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Math Help - Subspaces of Vector Spaces

  1. #1
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    Subspaces of Vector Spaces

    Let V be some vector space and S be a subset.

    a. V=T(3,3) and S is the set of invertible linear transformations.

    I know the V represents all linear transformations from R^3 -> R^3, and all linear transformations in S must be invertible, thus they must all be bijective. But I'm unsure how to go about this?
    I know the three conditions that establish a subspace - S contains the 0 vector, S is closed under addition, and S is closed under scalar multiplication. Just not sure how to proceed.
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    Re: Subspaces of Vector Spaces

    are you trying to show S is a subspace? Let's get that squared away first.
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    Re: Subspaces of Vector Spaces

    Yes, either proving it is or disproving it.
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    Re: Subspaces of Vector Spaces

    well you've outlined the conditions that need to be met for a subspace.

    Is 0 in your space? .. DOH. 0 isn't very invertible is it? (it's the zero matrix by the way. your space consists of 3x3 matrices)
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    Re: Subspaces of Vector Spaces

    Quote Originally Posted by turbozz View Post
    Yes, either proving it is or disproving it.
    Quote Originally Posted by turbozz View Post
    Let V be some vector space and S be a subset.

    a. V=T(3,3) and S is the set of invertible linear transformations.
    I know the V represents all linear transformations from R^3 -> R^3, and all linear transformations in S must be invertible, thus they must all be bijective. But I'm unsure how to go about this?
    I know the three conditions that establish a subspace - S contains the 0 vector, S is closed under addition, and S is closed under scalar multiplication. Just not sure how to proceed.
    What would such a vector look like?
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    Re: Subspaces of Vector Spaces

    I was unsure of what the 0 vector would be in T(2,2). Originally I thought it would be T(0), but I guess since every linear transformation can be expressed as T(x) = Ax, then the zero 3 by 3 matrix represents the zero matrix for T(2, 2). I assume this is true then for any vector space
    T(m, n), where the zero vector is the n by m matrix.
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  7. #7
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    Re: Subspaces of Vector Spaces

    any dimensions.. what is 0? It's the element such that (A + 0) = A

    There's only one matrix in the complex field that satisfies that and that's a matrix of all 0s.
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    Re: Subspaces of Vector Spaces

    Equivalently, T is the set of linear transformations from R^3 to R^3 so that "0" is the linear transformation such that 0(x, y, z)= (0, 0, 0). Such a linear transformation does not have an inverse because it is not "one to one".
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    Re: Subspaces of Vector Spaces

    Ok thanks and just to verify one last point, its fair to say that the dim(T(m,n)) =dim(R^n*m).
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    Re: Subspaces of Vector Spaces

    Quote Originally Posted by turbozz View Post
    Ok thanks and just to verify one last point, its fair to say that the dim(T(m,n)) =dim(R^n*m).
    If T must be invertible then T cannot be non-square. So m=n

    If T is invertible then dim(T(n,n)) = n = dim(R^n)
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    Re: Subspaces of Vector Spaces

    But if we just say vector space T(m, n) contains all linear transformations of the form T:R^m -> R^n (regardless of whether they are invertible or not) then dim(T(m,n)) =mn, right? And since dim(R^n*m) = nm, can't one infer the dimensions are the same?
    Note, when I write R^n*m I mean the vector space that contains all n by m matrices.
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    Re: Subspaces of Vector Spaces

    Quote Originally Posted by turbozz View Post
    But if we just say vector space T(m, n) contains all linear transformations of the form T:R^m -> R^n (regardless of whether they are invertible or not) then dim(T(m,n)) =mn, right? And since dim(R^n*m) = nm, can't one infer the dimensions are the same?
    Note, when I write R^n*m I mean the vector space that contains all n by m matrices.
    at best dim(T(m,n)) = min(m,n)
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    Re: Subspaces of Vector Spaces

    Quote Originally Posted by romsek View Post
    If T must be invertible then T cannot be non-square. So m=n

    If T is invertible then dim(T(n,n)) = n = dim(R^n)
    No, that's not true. In the vector space, T(n, n), of linear transformations (NOT necessarily invertible) from R^n to R^m, given a basis \{v_k\} for R^n, we can have take as basis the collection of transformations A_{ij} defined by A_{ij}(v_i)= v_j if but A_{ij}(v_k)= 0 for k\ne i. That has dimension n^2, not n. More generally, T(n, m), the set of linear transformations (NOT necessarily invertible) from R^n to R^m, given basis \{u_i} for R^n and basis [itex]\{v_j\}[/itex] for R^m, has basis the collection of transformations A_{ij} deined b A_{ij}(u_i)= v_j with A_{ij}(u_k)= 0 if k\ne i and so has dimension nm.
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  14. #14
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    Re: Subspaces of Vector Spaces

    I'm going to have to study this.

    Thank you for the correction.
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