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

1. ## 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?

2. Hi,
let X be a vector matrix,
then you can proove that {ABX, X}={AX, X}, what could help you

3. If A is from $R^n$ to $R^n$ 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?

4. I am trying to find a proof that, for B an invertible matrix, rank(AB) = rank(BA) = rank(A).
The rank of a matrix $A:V\to V$ is merely the dimension of the range of $A$, $\mathop{\textrm{Ran}}A$, which is the subspace of $V$ consisting of all vectors of the form $Au$ with $u\in V$.

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

Here are some hints:

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

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

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

2. It is unfortunate that $\mathop{\textrm{Ran}}BA$ is not the same as $\mathop{\textrm{Ran}} A$ 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 $A$, $\mathop{\textrm{Nul}}A$, is the set of all vectors $u$ in $V$ for which $Au=0$.

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

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

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

Now if $\mathop{\textrm{Nul}}BA$ and $\mathop{\textrm{Nul}}A$ 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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### rank of product of full rank matrices

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