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I have the following product matrix $XYZ$, with $X,Y,Z$ all $n\times n$ matrices. $X$ and $Z$ are unitary matrices, i.e., they are norm preserving: for every vector $v$, we have $\|Xv \| = \|v\|$.

I am trying to prove that $\| XYZ \| = \|Y\|$, with as norm chosen the operator norm.

Now I am sure that I can say $\|XYZ v \| = \|YZ v\|$, as the matrix $X$ is norm preserving for every vector (including the vector $YZ v$). But I am not sure if I can say that $\|XYZ v\| = \|XY v\|$. If the latter is also the case, I'm done.

Rodrigo de Azevedo
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Sven
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4 Answers4

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Just notice the following $$ \Vert XYZ \Vert = \sup_{\Vert x \Vert = 1} \Vert XYZ x\Vert = \sup_{\Vert x \Vert = 1} \Vert YZ x\Vert = \sup_{z = Zx, \Vert z \Vert = 1} \Vert Y z\Vert = \sup_{\Vert x \Vert = 1} \Vert Y x\Vert = \Vert Y \Vert.$$ The second equality is due to your argument above and the third due to the fact that the unit sphere is invariant under multiplication with $Z$.

Yaddle
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Another way to see it:

For any linear map $A$ on an inner product space we have that $\|A\|= \sup_{\|x\|= \|y\| = 1}\langle Ax,y\rangle$.

In our case:

$$\|XYZ\|= \sup_{\|x\|= \|y\| = 1}\langle (XYZ)x,y\rangle = \sup_{\|x\|= \|y\| = 1}\langle XY(Zx),X(X^*y)\rangle $$

$X$ and $Z$ are unitary and hence bijective so we can substitute $u = Zx$ and $X^{*}y$. Also, norm preserving gives $\|u\| = \|Zx\| = \|x\|$ and $\|v\| = \|X^*y\| = \|y\|$. Therefore:

$$\|XYZ\| = \sup_{\|u\|= \|v\| = 1}\langle Yu, v\rangle = \|Y\|$$

mechanodroid
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The equality $\Vert XYZv \Vert = \Vert XY v\Vert$ is surely not valid. Take for example in dimension $2$ with canonical basis $(e_1,e_2)$, $X=Id$, $Z$ the rotation of angle $\pi/2$ and $Y$ defined by $Ye_1=0$ and $Ye_2=e_1$ . Then $XYZe_1=e_1$ while $XYZe_2=0$.

mathcounterexamples.net
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The adjoint of an operator on a Hilbert space has the same norm as the original operator, so using unitarity of $X$ and $Z^*$, we have $\|XYZ\| = \|YZ\|=\|Z^*Y^*\|=\|Y^*\|=\|Y\|$.

shalop
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