unitary operator explanation

Question

the unitary operator definition is $ ⟨Tv,Tw⟩ = ⟨v, w⟩$ for every $v, w$ in $V.$
can you please explain the intuition and what the formal definition actually means? why unitary operator preserves the orthonormal basis? Can we infer from having unitary operator that we have eigenvectors? And why when $T^* = T^{-1}$, $T$ preserves the inner product and therefore preserves the the orthonormal basis and the length and distance? thank you

Answer 1

First of all, recall that the inner product gives us a generalization of the "dot-product" in $\Bbb R^n$. Just as with the dot-product, we can interpret the inner product as giving us the information about the "angle" between two vectors. In particular, because of the Cauchy-Schwarz inequality $|\langle u,v \rangle| \leq \|u\| \cdot \|v\|$, we can always find an angle $\theta$ for which $$ \langle u,v \rangle = \cos \theta \cdot \|u\|\,\|v\| \implies \theta = \cos^{-1} \left( \frac{\langle u,v \rangle}{\|u\| \, \|v\|}\right), $$ and we can think of this $\theta$ as the angle between the vectors $u$ and $v$. In particular: if $u,v$ are unit vectors (if $\|u\| = \|v\| = 1$), then $\langle u, v \rangle = \cos \theta$.

Note that $T$ is unitary if and only if for every $u,v \in V$, we have $\langle Tu,Tv \rangle = \langle u,v \rangle$ We make the following observations:

• We see that $\|Tu\|^2 = \langle Tu,Tu \rangle = \langle u,u \rangle = \|u\|^2$. In other words, $Tu$ has the same length as $u$. That is, $T$ preserves the length of its input.
• For every pair of unit-vectors $u,v,$ we have $\langle Tu, Tv \rangle = \langle u, v \rangle$; note that $Tu$ and $Tv$ are also unit vectors. So, we see that $T$ preserves the angle between two inputs.

With this established, it's clear why $T$ should preserve orthonormal bases: if each vector $u_1,\dots,u_n$ has length $1$, then the same applies for $Tu_1,\dots,Tu_n$. Similarly, if the angle between $u_1,u_2$ is $90^\circ$, then the same is true for $Tu_1,Tu_2$. Putting that together: if $u_1,\dots,u_n$ are pairwise-orthogonal unit vectors, then the same is true for $Tu_1,\dots,Tu_n$.

Geometrically, it turns out that every unitary transformation consists of a combination of "reflections" and "rotations". You should verify that rotations and reflections indeed satisfy the two properties listed above, i.e. they preserve both length and the angle between vectors.

Regarding eigenvalues: note that if $x \neq 0$ is an eigenvalue of $T$ with $Tx = \lambda x$, then it would necessarily follow that $$ \|x\|^2 = \langle x,x \rangle = \langle Tx,Tx \rangle = \langle \lambda x, \lambda x \rangle = \lambda \bar \lambda \langle x, x \rangle = |\lambda|^2 \|x\|^2. $$ In other words, the eigenvalue $\lambda$ must satisfy $|\lambda| = 1$. If $\lambda$ is real, then either we have $\lambda = 1$ (so that $x$ is a fixed point of $T$) or $\lambda = -1$ (so that $T$ reflects $x$ across the origin). Complex eigenvalues of magnitude $1$ correspond to "rotation" in a sense.

For the finite dimensional case, the spectral theorem tells us that every transformation $T$ can be diagonalized, so that we can describe the transformation completely in terms of its eigenvectors.

Regarding the fact that $T^* = T^{-1}$: it is difficult to get a generalized, geometric, intuitive sense for the meaning of the relationship between $T^*$ and $T$. I think the easiest way to think about this fact is via the proof. Note that by definition, we have $\langle x,Ty \rangle = \langle T^*x, y \rangle$. It follows that for every vector $u,v$, $$ \langle u, v \rangle = \langle Tu,Tv \rangle = \langle T^*Tu, v \rangle. $$ In other words, $\langle u, v \rangle$ is always the same as $\langle T^*Tu, v \rangle.$ This can only happen, however, if we always have $T^*Tu = u$. That is, $T^*T = I$.

If $V$ is finite dimensional, this is enough to conclude that $T^* = T^{-1}.$