5
$\begingroup$

Knowing that $A$ is equivalent to $B$ if there exists an invertible $m\times m$ matrix $P$ and an invertible $n\times n$ matrix $Q$ such that $PAQ = B$, how can I prove that $A$ and $B$ are equivalent iff $\text{rank}(A) =\text{rank}(B)$?

I've managed to solve the forward direction of the iff and am confident it is correct:

Suppose $A$ and $B$ are equivalent. Then, $PAQ = B$. Knowing this, we can assume $$ \text{rank}(PAQ) \leq \text{rank}(A) = \text{rank}(P^{-1} B Q^{-1}) \leq \text{rank}(B) $$ As $\text{rank}(PAQ) = \text{rank}(B)$, all inequalities must be equalities, so $\text{rank}(A) =\text{rank}(B)$.

I am not sure how to prove this statement in the reverse direction. I think that the invertible matrix theorem could be useful for this problem

$\endgroup$
0
$\begingroup$

For one way, the answer is correct.

The other way, let $A$ and $B$ be matrices which are equivalent.

Note that the rank of $A$ is equal to the rank of $B$, therefore the dimensions of the image of $A$ and the image of $B$ are the same. Let $k$ be the rank of $A$(and of $B$). Let $\{v_i\}_{i=1,...,k}$ and $\{w_i\}_{i=1,...,k}$ be bases for the images of $A$ and $B$ respectively. Note that $k \leq m,n$ by the fact that row rank equals column rank. So, we complete the bases of the images, to bases for $\mathbb R^m$, and to avoid confusion, the completed bases are $\{v_i\}_{i=1,...,n}$ and $\{w_i\}_{i=1,...,n}$.

Now, we want matrices $P,Q$ such that $PAQ = B$. Think of it this way : $Q$ rewrites the input vector of $B$ in a manner convenient for $A$. Then $A$ does its job on the rewritten vector which it finds easy to work with, and then $P$ rewrites the output of $A$ in a way which $B$ would have written it. That is the break up of the jobs of $P$ and $Q$ : they are one way translators from the input/output language of $A$ to the language of $B$ and vice-versa, if you like.

For each $v_i$, pick a single preimage $e_i$, and for each $w_i$, pick a preimage $f_i$. Now, $\{e_i\}_{i=1,...,k}$ and $\{f_i\}_{i=1,...,k}$ are linearly independent sets (check!) so they can be completed to bases of $\mathbb R^n$. Without confusion, we will call these bases as $\{e_i\}_{i=1,...,n}$ and $\{f_i\}_{i=1,...,n}$ respectively.

Now, what has to be thought here, is that $\{e_i\}$ is like $A$'s mother tongue and $\{f_i\}$ is like $B$'s mother tongue.

Therefore, the task of $Q$, when it receives a vector that is $B$'s mother tongue, is to convert it into $A$'s mother tongue. That leads to a very simple answer : $Q$ is the basis conversion matrix from $f_i$ to $e_i$. That is, $Q$ is the matrix of the unique linear transformation that satisfies $Q(f_i) = e_i$ for all $i = 1,...,n$.

Now that $Q$ has done its job, $A$ receives input in its mother tongue, so it outputs some vector whose entries are in the basis of $\{v_i\}$.

But $B$ outputs in the basis $\{w_i\}$! Obviously, it is clear that $P$, then, must be the unique linear transformation with $P(v_i) = w_i$.

So $Q$ and $P$ are just basis transformation matrices.

Finally, we can provide a proof that $PAQ = B$. Let $x \in \mathbb R^n$.

Then, $x = \sum_{i=1}^n x_if_i$, so $Bx = \sum_{i=1}^k x_iw_i$.

Alternately, by the nature of $Q$, $Qx = \sum_{i=1}^n x_ie_i$. Now, given what $A$ does, $AQx = \sum_{i=1}^k x_iv_i$, and then by what $P$ does, $PAQx = \sum_{i=1}^k x_iw_i$.

Hence $B = PAQ$. Since $P,Q$ are change of basis matrices, they are clearly invertible. They are also seen to be of the right dimension.

This proves the proposition.

$\endgroup$
  • $\begingroup$ I think completed basis should contain $m$ elements. $\endgroup$ – user371231 Oct 8 '18 at 4:42
  • $\begingroup$ Careful: You have to complete the $e$- and the $f$-bases not randomly but rather using the kernels of the maps. $\endgroup$ – darij grinberg Feb 9 at 23:47
  • $\begingroup$ Just realized it so I will make the edit $\endgroup$ – астон вілла олоф мэллбэрг Feb 10 at 7:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.