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Homework Help: Linear Algebra: Basis for the Null Space of a Matrix

  1. Oct 27, 2005 #1
    I think I understand how to do this, but I wanted to double check my work. I have to find the basis of the null space for the matrix:

    1 0 2
    0 0 7

    So I knew that the basis of the image had two dimensions and a null space of one. The first and third columns are linearly independent (or at least thats what I made them)... which means that the second column is a free variable... However, this is where I get confused. The second column is 0, so the basis of the null space is... (lets let the first column be x, the second column be y, and the third column be z) 0y = -1x -2z and 0y = -7z + 0x?

    Thus if I were to write it in matrix form it'd be something like:
    ??? This doesn't seem right though...

    I have another question regarding how to find the basis of a subspace. If its written as a matrix I think I can do it, however when its written as an equation I get slightly confused. There are two problems:

    A){(x,y,z) are elements of R^3: 7x - 3y + z = 0}
    I have to give a basis and the dimension of this equation.

    So what I did was:
    (7 -3 1) So, z = 3y - 7x. So there are two free variables and one dependent variable, thus the dimension is 2 and the Null space must be 1?

    In the same vein of thought:

    B){(x,y,z,w) are elements of R^4: x + 2y -z = 5x -4y + w -3z = 0}

    I'm not really sure how this equation should be interpreted... Should I just treat it as x + 2y - z = 0, and leave w as a free variable? Thus z = x + 2y, so the dimension is 1 and the Null Space is 3?

    I'm not very familiar with this site yet (or linear algebra), so sorry if my formatting is wrong, if you need something clarified or have any questions feel free to ask. Thanks.
  2. jcsd
  3. Oct 27, 2005 #2
    The 0's are a dead giveaway. What happens to the vector [0,1,0] when transformed by this matrix ?
    That's a good analysis.
    Treat them as two simultaneous equations in 4 variables. z = f(x,y) for some non-constant function f, so you get w = g(x,y) for some non-constant function g. Thus you have 2 free variables (x,y) that determine (w,z).
    Last edited: Oct 27, 2005
  4. Oct 27, 2005 #3
    Thanks a bunch! It makes alot of sense once I see it, it just takes me awhile to figure it out. I have another question.

    {p(x) is an element of P^3: p(3) = p(4)}
    Find the dimension and a basis

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

    Where p(5) - p(-3) = p prime (2)

    So I just have to plug in the values, which gives me a matrix, then solve for the free variables?

    Let me show you what I did:

    p(5) = 5 + b5 + c25 + d125 - (-3 + -b3 + c9 + -d27) = a0 + b + c4 + d8

    So simplifying it we get:

    a8 + b7 + c12 + d144 = 0

    So b = c(-12/7) + d(-144/7) + a(8/7)

    So there are 3 free variables, which are c, d, and a. Thus the dimension of the image is 1 and the dimension of the null space is 3?
    Last edited: Oct 27, 2005
  5. Oct 27, 2005 #4
    An easy way is to note that if (a,b,c,d) is a vector in P3 that satisfies the requirement, then a + b*3 + c*9 + d*27 = a + b*4 + ... . You can already see that this is familiar territory.
  6. Oct 27, 2005 #5
    Thanks for the reply, thats what I did, or at least meant to say before. It seems I posted a solution to a different problem. So far you've been an immense help, as most of this makes sense now, however I have a few last questions I'd like to ask.

    Find the dimension and basis of the image of L and the null space of L:

    1) L: P^2 -> R^1 where L(p) = the integral (p(x) + the derivative of p(x+1)) dx, evaluated from 1 to 3.

    Basically I first found a basis of P^2 and used that, so I chose, x^2, x, and 1 as my basis for P^2. I then completed the integral (been awhile since calc, so lets see if I did this right).

    L(1) = x
    L(x) = (x^2)/2 + x + 1
    L(x^2) = (x^3)/3 + (x+1)^2

    I then evaluated it for each of the values:

    3-1 = 2
    (9/2 + 3 + 1) - (1/2 + 1 + 1) = 6
    (9 + 16) - (1/3 + 4) = 62/3

    Its getting late so my brain is getting alittle confused, what does this result tell me again? Since its a linear map from P^2 to R^1, the largest the dimension of the image can be is 1, yet when I put the values in a matrix:


    The matrix tells me that the dimension of the image is 3... but this makes no sense! Should the matrix instead be written as:

    (2, 6, 62/3)

    Thus, the matrix has a dimension of 1 and two free variables, which makes sense. The only thing I am confused about is that 2, 6, and 62/3 are real number values, so how can I write one in terms of another (unless the equation goes something like 2a + 6b + 62/3c = 0, and I then write a in terms of b and c).

    One last question for the night:

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

    I wasn't entirely sure how to approach this problem, but what I decided was that this problem seems similar to a problem we went over earlier where P^3 gives us the equation:

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

    I then set up the problem and plugged in the values:

    2(a + 7b + c49 + d343) - 3(a + b6 + c36 + d216)

    and simplified to:

    -a -4b -10c + 38d

    Now this is where I'm confused, in the original equation 2p(7) - 3p(6) = 14. What happens to the 14? Since its the same type of unit as "a", does it get multiplied by a scalar "a"? That is to say, in the equation I simplified:

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

    So I would then get 0 on one side of the equation and say that the dimension is 1 and there are 3 free variables. In line with this question, when it says "Find all solutions to the following problem," since we're working with basis and Null Space's, how should I write my answer? What I mean is, I can easily just say that three free variables gives infinite possible answers, but I'm not sure that was what the question intended to ask (or rather it was, but it needs to be written correctly). So, assuming that the answer has to be written in terms of a basis, I can just use the equation as my basis for the Image, saying it has a dimension of 1 and the Null space is 3, and then just provide a basis for the Null Space?

    In any case, thanks for all the help, I really learned alot.
  7. Oct 27, 2005 #6
    Immediately, we know that the dimension of the image of L will either be 0 or 1. There are no other choices for R1. :wink:
    Now that we know that the image space is not 0-dimensional (some polynomial integrated over that interval will be non-zero), we can choose any vector of R1, like 1, and use that as our basis without any more work needed! :biggrin:
    It is getting late. :smile: The fact that the values of the images of all the basis vectors are all scalar multiples of each other should tell you that the image space is 1-dimensional. Putting them in a matrix doesn't make any sense. Use the values you got from integrating the basis to find the form of polynomials that integrate to 0. For example, 2 and 6 are the images of (1,0,0) and (0,1,0), respectively. Therefore, the image of -3*(1,0,0) + 1*(0,1,0) = -6 + 6 = 0, so that vector can be used as your first basis vector in the null space.
    The dangling phrases above aren't anything related to the set you want. You should be looking at the equation 2(a + 7b + c49 + d343) - 3(a + b6 + c36 + d216) = 14, which specifies the vectors (a,b,c,d) that qualify for the set in question. Since there is only one equation, you should see that there will be 3 free variables.
    Last edited: Oct 27, 2005
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