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Let $V,W$ be complex inner product spaces. Suppose $T: V \to W$ is a linear map, then we define $$\|T\|:=\sup\{\|Tv\|_{W}:\|v\|_{V}=1\}$$ where $\|v\_{V}\|:=\sqrt{\langle v,v\rangle}$ and $\|Tv\|_{W}:=\sqrt{\langle Tv,Tv\rangle}$.

Question: Suppose $U_1,\ldots,U_k$ and $V_1,\ldots,V_k$ are $n {\times} n$ unitary matrices. Show that $$\|U_1\cdots U_k-V_1\cdots V_k\| \leq \sum_{i=1}^{k}\|U_i-V_i\|$$

I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!

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1 Answer 1

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For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$: $$ \begin{array}{ll} & ||U_1 U_2 - V_1 V_2||\\ \\ = & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\\ \\ = & || ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \\ \\ \leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\\ \end{array} $$ (The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_{2}|| = 1$, so $$ || ( U_1 - V_1) U_2 || \leq || U_1 - V_1 ||. $$ A similar bound obtains for $||V_1 (U_2 - V_2) ||$.

This should give you enough "building blocks".:)

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  • $\begingroup$ Thank you so much! $\endgroup$
    – bbw
    Commented Mar 15, 2019 at 23:36
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    $\begingroup$ You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:) $\endgroup$
    – avs
    Commented Mar 16, 2019 at 0:05

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