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Math Help - Linear Spaces

  1. #1
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    Linear Spaces

    1) If v is any vector in R^n, what are the possible dimensions of the space of all n x n matrices A such that Av = 0?

    2) Find a basis of each of the following linear spaces, and thus determine their dimensions.
    a) {f in P4: f is even}
    b) {f in P4: f is odd}
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  2. #2
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    Generally speaking, one equation reduces the number of "free variables", and so the dimension, by 1. When you multiply a matrix, A, by the fixed vector, v, and set it equal to 0: Av= 0, each row of the matrix is one equation. Thus while the dimension of the space of all n by n matrices is n^2 the fact that Av= 0 gives n equations, reducing the dimension to n^2- n= n(n-1).

    P4 is the set of polynomials of degree 4 or less? Then your first space consists of all such polynomials that have only even powers: \{ax^4+ bx^2+ c\}. The second is all such polynomials that have only odd powers: \{ax^3+ bx\}. It should be easy to find bases for those.
    Last edited by HallsofIvy; October 10th 2009 at 07:10 AM.
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    Quote Originally Posted by HallsofIvy View Post
    Generally speaking, one equation reduces the number of "free variables", and so the dimension, by 1. When you multiply a matrix, A, by the fixed vector, v, and set it equal to 0: Av= 0, each row of the matrix is one equation. Thus while the dimension of the space of all n by n matrices is n^2 the fact that Av= 0 gives n equations, reducing the dimension to [tex]n^2- n= n(n-1)[tex].

    @HallsofIvy - Would you please explaining this - or suggesting somewhere from where I can read it up? Question I have is
    " one equation reduces the number of "free variables", and so the dimension, by 1" - By this I guess you mean the dimension of the null-space of the equations. How are you relating that to dimension of the matrix space? " n^2- n= n(n-1)."
    Thanks
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    For example. Find the dimension of the subspace of R^4, <u, v, w, x>, satisfying 3u- 2v+ 3w- x= 0. (Yes, we can consider it the null space of the operator that maps any <u, v, w, x> to 3u- 2v+ 3w- x in R^1.) Since that is a single equation I know immediately that the dimension of this subspace is 4- 1= 3. In fact, I can solve for x, x= 3u- 2v+ 3w, so I can write any member of that subspace as <u, v, w, 3u- 2v+3w>= u(1, 0, 0, 3>+ v<0, 1, 0, -2>+ w<0, 0, 1, 3> showing that {<1, 0, 0, 3>, <0, 1, 0, -2>, <0, 0, 1, 3>} is a basis.

    If I were talking about the subspace satifying both that and the equation u+ v- w+ x= 0 (Which we can think of as the null space for the linear transformation that takes <u, v, w, x> to <3u-2v+3w-x. u+v-w+x> in R^2. Any subspace is the null space of some linear transformation!), I can see that there are two equations and, once I have checked that they are independent, know that this subspace will have dimension 4- 2= 2. If I add the two equations, I eliminate x and get 4u- v+ 2w= 0. I can solve that for v: v= 4u+ 2w. Putting that into the first equation, 3u-2(4u+2w)+ 3w- x= -5u- w- x= 0 so x= -5u- w. That means that any member of this subspace can be written as <u, 4u+2w, w, -5u- w>= u<1, 4, 0, -5>+ v<0, 2, 1, -1> which means that {<1, 4, 0, -5>,<0, 2, 1, -1>} is a basis for the subspace.
    Last edited by HallsofIvy; October 10th 2009 at 10:45 AM.
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    Quote Originally Posted by HallsofIvy View Post
    For example. Find the dimension of the subspace of R4, <u, v, w, x>, satisfying 3u- 2v+ 3w- x= 0. (Yes, we can consider it the null space of the operator that maps any <u, v, w, x> to 3u- 2v+ 3w- x in R^1.) Since that is a single equation I know immediately that the dimension of this subspace is 4- 1= 3. In fact, I can solve for x, x= 3u- 2v+ 3w, so I can write any member of that subspace as <u, v, w, 3u- 2v+3w>= u(1, 0, 0, 3>+ v<0, 1, 0, -2>+ w<0, 0, 1, 3> showing that {<1, 0, 0, 3>, <0, 1, 0, -2>, <0, 0, 1, 3>} is a basis.

    If I were talking about the subspace satifying both that and the equation u+ v- w+ x= 0 (Which we can think of as the null space for the linear transformation that takes <u, v, w, x> to <3u-2v+3w-x. u+v-w+x> in R^2. Any subspace is the null space of some linear transformation!), I can see that there are two equations and, once I have checked that they are independent, know that this subspace will have dimension 4- 2= 2. If I add the two equations, I eliminate x and get 4u- v+ 2w= 0. I can solve that for v: v= 4u+ 2w. Putting that into the first equation, 3u-2(4u+2w)+ 3w- x= -5u- w- x= 0 so x= -5u- w. That means that any member of this subspace can be written as <u, 4u+2w, w, -5u- w>= u<1, 4, 0, -5>+ v<0, 2, 1, -1> which means that {<1, 4, 0, -5>,<0, 2, 1, -1>} is a basis for the subspace.
    Hi - Thanks for a detailed reply. I will read it carefully and post any further questions plz.
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