 = 0, and the second row stands for b2 - 2b3 + b4 = 0,

    Solve the first equation for b1 and the second equation for b2. b3 and b4 are free variables. If you write the four equations, you'll see that <b1, b2, b3, b4> can be written as a linear combination of two vectors.
     
  10. Mar 28, 2014 #9
    So
    b1 = 3b3 - 2b4 = 0,
    b2 = 2b3 - b4=0
    b3=free
    b4=free

    So

    ##b_{3}\begin{bmatrix}3 \\2 \\1 \\0 \end{bmatrix}+b_{4}\begin{bmatrix}-2 \\-1 \\0 \\1 \end{bmatrix}##

    or

    ##\begin{bmatrix}3 \\2 \\1 \\0 \end{bmatrix},\begin{bmatrix}-2 \\-1 \\0 \\1 \end{bmatrix}##
     
    Last edited: Mar 28, 2014
  11. Mar 28, 2014 #10

    Mark44

    Staff: Mentor

    Bingo! That's exactly what I have.
     
  12. Mar 28, 2014 #11
    And that's the basis?

    Thanks for your help!
     
  13. Mar 28, 2014 #12

    Mark44

    Staff: Mentor

    Those two vectors are a basis for the range of L.
     
  14. Mar 28, 2014 #13
    Can you also check my answers for these questions?

    Find the basis for the row space and column space.

    The Basis of the Row Space is the first 2 rows (row vectors?) from the RREF form and the column space is the first two columns (column vectors?) of the original matrix.
     
  15. Mar 28, 2014 #14

    Mark44

    Staff: Mentor

    Sounds about right.
     
  16. Mar 28, 2014 #15
    How would I go about finding the basis of the kernel?
     
  17. Mar 28, 2014 #16

    Mark44

    Staff: Mentor

    Solve the matrix equation Ax = 0, where A is the matrix of your transformation. Operationally, you could have an augmented matrix, with the last column all zeroes, but you can omit it since it cannot change. Get the matrix in RREF form and read off the values for x1, etc., exactly as you did when you found the range. This time, though, the solution will represent the vectors in R5 that get mapped to the 0 vector in R4.

    The important part to remember is that you're finding the solutions to Ax = 0. Since there will be an infinite number of them, you want to present the kernel as a set of vectors that form a basis for it.
     
  18. Mar 28, 2014 #17
    So x1=x3+2x4+3x5, x2=-2x3-3x4-4x5.

    Would the solution be
    Basis of kernel = {##\begin{bmatrix}0 \\0 \\1 \\2 \\3 \end{bmatrix},\begin{bmatrix}0 \\0 \\-2 \\-3 \\-4 \end{bmatrix}##}
     
  19. Mar 28, 2014 #18

    Mark44

    Staff: Mentor

    Yes.
    No.

    This is the standard trick, so it pays to learn it.
    Write the five equations.
    Code (Text):

    x[SUB]1[/SUB] = x[SUB]3[/SUB] + 2x[SUB]4[/SUB] + 3x[SUB]5[/SUB]
    x[SUB]2[/SUB] = -2x[SUB]3[/SUB] - 3x[SUB]4[/SUB] - 4x[SUB]5[/SUB]
    x[SUB]3[/SUB] = x[SUB]3[/SUB]
    x[SUB]4[/SUB] =        x[SUB]4[/SUB]
    x[SUB]5[/SUB] =             x[SUB]5[/SUB]
    If you look at this just right, you should see that a vector x in the kernel is a linear combination of three vectors.
     
  20. Mar 28, 2014 #19
    So
    Basis of kernel = {##\begin{bmatrix}1 \\-2 \\1 \\0 \\0 \end{bmatrix},\begin{bmatrix}2 \\-3 \\0 \\1 \\0 \end{bmatrix},\begin{bmatrix}3 \\-1 \\0 \\0 \\1 \end{bmatrix}##}

    With those values being from the coefficients of x3, x4, and x5 respectively.
     
  21. Mar 28, 2014 #20

    Mark44

    Staff: Mentor

    The 3rd vector isn't in the kernel. The other two are. You have a mistake in your row reduction or copied a number incorrectly.
     
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