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Consider a linear operator $V$ between two Hilbert spaces $H_A$ and $H_B$, we say that $V:H_A\rightarrow H_B$ is an isometry if $$ V^*V=\mathbb{1}_A$$

if $V$ is an isometry, does it always hold that $VV^*=\mathbb{1}_B$? In other words, is the adjoint on an isometry an isometry? If not, does it in finite dimension? If not even in finite dimension, does anyone have a finite dimensional counterexample?

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It will hold in finite dimensional spaces if $\dim(H_A) = \dim(H_B)$. It will not, however, hold more generally.

If we take $H_A = \Bbb C^2$ and $H_B = \Bbb C^3$, then the map $$ V(x,y) = (x,y,0) $$ is an isometry. Its adjoint $$ V^*(x,y,z) = (x,y) $$ is not an isometry.

For an infinite dimensional example with $H_A = H_B$, we can take the right shift operator over $\ell^2$.

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  • $\begingroup$ Thank you, math doesn't forgive I guess, does it hold in finite dimension that the restriction of $VV^*$ to a space of the same dimension of $H_A$ is the identity? It holds for the example you provided $\endgroup$ – user438666 Feb 4 at 13:45
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    $\begingroup$ @user438666 Yes, and you can say a bit more: $VV^*$ will always be the orthogonal projection onto the image of $V$ in $H_B$. $\endgroup$ – Omnomnomnom Feb 4 at 14:10

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