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Math Help - null space and rank

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    null space and rank

    let A be mxn matrix and B be the transpose of A. let P=BA. prove that A and P have the same null space and the same rank.
    please help.
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    Quote Originally Posted by Kat-M View Post

    let A be mxn matrix and B be the transpose of A. let P=BA. prove that A and P have the same null space and the same rank.
    please help.
    we only need to prove that the null space of A and P are equal, because then by the rank-nullity theorem they must also have the same rank.

    if Ax=0, then Px=0 and hence \text{null}(A) \subseteq \text{null}(P). conversely, if x \in \text{null}(P), then Px=A^TAx=0, i.e. Ax \in \text{null}(A^T)=(\text{col}(A))^{\perp}. hence Ax \in \text{col(A)} \cap (\text{col}(A))^{\perp}=(0). \ \Box
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    Quote Originally Posted by NonCommAlg View Post
    if x \in \text{null}(P), then Px=A^TAx=0, i.e. Ax \in \text{null}(A^T)=(\text{col}(A))^{\perp}. hence Ax \in \text{col(A)} \cap (\text{col}(A))^{\perp}=(0). \ \Box
    Alternatively, if Px=A^{\textsc t}Ax=0 then (Ax)^{\textsc t}(Ax)= x^{\textsc t}A^{\textsc t}Ax = 0 and so Ax = 0.
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    Quote Originally Posted by Opalg View Post

    Alternatively, if Px=A^{\textsc t}Ax=0 then (Ax)^{\textsc t}(Ax)= x^{\textsc t}A^{\textsc t}Ax = 0 and so Ax = 0.
    that's certainly an easier way!
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    Quote Originally Posted by NonCommAlg View Post
    we only need to prove that the null space of A and P are equal, because then by the rank-nullity theorem they must also have the same rank.

    if Ax=0, then Px=0 and hence \text{null}(A) \subseteq \text{null}(P). conversely, if x \in \text{null}(P), then Px=A^TAx=0, i.e. Ax \in \text{null}(A^T)=(\text{col}(A))^{\perp}. hence Ax \in \text{col(A)} \cap (\text{col}(A))^{\perp}=(0). \ \Box

    i am trying to see why \text{null}(A^T)=(\text{col}(A))^{\perp}.. could you also help me with this?
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    Then look at Opalg's response.
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