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Math Help - Quick questions about matricies/subspaces

  1. #1
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    Quick questions about matricies/subspaces

    Hi.

    Just two quick questions.

    i) When solving a linear system and you've reduced the matrix to reduced row echelon form. For example:

    1 -2 0 5 0
    0 0 1 2 0
    0 0 0 0 1
    0 0 0 0 0

    How do you determine what the free variables are in this equation or in any equation in general? Is it always the ones that don't have coefficients of 1?

    ii) let a b c be scalars such that abc != 0.

    prove that ax + by + cz = 0 is a subspace of R3.

    Now the conditions to be a subspace is such that the subset has to be closed under vector addition and scalar multiplication.

    now the book says that ax+by+cz is a solution to the homogenous system Ax = 0
    In this case A would be the row vector a, b, c. and x would be the column vector
    x, y ,z.

    Now I'm a bit confused as to how you can describe the solution as a plane of ax + by + cz and how can you use that to prove that its a subspace of R3.
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    (i) I'm a little confused by the matrix you've exhibited there. What size is the original problem? A nonhomogeneous 4 x 4? Or a homogeneous 4 x 5?

    In general, the columns that don't have leading 1's in the rref (the second and fourth, in your case) correspond to the variables that will be the arbitrary parameters of your solution.

    [EDIT]: See HallsofIvy's post for a correction here. I would say that the way I do back substitution, my statement stands.

    (ii) You need also to show that the set is nonempty. Easiest way is usually to show that the zero vector is in your set (which is necessary, anyway, to have closure under scalar multiplication and vector addition). The zero vector is in the set. Why?

    Take scalar multiplication. Start with the vector \langle x,y,z\rangle in your set. That implies ax+by+cz=0. Multiply the vector by an arbitrary scalar r. Is the result still in the set?

    Take vector addition. Start with two vectors \langle x_{1},y_{1},z_{1}\rangle and \langle x_{2},y_{2},z_{2}\rangle that are both in the set. That implies two conditions. Add the two vectors together and check if the new vector still satisfies the necessary condition.

    Using the plane condition probably amounts to recognizing that the equation for the set is also a dot product:

    \mathbf{n}\cdot\mathbf{x}=0, where

    \mathbf{n}=\langle a,b,c\rangle and

    \mathbf{x}=\langle x,y,z\rangle.

    This is also the equation for a plane.

    You can use the properties of the dot product to prove the necessary result. Do you see how?
    Last edited by Ackbeet; February 28th 2011 at 02:50 AM. Reason: Correction in HallsofIvy's post.
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  3. #3
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    Quote Originally Posted by Kuma View Post
    Hi.

    Just two quick questions.

    i) When solving a linear system and you've reduced the matrix to reduced row echelon form. For example:

    1 -2 0 5 0
    0 0 1 2 0
    0 0 0 0 1
    0 0 0 0 0

    How do you determine what the free variables are in this equation or in any equation in general? Is it always the ones that don't have coefficients of 1?
    The way that is written, there are NO free variables because that system is inconsistent. Taking your variables to be x, y, z, w, that augmented matrix corresponds to the equations x- 2y+ 5w= 0, z+ 2w= 0, and 0= 1. That last is, of course, impossible to satisfy.

    But, even in a consistent system, there is no way to say that certain variables are free variables. Which variables are "free" depends upon how you solve the equatons. If the last column were all "0"s then we would have x- 2y+ 5w= 0 and z+ 2w= 0. We can solve for z= -2w and x= 2y- 5w so that if we know w and y, we can find x and z. There w and y are the "free variables" because we are free to choose them to be whatever we want. But we could also have solved for w= -z/2 and then solve x- 2y- 5z/2= 0 for y= x/2- 5z/4. Then x and z are the free variables. It is the number of free variables that does not change with how we solve the equations, not which variables are the free variables.

    ii) let a b c be scalars such that abc != 0.

    prove that ax + by + cz = 0 is a subspace of R3.
    You mean "the set of all (x, y, z) such that ax+ by+ cz= 0 is a subspace of R3".

    Now the conditions to be a subspace is such that the subset has to be closed under vector addition and scalar multiplication.

    now the book says that ax+by+cz is a solution to the homogenous system Ax = 0
    In this case A would be the row vector a, b, c. and x would be the column vector
    x, y ,z.

    Now I'm a bit confused as to how you can describe the solution as a plane of ax + by + cz and how can you use that to prove that its a subspace of R3.
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  4. #4
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    Hi.
    Thanks for the reply. I realized i made a major mistake and forgot to put down what each row equals to in each of those equations. There are 5 variables. x1-x5 or x y z w v if you may. Here is the correction.

    1 -2 0 5 0|4
    0 0 1 2 0|-7
    0 0 0 0 1|1
    0 0 0 0 0|0

    Quote Originally Posted by Ackbeet View Post
    (i) I'm a little confused by the matrix you've exhibited there. What size is the original problem? A nonhomogeneous 4 x 4? Or a homogeneous 4 x 5?

    In general, the columns that don't have leading 1's in the rref (the second and fourth, in your case) correspond to the variables that will be the arbitrary parameters of your solution.

    [EDIT]: See HallsofIvy's post for a correction here. I would say that the way I do back substitution, my statement stands.

    (ii) You need also to show that the set is nonempty. Easiest way is usually to show that the zero vector is in your set (which is necessary, anyway, to have closure under scalar multiplication and vector addition). The zero vector is in the set. Why?

    Take scalar multiplication. Start with the vector \langle x,y,z\rangle in your set. That implies ax+by+cz=0. Multiply the vector by an arbitrary scalar r. Is the result still in the set?

    Take vector addition. Start with two vectors \langle x_{1},y_{1},z_{1}\rangle and \langle x_{2},y_{2},z_{2}\rangle that are both in the set. That implies two conditions. Add the two vectors together and check if the new vector still satisfies the necessary condition.

    Using the plane condition probably amounts to recognizing that the equation for the set is also a dot product:

    \mathbf{n}\cdot\mathbf{x}=0, where

    \mathbf{n}=\langle a,b,c\rangle and

    \mathbf{x}=\langle x,y,z\rangle.

    This is also the equation for a plane.

    You can use the properties of the dot product to prove the necessary result. Do you see how?
    Bit confused about the answer to ii.

    How does just taking (x,y,z) as a vector in the subset imply ax+by+cz = 0?

    I get the part with scalar multiplication and vector addtion. The resultant vectors will be of the form (x,y,z) which will result in closure under vector addition and multiplication.

    I still don't understand how you can use the dot product to prove it though. The dot product makes sense n.x = 0.



    Also one last question. Does the zero vector have to be in EVERY subspace? For example here is another easy question
    determine whether the indicated subset is a subspace of the Euclidian space Rn.

    {[x, x+1] | x E R}

    now in this example, the zero vector can never be in this subset because no matter what value of x,you can never end up with [0,0]. Is that proof enough that it is not a subspace?
    Taking vector addition and scalar multiplication, you will always end up with a vector of the form [x,x+1] though.
    Last edited by Kuma; February 28th 2011 at 05:53 AM.
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  5. #5
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    Ok, so typically, the way I do back substitution, I'd get that v=1. Then I'd have z + 2w = -7. Let s = w be a parameter. Then, z = -7 - 2s. Next, you'd have x - 2y + 5w = 4, so you'd get x = 4 + 2y - 5w. Let t = y be a parameter. Then you'd end up with x = 4 + 2t - 5s. That's the way I do back substitution, and you can see that the columns without leading 1's end up corresponding to variables that are arbitrary parameters in the final solution.

    Make sense?
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  6. #6
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    Quote Originally Posted by Ackbeet View Post
    Ok, so typically, the way I do back substitution, I'd get that v=1. Then I'd have z + 2w = -7. Let s = w be a parameter. Then, z = -7 - 2s. Next, you'd have x - 2y + 5w = 4, so you'd get x = 4 + 2y - 5w. Let t = y be a parameter. Then you'd end up with x = 4 + 2t - 5s. That's the way I do back substitution, and you can see that the columns without leading 1's end up corresponding to variables that are arbitrary parameters in the final solution.

    Make sense?
    Yup, but that's what I'm wondering about is why the choice of arbitrary variables. So anything without a coefficient of 1 would become an arbitrary variable? For example you set s = w in the equation. Why can't you just set s = y and set w = -1/2y - 7/2. I think I'm missing something here.
    What about in the case where there are no columns with leading 1's? Is that even a possible case?
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  7. #7
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    Re-read HallsofIvy's Post # 3. The total number of arbitrary parameters will be the same. But it might look different depending on how you assign them. The actual vectors that you get as solutions, though, will definitely be the same no matter what it looks like.

    If after reducing to row-echelon form, there are no columns with leading 1's, then your matrix is the zero matrix. You'll have two possibilities, depending on whether the system is homogeneous or not. If the system is homogeneous, then any vector whatsoever of the correct dimensions will solve the system. If the system is non-homogeneous, then no vector solves the system.
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  8. #8
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    Quote Originally Posted by Ackbeet View Post
    Re-read HallsofIvy's Post # 3. The total number of arbitrary parameters will be the same. But it might look different depending on how you assign them. The actual vectors that you get as solutions, though, will definitely be the same no matter what it looks like.

    If after reducing to row-echelon form, there are no columns with leading 1's, then your matrix is the zero matrix. You'll have two possibilities, depending on whether the system is homogeneous or not. If the system is homogeneous, then any vector whatsoever of the correct dimensions will solve the system. If the system is non-homogeneous, then no vector solves the system.
    ooh ok. So the number of free variables does not change in the end but does it matter which ones we determine are free? It seems that it's totally our choice.
    Thanks
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    You have some choice. In your particular system, you can't choose what v is. But which of the other variables are chosen to be parameters is up to you. I would recommend using some consistent process.
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  10. #10
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    To take a very simple example, the single equation in two unknows, x- y= 0, has an infinite number of solutions: choose any value for x and take y= x. There I am thinking of x as the "free variable". But obviously, I could as easily choose y to be anything and the x= y. Now, y is the "free variable". I have exactly one free variable either way but which variable it is depends upon how we want to write the solution. Of course, if we think of x and y as representing the vector <x, y>, then the solution space is spanned by the vector <1, 1>.
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