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Thread: 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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    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 $\displaystyle f: \ker(ST) \longrightarrow \ker S$ by $\displaystyle f(x)=T(x),$ for all $\displaystyle x \in \ker(ST)$ and apply the rank-nulity theorem to $\displaystyle f$.

    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:

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

    ohh...
    So am I right to say about $\displaystyle rank(AB) \leq min(rank(A), rank(B))$? That's the rank of $\displaystyle AB$ would always be less than or equals to the minimum rank between rank(A) and rank (B)?
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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
    ohh...
    So am I right to say about $\displaystyle rank(AB) \leq min(rank(A), rank(B))$? That's the rank of $\displaystyle AB$ 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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