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Homework Help: Linear Algebra Questions

  1. Oct 30, 2005 #1
    I posted before about Null Spaces and, after some review, I think I have a grasp on it. However I have a few more general questions, so I thought I'd start a new thread (although there may be some redundancy)

    Problem 1:
    Find the dimension and a basis for the Row, Column, and Null Spaces.

    1 0 0 (-5/36) (101/36) (-11/3)
    0 1 0 (-1/9) (4/9) (2/3)
    0 0 1 (5/6) (1/6) 0

    The solution I arrived at was:

    Dimension of Column and Row Spaces = 3
    Dimension of the Null Space = 3

    Basis for the Row Space (Above Matrix)
    Basis for the Column Space:

    1 0 0
    0 1 0
    0 0 1

    Basis for the Null Space:
    (-5/36) (101/36) (-11/3)
    (-1/9) (4/9) (2/3)
    (5/6) (1/6) 0
    1 0 0
    0 1 0
    0 0 1

    I was really just hoping someone could check this problem for me, as I think I did it right, but I'm not entirely sure.

    Problem 2:
    Find the Dimension of the Image and the Null Space, Find a Basis for the Image and the Null Space.

    L: V -> V where L(f) = Second deriv(f) + f
    Vector Space is spanned by Cos(x), Sin(x), xCos(x), xSin(x)

    Ok, so I used those as my basis, and the transformed equations turn into:

    Cos(x) = -sin(x) + sin(x), so it equals 0
    Sin(x) = -cos(x) + cos(x), so this too equals 0
    xsin(x) = cos(x) + cos(x) - xsin(x) + xsin(x) = 2cos(x)
    xcos(x) = -sin(x) - sin(x) + xcos(x) - xcos(x) = -2sin(x)

    Ok, so I got this far, since cos and sin are linearly independant of each other, the Dimension of the Image is 2, and the Dimension of the Null Space is 2 (we started with 4 linearly independent vectors). However, this is where I get confused, the basis for the image is:

    2cos(x) 0
    0 -2sin(x) (??)

    and the basis for the Null Space is where 2cos(x) - 2sin(x) = 0? So, the null space is where Ax = b is Ax = 0, so the solution would be...??:

    2cos(x) - 2sin(x) = 0
    cos(x) - sin(x) = 0
    cos(x) = sin(x)

    Whenever Cos(x) is equivalent to sin(x)?? ... So my basis for the null space would be the two occasions within 6pi radians where cos(x) = sin(x)? How would I write this in matrix/vector notation?

    Problem 3:

    Find all solutions to the following linear problems:
    All polynomials p in P^3 for which 2p(7) - 3p(6) = 14

    So I set p(x) = a + bx + cx^2 + dx^3

    I then did 2p(7) - 3p(6) = 14 and got:

    a = 38d - 10c - 4b - 14

    Ok, so there are 3 free variables, and one dependant variable, so there are infinite solutions, however my question is how to write that. Basically I have to write this solution as an equation of vectors, which I am unsure how to do unless I can just write is as:

    <38d - 10c -4b -14, b , c , d> --->
    14<-1, 0, 0, 0> + b<-4, 1, 0, 0> + c<-10, 0, 1, 0> + d<38, 0, 0, 1> ???

    Problem 4:
    Last one!

    Find all the solutions to the following linear equation:
    All functions in V for which: second deriv(f) + f = 3sin(x), where V is the vector space defined in Problem 2 (see above).

    We have a basis from before, that was figured out to be:

    2cos(x) 0
    0 2sin(x)

    Here's what I did, I'm not entirely sure this makes sense.

    We set the equation 2cos(x) - 2sin(x) = 3sin(x). So the equation has solutions whenever 2cos(x) = 5sin(x)

    So the solutions would be all multiples of the appropriate x value for this (I used a calculator, got something like .38radians and 3.522 radians). This is all well and good, however, are these numbers what I was looking for, and furthermore, how do I write the solutions properly (in matrix/vector form or something like that).

    I know I've been posting alot, but I think I'm slowly piecing together the basics of linear algebra, thanks for all the help
  2. jcsd
  3. Oct 30, 2005 #2
    Basically I figured out the answer to question 1, however my main problem is finding out how to write the basis for problem 2 as vectors. That, and I'm still working on problems 3 and 4, any help is greatly appreciated!
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