1
$\begingroup$

I'm working on some practice problems, and I'd like some feedback.

  1. We have a n x n square matrix $A$ with rank $r$. There are other matrices $B$ and $C$ and $AB=AC$. I'm asked to find the maximal rank of $B-C$.

So in the case where $B = C$, the max rank would be 0. Now when is $B =/= C$? If A is zero, then AB = AC and r = 0. In that case, C could be 0 and B could be $I_n$, which would give a max rank of n.

I believe that makes sense, yeah?

  • 0
    $\det(AB) = \det(A)\det(B)$. If $A,B$ are invertible then $\det(A),\det(B) \neq 0$. What can you say about $\det(AB)$ then?2011-12-12
  • 0
    You have what looks to be several questions slapped together. You may have more success getting an answer if you stick to one question per post.2011-12-12
  • 0
    Ahh that does it!! Thanks a ton!2011-12-12
  • 0
    @BillCook - There are 4 distinct questions, but I didn't want to make 4 different topics since the subject matter was mostly the same. Would it be better to post a thread for each question?2011-12-12
  • 0
    Up towards the top, you use $A^{-1}$ to conclude $B=C$. I don't think you can assume $A^{-1}$ exists (from your problem statement). $A^{-1}$ will only exist if $r=\mathrm{rank}(A)=n$ ($A$ is full rank). Next, $B$ and $B^2$ can have different ranks [nilpotent matrices give easy examples].2011-12-12
  • 0
    If the null space is 4 dimensional, you are correct, $A$ times anything is zero. Now consider $A$ times $[1,0,0,0]^T$. This peels off the first column of $A$. Thus the first column of $A$ is all zero's. Now repeat using $[0,1,0,0]^T$ etc.2011-12-12
  • 0
    The next two look ok. No the zero matrix is not invertible so you've found the right counterexample.2011-12-12
  • 0
    For the final one. If $AB$ is singular, then either $A$, $B$, or both matrices must be singular (if both were invertible, their product $AB$ would be invertible). Therefore, either the rank of $A$ or $B$ (or both) is less than 4. So their ranks cannot add to 8.2011-12-12
  • 0
    Thanks for your feedback Bill! I took off the other questions so there's only one here now. So if you assume that A is not invertible, then what's a way to "mathematically" solve the problem? I believe my example holds up, but is there a way to algebraically show what the max rank would be?2011-12-12
  • 0
    The question is asking, what if all you know is that $A$ has rank $r$, and you don't know what $r$ is - with that, and $AB=AC$, what is the maximal possible rank of $B-C$?2011-12-12
  • 0
    I don't think it's a good idea to change the question like this. The answers seem to have no connection to the question being asked, now.2011-12-12
  • 0
    @Dylan Moreland - Should I delete this and ask a new question then?2011-12-12
  • 0
    @JohnDoe for future reference: be careful when you edit questions to keep the original question intact. That way comments and answers still make sense after it is changed. At this point, I wouldn't recommend deleting the question but you might want to put the original question back up (above or below your new question). In the future, when you find a question is flawed as asked (this happens fairly often), if it got answered leave it alone. Then post a new (corrected) question separately.2011-12-13

2 Answers 2

1

I'll expand on what I wrote in the comments.

First, as your problem is stated, your solution is correct. For any matrix $A$, $AI_n=AI_n$ and $I_nI_n=I_n$ which has rank $n$. Since the rank of an $n \times n$ matrix cannot exceed $n$. The maximal rank of such a product of matrices, $BC$, is $n$. Rank of $A$ is not really relevant which makes me think you've possibly miss-copied part of this problem.

Your second attempt doesn't work. You cannot use $A^{-1}$ to cancel off $A$ in the equation $AB=AC$ since $A^{-1}$ exists only if you make the additional assumption $\mathrm{rank}(A)=r=n$.

Next, it is not generally the case that $\mathrm{rank}(B)=\mathrm{rank}(B^2)$. Consider for example:

$$ B = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$

Notice that $\mathrm{rank}(B)=1$ whereas $B^2=0$ so $\mathrm{rank}(B^2)=0$.

Edit: For the new problem.

$AB=AC$ is equivalent to $A(B-C)=0$. Thus the rank of $B-C$ is bounded above by the number of linearly independent solutions of $Ax=0$ (this is the nullity of $A$: $n-r$). To the max rank of $B-C$ is less than or equal to $n-r$. To see that we can achieve this bound: Let $B$ be a matrix whose columns form a basis for $N(A)$ plus some zero columns to pad out $B$ to it's square. Then $AB=0$ (since $B$ is made of zero columns and elements of the nullspace of $A$). So, letting $C=0$, we get $AB=0=A0=AC$ where $B-C=B-0=B$ whose rank is the nullity of $A$: $n-r$. Thus the max rank of $B-C$ is $n-r$.

  • 0
    I did follow what you were saying in the comments, but I realized I copied it incorrectly. The question asks for max rank of B-C, as opposed to BC.2011-12-12
  • 0
    Ah, that's quite a different question :)2011-12-12
  • 0
    I updated my question with my reasoning. Does that make sense?2011-12-12
  • 0
    What you've said is more or less true, but it does not help solve the problem. If $B=C$, then, yes, $\mathrm{rank}(B-C)=\mathrm{rank}(0)=0$. But the problem is: "Given $A$, what is the biggest you can make the rank of $B-C$?" So choosing $B=C$ isn't helpful. If $B \not= C$, you cannot conclude that $A=0$. It possible to have $AB=AC$ with $B \not= C$ and $A \not=0$. Singular matrices do not share the cancellation properties that the real numbers enjoy.2011-12-13
0

Hint for new question

If $M$ and $N$ are two square matrices, what can you say about the dimension of the kernel of $MN$, in terms of the dimension of the kernels of $M$ and of $N$?

Then, how is the rank of a matrix related to the dimension of its kernel?

Hints for original questions

Here are some hints.

  1. You can't cancel $A$ unless it has full rank (n). So you will need a different approach to the question. There are several plausible guesses for the maximum rank of $BC$, including $r$, $n-r$, and $n$. Suppose the maximum rank is $

  2. Your idea makes sense. What part are you having trouble with? Showing that if $Ax=0$ for all $x$, then $A=0$? Can you think of a nice $x$ so that if $Ax=0$, the whole first column of $A$ must be zero? What about the other columns?

  3. Your counterexample doesn't work because $A+B = \begin{bmatrix}1 &1\\1 &1\end{bmatrix}$ is not invertible. However, I'm sure you'll be able to modify your counterexample slightly to fix this problem.

  4. Yep, that disproves it.There is no such thing as a matrix being "technically invertible" -- a square matrix is invertible if and only if it has full rank, or equivalently, iff its determinant is nonzero. What is the determinant of the zero matrix? What is its rank?

  • 0
    Thanks for your reply. I did adjust my solutions to 2-4 so they make sense. I've also updated q1, so could you take a look at it? I'll write down what I came up with as well.2011-12-12