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Math Help - Dimension of null space of two matrix multiplication

  1. #1
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    Dimension of null space of two matrix multiplication

    let S and T be linear transformation on n-dimensional vector space
    why is dim ker(ST) less than dim ker(S) + dim ker(T)
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  2. #2
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    Re: Dimension of null space of two matrix multiplication

    Quote Originally Posted by parklover View Post
    let S and T be linear transformation on n-dimensional vector space
    why is dim ker(ST) less than dim ker(S) + dim ker(T)
    define f: \ker(ST) \longrightarrow \ker S by f(x)=T(x), for all x \in \ker(ST) and apply the rank-nulity theorem to f.

    Spoiler:
    if you tried and still couldn't finish the proof, see the proof of the "claim" in my blog.
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    Re: Dimension of null space of two matrix multiplication

    Actually, would it be always less than? I always thought it could be less than or equals, am I right on this?
    Say like if matrix A times B, then rank(AB) <= min(rank(A), rank(B)), is this right? And then apply this same idea to the nullspace too, would it all come out right?
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Dimension of null space of two matrix multiplication

    Quote Originally Posted by xEnOn View Post
    Actually, would it be always less than?
    The equality is possible:

    \dim \ker (SI)=\dim \ker (S)=\dim \ker (S)+0=\dim \ker (S)+\dim \ker (I)
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    Re: Dimension of null space of two matrix multiplication

    ohh...
    So am I right to say about rank(AB) \leq min(rank(A), rank(B))? That's the rank of AB would always be less than or equals to the minimum rank between rank(A) and rank (B)?
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: Dimension of null space of two matrix multiplication

    Quote Originally Posted by xEnOn View Post
    ohh...
    So am I right to say about rank(AB) \leq min(rank(A), rank(B))? That's the rank of AB would always be less than or equals to the minimum rank between rank(A) and rank (B)?
    Yes, you are right.
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