I have a question about projection in matrix algebras over the complex number, that I can not solve..

Let p be a matrix in some $M_n(\mathbb C)$ and suppose that p is almost self-adjoint (i.e. $||p-p^*||<\epsilon$) and almost idempotent (i.e. $||p-p^2||<\epsilon$). Is there a bound on the distance between $p$ and a projection? In particular is there $k\in\mathbb N$ and $q=q^*=q^2\in M_n(\mathbb C)$ such that $||q-p||<k\epsilon$ where $k$ does NOT depend on $\epsilon$?

Thanks a lot!


2 Answers 2


As all norms on $M_n(\mathbb C)$ are equivalent, we can use any norm we please. I will use the norm $\|a\|=\sum_{i,j}|a_{ij}|$.

Let $p_1$ be the selfadjoint matrix given by $$ (p_1)_{ij}=\begin{cases}p_{ij},&\mbox{ if }i< j\\ \\ \mbox{Re}\,p_{jj},&\,\mbox{ if }i=j \\ \\ \overline{p_{ji}},&\mbox{ if }i>j\end{cases} $$ Note that $p_1$ is selfadjoint, $\|p_1-p\|\leq\|p^*-p\|<\varepsilon$, and $$\|p_1^2-p_1\|\leq\|p_1^2-p_1p\|+\|p_1p-p^2\|+\|p^2-p\|+\|p-p_1\|\leq(\|p_1\|+\|p\|+2)\,\varepsilon=(2+\|p\|)\,\varepsilon.$$

Also, from $\|p-p^2\|<\varepsilon$ we get $$ \|p\|^2\leq\|p\|+\varepsilon, $$ which shows that $\|p\|\leq1/2+\sqrt{\varepsilon+1/4}\leq 1+\sqrt\varepsilon$. So our first estimate becomes $$ \|p_1^2-p_1\|\leq(3+\sqrt\varepsilon)\,\varepsilon. $$

Since $p_1$ is selfadjoint, we can diagonalize it: $p_1=vdv^*$, with $v$ unitary and $d$ diagonal with real entries. Since $v$ is unitary, $\|v\|\leq n^2$ (just using that all entries in $v$ are of absolute value at most one) and so $$ \|d^2-d\|=\|v^*(p_1^2-p_1)v\|\leq n^4\|p_1^2-p_1\|\leq n^4(3+\sqrt\varepsilon)\,\varepsilon. $$ This shows that each eigenvalue of $p_1$ is either near $0$ or near $1$ when $\varepsilon$ is small. Concretely, $$ \frac{1-\sqrt{1+4n^4(3+\sqrt\varepsilon)\,\varepsilon}}2\leq|d_{jj}|\leq\frac{1-\sqrt{1-4n^4(3+\sqrt\varepsilon)\,\varepsilon}}2 $$ or $$ \frac{1-\sqrt{1+4n^4(3+\sqrt\varepsilon)\,\varepsilon}}2\leq|1-d_{jj}|\leq\frac{1-\sqrt{1-4n^4(3+\sqrt\varepsilon)\,\varepsilon}}2. $$ Using that if $a<|d|<b$ for $b>0$, $a<0$ then $|d|<b-a$, we get $$ |d_{jj}|\leq8n^4(3+\sqrt\varepsilon)\,\varepsilon,\ \mbox{ or }|1-d_{jj}|\leq8n^4(3+\sqrt\varepsilon)\,\varepsilon $$ (using that $\sqrt{1+a}-\sqrt{1-a}\leq2a$ for $a>0$).

So let $e$ be a diagonal matrix with $e_{jj}=0$ if $|d_{jj}|\leq1/2$, $e_{jj}=1$ if $|d_{jj}|>1/2$. Let $q=vev^*$. Then $q$ is a selfadjoint projection and $$ \|q-p\|\leq\|q-p_1\|+\|p_1-p\|\leq n^4\|e-d\|+\varepsilon\leq n^4\,n8n^4(3+\sqrt\varepsilon)\,\varepsilon=8n^6(3+\sqrt\varepsilon)\,\varepsilon. $$

Finally, to answer the question: if a bound is prescribed on $\varepsilon$ (which is basically the same as prescribing a bound on the norm of $p$), the required $k$ can be found. For example, if $\varepsilon\leq1$, then we can take $k=8n^6\times4=32n^6$.

  • $\begingroup$ Everything is correct, but $p_1$ could be not self-adjoint (maybe $p_{i,i}$ it is not a real number, could be $1+\frac{\epsilon}{100000}i$ for what we know). $\endgroup$ Sep 27, 2013 at 7:50
  • $\begingroup$ Good point! I have corrected the definition of $p_1$. $\endgroup$ Sep 27, 2013 at 10:03

there is a simple trick: let $x=2p-1$, so that $x^2$ is almost $1$. More precisely, $x^2=1+4(p^2-p)$, i.e. $||x^2-1||<4\epsilon$. We can thus compute $(x^2)^{-1/2}$ using the Taylor (i.e. binomial) series for $(1+t)^{-1/2}$ (we plug in $t=4(p^2-p)$) provided $4\epsilon<1$. For $\epsilon$ small enough, $||(x^2)^{-1/2}-1||<2^+\epsilon$, where $2^+$ is any number $>2$. Let now $y=x(x^2)^{-1/2}$, so that $y^2=1$. This implies that $q$ defined by $y=2q-1$ satisfies $q^2=q$. Notice that $q-p=(x-y)/2=y((x^2)^{1/2}-1)/2$.

To make $q$ self-adjoint, just replace $p$ with $p'=(p+p^*)/2=p+(p^*-p)/2$, which is self-adjoint and close to the original $p$. This will make $x'$, $y'$, $q'$ also self-adjoint. To get an estimate on $||q'-p||$ notice that: $||p||^2<||p||+\epsilon$, i.e. $||p||<1+\epsilon$, which implies $||p'^2-p'||<(3/2)^+\epsilon$, hence $||(x^2)^{-1/2}-1||<3^+\epsilon$. As $y'^2=1$ and $y'$ is self-adjoint, $||y'||=1$, so finally $||q'-p||<(3/2)^+\epsilon$. Notice that this estimate doesn't depend on $n$ and works in infinite dimension.

  • $\begingroup$ Nice trick! Were did you learned it? $\endgroup$
    – Norbert
    Sep 26, 2013 at 22:31
  • $\begingroup$ I don't see how you guarantee that your $q$ is selfadjoint. $\endgroup$ Sep 26, 2013 at 22:37
  • $\begingroup$ I do not see how to guarantee that $x$ is a positive matrix (in terms of element of a C*-algebra), hence I do not see why you are allowed to talk about $(x^2)^{-1/2}$ (that is, basically, $x$, if $x$ is positive, or $-x$) since $x$ is not self-adjoint $\endgroup$ Sep 27, 2013 at 7:53
  • $\begingroup$ @JohnSmith I edited my answer to make it clearer. $\endgroup$
    – user8268
    Sep 27, 2013 at 9:42
  • $\begingroup$ @Norbert I saw it in Fedosov's book on deformation quantization $\endgroup$
    – user8268
    Sep 27, 2013 at 9:43

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