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Let $X$ be a normed vector space. Let $T,S$ be bounded linear operators such that $T^2=T,S^2=S,ST=TS$. Show either $T=S$ or $\|T-S\|\geq 1$.

My observations:

$1\leq\|S\|,1\leq \|T\|$

$\|T-S\|=\|T^2-S^2\|\leq\|T-S\|\|T+S\|$ Thus $1\leq \|T+S\|$

I am not sure how to proceed from here.

2132123
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2 Answers2

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We consider two cases:

  1. $\ker T \neq \ker S$. Then WLOG we may assume that there exists $x \in \ker S \setminus \ker T$. Then $Tx \neq 0$ while $Sx = 0$. So $$ (T-S)(Tx) = T^2x - STx = Tx - TSx = Tx, $$ and so, $ \| T - S \| \geq 1$.

  2. $\ker T = \ker S$. We note that $\operatorname{im}(1-T) \subseteq \ker T = \ker S$. So, for any $x$, $$ Sx = S((1-T)x + Tx) = S\underbrace{(1-T)x}_{\in \ker S} + STx = STx. $$ By the same argument, we have $Tx = TSx = STx$ for all $x$. Therefore $T = S$.

Sangchul Lee
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Both of your matrices are projection matrices (which can always be diagonalized), and given that they commute it follows they can be simultaneously diagonalized, i.e. $$S = Q D_S Q^{-1}, ~ T = Q D_T Q^{-1}$$ for some invertible $Q$ and diagonal matrices $D_S$ and $D_T$. Since $S$ and $T$ are both projections, the diagonal entries of $D_S$ and $D_T$ are either zero or one. It thus follows that unless $S = T$, $S - T$ has an eigenvalue of absolute value at least $1$, since $$T - S = Q E Q^{-1} = Q (D_T - D_S) Q^{-1}$$ since the diagonal entries of $E$ (which are the eigenvalues of $S - T$) can only be $0$ (if the corresponding entries in $D_S$ and $D_T$ were the same, or either $-1$ or $1$ (if they were different).

Now we know that the operator (spectral) norm is the largest singular value, which must be greater than the magnitude of all of the eigenvalues. This is straightforward, as $\|A \vec{v} \| = |\lambda|$ for any unit eigenvector $\vec{v}$ of eigenvalue $\lambda$. Hence, $$\|T - S\| = \max_{\|\vec{v}\| = 1} \|(T - S) \vec{v} \| \geq 1$$ unless $S = T$. $\square$

paulinho
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