Existence of unitary matrix $U$ such that $UEU^* =F$ after knowing $E$ and $F$ have same eigenvalues

Question

Let $M_n(\mathbb C)$ be the algebra of all $n \times n$ matrices of complex entries. Suppose $E, F \in M_n(\mathbb C)$ are projections, i.e., $E,F$ are selfadjoint matrices so that $E = E^2$ and $F = F^2$. Assume that $||E - F || \le \frac{1}{2}$, where $|| · ||$ is the operator norm. Show that there is a unitary matrix $U$ such that $UEU^* =F.$ $$$$ Selfadjoint matrices can only have eigenvalue $0$ or $1$. I think the operator norm condition can infer that $E$ and $F$ have the same set (counting numbers) of eigenvalues by using $\rho(E-F)\le||E - F || \le \frac{1}{2}$. However, is proving the eigenvalue set is sufficient for this problem? I'm unsure about this, and I'm asking for confirmations here.

Answer 1

Let $r=\operatorname{rank}(E)$ and $s=\operatorname{rank}(F)$. If $r>s$, then $\operatorname{rank}(E)+\operatorname{nullity}(F)=r+n-s>n$. Hence there is some nonzero vector $x\in\operatorname{range}(E)\cap\ker(F)$. But then $\|(E-F)x\|=\|Ex\|=\|x\|$, which is a contradiction to the assumption that $\|E-F\|<1$. A similar contradiction arises if $r<s$. Thus $E$ and $F$ must have the same rank $r$. It follows that both matrices are unitarily similar to $I_r\oplus0$. Now the result follows.

Answer 2

Since $E$ is a projection, we have $V=\text{Im}E\bigoplus\text{Ker}E$. Let $v_1\in\text{Im}E$ and $v_2\in\text{Ker}E$. We have $E(v_1)=v_1$ and $E(v_2)=0$.

We also know that $E$ is self-adjoint. This means that $\langle v_1,v_2\rangle=\langle Ev_1,v_2\rangle=\langle v_1,Ev_2\rangle=\langle v_1,0\rangle=0$. In other words, $\text{Im}E$ is orthogonal to $\text{Ker}E$.

Therefore we can write an orthonormal basis for $V$ as $w_1,\ldots,w_m,u_1,\ldots,u_{n-m}$, where $m=\text{rank}E$.

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A similar procedure can be done, of course, for $F$. Let the orthonormal basis made after $F$ be $w'_1,\ldots,w'_{m'},u'_1,\ldots,u'_{n-m'}$.

Now, if $m=m'$, then by switching the basis from $u$ and $w$ to $u'$ and $w'$, you get a unitary matrix $U$ such that $UEU^* =F$.

The only question is to prove that the condition $||E - F || \le \frac{1}{2}$ implies $m=m'$.

To prove this by contradiction, suppose without loss of generality that $m>m'$. This would mean that $m+(n-m')>n$. Therefore, $\text{Im}E\bigcap\text{Ker}F\neq\{0\}$. Let $j\in\text{Im}E\bigcap\text{Ker}F$ with $j\neq 0$. Then $(E-F)j=Ej-Fj=j-0=j$. This would imply that $||E - F || \ge 1$: contradiction.