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Thread: Preservation of the rank when multiplied by a full rank matrix.

  1. #1
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    Preservation of the rank when multiplied by a full rank matrix.

    Hello all,

    I am trying to find a proof that, for B an invertible matrix, rank(AB) = rank(BA) = rank(A).

    I know I should probably play with the null space of A, AB and BA but I cannot find the right approach to this problem.

    Do you have any hint for me, or online notes with the property proven?
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  2. #2
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    Hi,
    let X be a vector matrix,
    then you can proove that {ABX, X}={AX, X}, what could help you
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  3. #3
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    If A is from $\displaystyle R^n$ to $\displaystyle R^n$ then its rank is n- nullity(A) so, yes, looking at the null space is one way to go. If you can show that AB, BA and A all have the same nullity then they will all have the same rank.

    In particular, if u is in the null space of A, the Au= 0 so, of course, BAu= B0= 0 for any matrix B. That is, the null space of A is always a subspace of the null space of BA. Now, the other way. Suppose BAu= 0. The B(Au)= 0 and since B is invertible, Au= 0 so we have that the null space of BA is a subspace of the null space of A. That proves that A and BA have exactly the same null spaces so of course they are of the same dimension.

    Now, suppose u is in the null space of AB. That is, suppose ABu= A(Bu)= 0. What can you say about u?
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  4. #4
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    I am trying to find a proof that, for B an invertible matrix, rank(AB) = rank(BA) = rank(A).
    The rank of a matrix $\displaystyle A:V\to V$ is merely the dimension of the range of $\displaystyle A$, $\displaystyle \mathop{\textrm{Ran}}A$, which is the subspace of $\displaystyle V$ consisting of all vectors of the form $\displaystyle Au$ with $\displaystyle u\in V$.

    You could therefore prove that $\displaystyle \mathop{\textrm{Ran}}AB$ and $\displaystyle \mathop{\textrm{Ran}}BA$ have the same dimension as $\displaystyle \mathop{\textrm{Ran}}A$ if $\displaystyle B$ is invertible.

    Here are some hints:

    1. Show that $\displaystyle \mathop{\textrm{Ran}} AB=\mathop{\textrm{Ran}}A$.

    First off, if $\displaystyle u\in\mathop{\textrm{Ran}} AB$ then $\displaystyle u=ABv$ for some $\displaystyle v$ and so $\displaystyle u\in\mathop{\textrm{Ran}} A$. Thus $\displaystyle \mathop{\textrm{Ran}} AB\subseteq\mathop{\textrm{Ran}} A$.

    Now prove that $\displaystyle \mathop{\textrm{Ran}} A\subseteq\mathop{\textrm{Ran}} AB$ in similar fashion (needs $\displaystyle B$ to be invertible here).

    2. It is unfortunate that $\displaystyle \mathop{\textrm{Ran}}BA$ is not the same as $\displaystyle \mathop{\textrm{Ran}} A$ in general, so a similar proof to the above fails in this case. However there is another result that saves the day.

    Recall the null space of $\displaystyle A$, $\displaystyle \mathop{\textrm{Nul}}A$, is the set of all vectors $\displaystyle u$ in $\displaystyle V$ for which $\displaystyle Au=0$.

    Well, it turns out that $\displaystyle \mathop{\textrm{Nul}}BA=\mathop{\textrm{Nul}}A$ if $\displaystyle B$ is invertible (prove it!)

    Now there is a very well-known theorem which states that the dimensions of $\displaystyle \mathop{\textrm{Ran}}A$ and $\displaystyle \mathop{\textrm{Nul}}A$ add up to the dimension of $\displaystyle V$.

    The same is true of $\displaystyle \mathop{\textrm{Ran}}BA$ and $\displaystyle \mathop{\textrm{Nul}}BA$ of course.

    Now if $\displaystyle \mathop{\textrm{Nul}}BA$ and $\displaystyle \mathop{\textrm{Nul}}A$ are the same then they have the same dimension. It follows that ...

    Hold on, someone's at the door. I'll just go answer it. Back in a mo.

    (Dum dee doo, my eyes are old and bent, etc ...)

    Hello, can I help you? ... I said, what d'you want? Wait ... What ... No, don't ... HEEEEEEELP ...
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