Unitary Matrices and the Hermitian Adjoint

Question

I saw in a definition for unitary matrices, that for a complex matrix being unitary if $M: \mathbb{C}^{n} \rightarrow \mathbb{C}^{n}$ is unitary, or:

$\langle Mv, Mw \rangle = \langle v,w \rangle$ $\forall v,w \in \mathbb{C}^{n}$

Then, an equivalent definition was that $M$ is unitary if and only if $MM^{*}=\mathrm{Id}$. The proof I saw went as follows (can take the standard basis since the inner product is linear):

$\langle Me_{i}, Me_{j} \rangle = \langle e_{i},e_{j} \rangle = \delta_{ij}$

Since $Me_{i}$ is the $i$-th column of $M$, it follows $\langle Me_{i}, Me_{j} \rangle = \langle M^{*}Me_{i}, e_{j} \rangle$ is the $ij$-th entry of $M^{*}M$. However, the point I don't understand is why would this inner product give us such $ij$-th entry of the matrix. Are we assuming that this inner product is the standard inner product on $\mathbb{C}^{n}$? Or what would be the more precise definition of an unitary matrices that justifies this step?

Thanks for the help.

Answer 1

If I get your question correctly your basic doubt arises from converting a linear operator given in dirac notation to its matrix notation with respect to some basis. Let $A$ be a linear operator $A:V \to W$ and let the orthonormal basis for hilbert spaces $V$ and $W$ be respectively $\{v_1,v_2..v_m\}$ and $\{w_1,w_2..w_n\}$ respectively then the operator can be defined as $$A|v_i\rangle = \sum_i A_{ij}|w_i\rangle......(1)$$ here $A_{ij}$ are the entries of matrix representation of $A$ in input and output basis $\{v\}$ and $\{w\}$ respectively. Why is it so ? you can have a look for detailed explanation here Matrix Representation for Linear Operators in Some basis or prove it yourself.

Now according to the completeness relation if I have a hilbert space $V$ with an orthonormal basis $\{i\}$ then $\sum |i \rangle \langle i|=I_v$ ( identity operator for hilbert space $V$ ). So using the definition of linear operator and completeness relation you can write $$A=\sum_{ij} \langle w_j|A|v_i\rangle |w_j\rangle \langle v_i|......(2)$$ Finally by comparing ( comparing $(1)$ and $(2)$ ) it with previous notation it is easy to see that $A_{ji}=\langle w_j|A|v_i\rangle$. Coming back to your example if we replace $w_j$ by $e_i$ and $v_i$ by $e_j$ we get $A_{ij}=\langle e_i|A|e_j\rangle$ ( and in your case $A=MM^*$ ). I hope I answered your question.

Answer 2

The inner product you're considering is defined by $$ \langle v,w\rangle=v^*w $$ (or $w^*v$, but it's immaterial, do the necessary changes if this is the case).

Suppose $\langle Mv,Mw\rangle=\langle v,w\rangle$ for every $v,w$. This means $$ (Mv)^*(Mw)=v^*w $$ or $$ v^*(M^*Mw)=v^*w $$ so $$ v^*(M^*Mw-w)=0 $$ Since this holds for every $v$, we have that $M^*Mw-w=0$ for every $w$ and this is the same as $(M^*M-I)w=0$, so $M^*M-I$ is the zero matrix.

Conversely, if $M^*M=I$, we clearly have $$ \langle Mv,Mw\rangle=(Mv)^*(Mw)=v^*(M^*M)w=v^*w=\langle v,w\rangle $$

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Whenever you do $\langle Ae_i,e_j\rangle$ where $A$ is a Hermitian matrix, you're doing $e_iAe_j$: now $Ae_j$ is the $j$-th column of $A$, and multiplying by $e_i$ produces the coefficient in the $i$-th row. Hence we get the $(i,j)$ coefficient of $A$.

Finally, note that $M^*M$ is Hermitian.