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Let's first focus on $k$-by-$k$ matrices. We know that rank is a continuous function for idempotent matrices, so when we have, say, $\operatorname{rank}(A)>\operatorname{rank}(B)+1$, the two matrices cannot be close in norm topology.

But I wonder whether there is an explicit lower bound of the distance between two idempotent matrices in terms of their difference in their ranks.


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No, the rank isn't continuous, for example for $A = 0$ and $B = \epsilon \cdot \mathrm{Id}$ we have $\|A - B\| = \epsilon$, but $\mathrm{rank}\, A = 0$, $\mathrm{rank}\, B = n$. – martini Jun 20 '12 at 15:41
It's impossible to have a continuous function from a connected space (matrices under norm topology) to the totally disconnected $\mathbb Z$, except for constants. – Erick Wong Jun 20 '12 at 15:44
On the other hand, the rank is lower semi-continuous. – Najib Idrissi Jun 20 '12 at 17:02
Further on @N.I's comment: a matrix $A$ of rank $r$ has an $r \times r$ submatrix $C$ that is invertible. If $\|B - A\| < 1/\|C^{-1}\|$ then the corresponding submatrix of $B$ is invertible, so that $B$ has rank $\ge r$. This is for any (sub-multiplicative) matrix norm for which the standard projections of $k \times k$ matrices to $r \times r$ matrices are contractions. – Robert Israel Jun 20 '12 at 17:33
@martini I have edited the question and now we focus on idempotent matrices. Thanks! – Hui Yu Jun 20 '12 at 17:56
up vote 3 down vote accepted

The rank of an idempotent matrix is its trace, and $|\text{Tr}(A-B)| \le k \|A-B\|$, so if $\|A-B\| < 1/k$ they must have the same rank.

EDIT: Actually if $\|A - B\| < 1$ (where $\|\cdot\|$ is an operator norm) they must have the same rank. Suppose $\text{rank}(A) < \text{rank}(B)$ and $B$ is idempotent. Let $V = \text{Ran}(A)$ and $W = \text{Ran}(B)$. Then the restriction of $A$ to $W$ maps $W$ into $V$. Since $W$ has higher dimension than $V$, this map can't be one-to-one, so there is some nonzero $w \in W$ such that $Aw = 0$. But $Bw = w$, so $(B-A)w = w$ and $\|B-A\| \ge 1$.

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Could you explain why the trace is bounded by the dimension times the norm? Thanks! – Hui Yu Jun 20 '12 at 18:19
I got it. Thanks! – Hui Yu Jun 20 '12 at 18:23

martini's example is the first that come to mind, and it shows that rank is not continuous. One possible criticism is that the norms are small, so maybe there's a "normalized" version that works. But it's not the case: consider $$ A=\begin{bmatrix}1&0\\ 0&0\end{bmatrix}, \ \ \ \ B=\begin{bmatrix}1&0\\ 0&\varepsilon\end{bmatrix} $$

Then $\text{rank}(A)=1$, $\text{rank}(B)=2$, $\|A\|=1$, $\|B\|=1+\varepsilon$, and $\|A-B\|=\varepsilon$.

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Thanks! I changed the problem a little bit. Now we focus on idempotent matrices. – Hui Yu Jun 20 '12 at 17:59

Here is a generalization of a special case, namely the case of self-adjoint idempotents.

Suppose that $p$ and $q$ are self-adjoint idempotents (projections) in a C*-algebra $A$. If $\|p-q\|<1$, then there is a continuous path of projections in $A$ from $p$ to $q$. If $A$ is unital, this implies that $p$ and $q$ are unitarily equivalent. In general, it implies that there is a partial isometry $v\in A$ such that $v^*v=p$ and $vv^*=q$. This is shown in Chapter 2 of Rørdam et al.'s An introduction to K-Theory for C*-algebras

In the case where $A=M_n(\mathbb C)$, this tells us that $\|p-q\|<1$ implies that $\mathrm{rank}(p)=\mathrm{Tr}(v^*v)=\mathrm{Tr}(vv^*)=\mathrm{rank}(q)$, a special case of Robert Israel's answer.

On a related note, it isn't too hard to show that if $p$ and $q$ are arbitrary projections in a C*-algebra, then $\|p-q\|\leq 1$. More generally, if $a\geq 0 $ and $b\geq 0$ in a C*-algebra, then $\|a-b\|\leq \max\{\|a\|,\|b\|\}$. In particular, if $p$ and $q$ are self-adjoint idempotent matrices of different ranks, then $\|p-q\|=1$.

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Although I know little about K-Theory, but your answer is extremely surprising and beautiful. Unfortunately I have picked Robert's answer. But thanks! – Hui Yu Jun 21 '12 at 0:12
@HuiYu: But there isn't any K-theory required, just some facts about projections in C*-algebras that also happen to be useful in K-theory (and of course I have left out the proofs). I think that Robert's answer better answers your question; mine seems like overkill in a way, and it doesn't address the general idempotent case. On the other hand, for future reference, you can change which answer is "accepted" whenever you please. – Jonas Meyer Jun 21 '12 at 1:56
All right. The problem is I do not quite understand how you prove the existence of projections connecting $p$ and $q$. But I guess I will check the book today. Thanks – Hui Yu Jun 21 '12 at 13:10

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