# Is the adjoint of an isometry an isometry?

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?

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 $$\ell^2$$.
• 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 – user438666 Feb 4 at 13:45
• @user438666 Yes, and you can say a bit more: $VV^*$ will always be the orthogonal projection onto the image of $V$ in $H_B$. – Omnomnomnom Feb 4 at 14:10