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.

  • 1
    $\begingroup$ Yes we are assuming the inner product is the standard inner-product on $\mathbb{C}^{n}$. The first definition defines unitary matrices with respect to a given inner product. So it is actually independent of the matrix representation. The second one however is a definition in terms of matrices and the standard inner product. The reason they are equivalent is because any (nondegenrate) inner product on $\mathbb{C}^{n}$ is given by a positive definite Hermitian matrix which by spectral theorem is a product of two unitary matrices so you are not losing any information by choosing a inner product. $\endgroup$ – DBS Apr 16 '15 at 22:34

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 $$ \langle Mv,Mw\rangle=(Mv)^*(Mw)=v^*(M^*M)w=v^*w=\langle v,w\rangle $$

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.