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I've got a very interesting problem and wondering if my idea for the solution is right and complete. Please, take a look.

Problem

Given two matrices $A$ and $B$ such that $A^2=A$ and $B^2=B$. Prove that $\operatorname{rank}A=\operatorname{rank}B$ if the matrix $I - (A+B)$ is invertible.

Solution

$A^2=A$

$B^2=B$

Let's subtract the second equation from the first one.

$A^2 - B^2 = A - B$

$(A - B)(A+B) = A - B$

$(A - B)(A+B) - (A-B)= 0$

$(A - B)((A+B) - I))= 0$

$(A - B)(I - (A+B))= 0$

Since $(I - (A+B))$ is invertible, let's multiply the above equation by $(I - (A+B))^{-1}$ from the right.

$(A - B)(I - (A+B))(I - (A+B))^{-1}= 0$

$A - B = 0$

Hence, $A = B$ which means that $rankA = rankB$.

Bernard
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Roman Dryndik
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2 Answers2

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As the comments explain, your current approach is wrong.

For one correct approach, try a proof by contrapositive. Suppose $A$ and $B$ do not have the same rank. It follows that we have either $$ \dim \ker (A) + \dim \operatorname{im}(B) > n $$ or $$ \dim \ker (B) + \dim \operatorname{im}(A) > n $$ where $\ker(A)$ denotes the kernel (nullspace) of $A$ and $\operatorname{im}(A)$ denotes the image (column space) of $A$.

If the first condition holds, then $\ker(A) \cap \operatorname{im}(B)$ contains a non-zero vector $x$. This vector $x$ satisfies $Ax = 0$ and $Bx = x$. It follows that $$ (I - (A + B))x = x - Ax - Bx = x - 0 - x = 0. $$ That is, $I - (A + B)$ has a non-trivial kernel. Thus, $I - (A + B)$ is not invertible.

If the second condition holds, we similarly conclude that $I - (A + B)$ has a non-trivial kernel and so cannot be invertible.

So, we have proven the desired result: if $I - (A + B)$ is invertible, then $A$ and $B$ must have the same rank.

A slicker approach is as follows: note that $$ (I - (A + B))A = A - A^2 - BA = -BA\\ B(I - (A + B)) = B - BA - B^2 = -BA $$ That is, we have $$ (I - (A + B))A = B(I - (A + B)). $$ If $(I - (A + B))$ is invertible, then multiply both sides by $(I - (A + B))$ from the left to conclude that $$ A = (I - (A + B))^{-1}B(I - (A + B)). $$ Since $A$ and $B$ are similar, we conclude that they have the same rank.

Ben Grossmann
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  • Thanks, @Omnomnomnom for the answer. I've got 2 questions related to both solutions. 1) If $x$ is a non-zero vector from $im(B)$ why it satisfies $Bx = x$?; as far as I understand mapping $B$ is $B: X \rightarrow Y$ where $X = Y$ is not necessary true? 2) Didn't get why you can conclude $A = B$ from the fact that $A = (I - (A + B) )^{-1} B (I - (A + B))$. – Roman Dryndik Jul 13 '19 at 14:29
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    For 1: if $x$ is in the image of $B$, then $x=By$ for some vector $y$. It follows that $$Bx= B^2y = By = x$$ – Ben Grossmann Jul 13 '19 at 15:20
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    If $B$ is a mapping $B:X \to Y$ where $X$ and $Y$ are distinct, then $B^2$ no longer makes sense. – Ben Grossmann Jul 13 '19 at 15:22
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    For 2: I don’t conclude that $A=B$, I conclude that $A$ and $B$ have the same rank because they are similar matrices. – Ben Grossmann Jul 13 '19 at 15:26
  • Ok, now it's pretty clear. Thanks, @Omnomnomnom. – Roman Dryndik Jul 13 '19 at 15:29
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Recall that if $P$ is a projection matrix (matrix such that $P^2 = P$), it holds $\operatorname{rank} P = \operatorname{Tr} P$.

Notice that $A$ is a projection so $I-A$ is also a projection. It follows $$\operatorname{rank}(I-A) = \operatorname{Tr}(I-A) = n-\operatorname{Tr} A = n-\operatorname{rank} A$$

Now since rank is subadditive, we have

$$n = \operatorname{rank}(I-(A+B)) \le \operatorname{rank}(I-A) + \operatorname{rank}(-B) = n-\operatorname{rank} A + \operatorname{rank}B$$ so $\operatorname{rank} A \le \operatorname{rank}B$.

By symmetry it follows $\operatorname{rank} B \le \operatorname{rank}A$ so we have equality.

mechanodroid
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