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Show that every matrix $A \in M_{n,k}(\mathbb{C})$ whose columns are orthonormal vectors in $M_{n1}(\mathbb{C})$ can be supplemented with additional n-k columns to an unitary matrix $U \in M_{n}(\mathbb{C})$

$A = \begin{bmatrix} a_{11} & a_{12} & ... & a_{1k} \\ a_{21} & a_{22} & ... & a_{2k} \\ ... & ... & ... & ... \\ a_{n1} & _{n2} & ... & a_{nk} \end{bmatrix}$

Now, I can mark the columns with $c[i]$ for example where
$c[i] = \begin{bmatrix} a_{1i} \\ a_{2i} \\ ... \\ a_{ni} \end{bmatrix} ; i \in {1, ..., k}$

We know that $|| c[i] || = \langle c[i], c[i] \rangle = 1$ and $\langle c[i], c[j] \rangle = 0 \space \forall i \ne j$

Since there is an bijection between the spaces $M_{n,1}(\mathbb{C})$ and $C^n$ we can view the set $\{ c[1], c[2], ... , c[k] \}$ as an orthonormal set in the space $C^n$ and by applying the Gram-Schmidt process we can get an orthonormal base for $C^n ... \{ c[1], c[2], ... , c[k], v[k+1], v[k+2], ... , v[n] \}$

And by supplementing the vectors $v[k+1], ..., v[n]$ as columns to the original matrix $A$ we get:

$A' = \begin{bmatrix} a_{11} & a_{12} & ... & a_{1k} & v_{1,k+1} & ... & v_{1n} \\ a_{21} & a_{22} & ... & a_{2k} & v_{2,k+1} & ... & v_{2n} \\ a_{31} & a_{32} & ... & a_{3k} & v_{3,k+1} & ... & v_{3n} \\ a_{41} & a_{42} & ... & a_{4k} & v_{4,k+1} & ... & v_{4n} \\ ... & ... & ... & ... & ... & ... & ... \\ a_{n1} & _{n2} & ... & a_{nk} & v_{n,k+1} & ... & v_{nn} \end{bmatrix}$

$(A')^* = \begin{bmatrix} \overline{a_{11}} & \overline{a_{21}} & ... & \overline{a_{k1}} & \overline{v_{k+1,1}} & ... & \overline{v_{n1}} \\ \overline{a_{12}} & \overline{a_{22}} & ... & \overline{a_{k2}} & \overline{v_{k+1,2}} & ... & \overline{v_{n2}} \\ \overline{a_{13}} & \overline{a_{23}} & ... & \overline{a_{k3}} & \overline{v_{k+1,3}} & ... & \overline{v_{n3}} \\ \overline{a_{14}} & \overline{a_{24}} & ... & \overline{a_{k4}} & \overline{v_{k+1,4}} & ... & \overline{v_{n4}} \\ ... & ... & ... & ... & ... & ... & ... \\ \overline{a_{1n}} & \overline{a_{2n}} & ... & \overline{a_{kn}} & \overline{v_{k+1,n}} & ... & \overline{v_{nn}} \end{bmatrix}$

And by multiplying the matrices, we notice that we can write the elements of the matrix as:

$(A')^*A' = \begin{bmatrix} \sum_{i=1}^na_{i1}\overline{a_{i1}} & \sum_{i=1}^na_{i1}\overline{a_{i2}} & ... & \sum_{i=1}^na_{i1}\overline{v_{in}} \\ \sum_{i=1}^na_{i2}\overline{a_{i1}} & \sum_{i=1}^na_{i2}\overline{a_{i2}} & ... & \sum_{i=1}^na_{i2}\overline{v_{in}} \\ ... & ... & ... & ... \\ \sum_{i=1}^na_{in}\overline{a_{i1}} & \sum_{i=2}^na_{in}\overline{a_{i2}} & ... & \sum_{i=1}^na_{in}\overline{v_{in}} \end{bmatrix}$

The standard inner product for $\mathbb{C}^n$ is $\sum_{i=1}^n \alpha_i \overline{\alpha_i}$ and since the elements of the matrix are summations which consist of multiplications of elements of the corresponding columns which form vectors which are part of an orthonormal base of $C^n$, we can apply the standard inner product for the summations/elements of the matrix $\implies (A')^*A' = I $

But the following multiplication remains:

$A'(A')^* = \begin{bmatrix} \sum_{i=1}^na_{1i}\overline{a_{1i}} & \sum_{i=1}^na_{1i}\overline{a_{2i}} & ... & \sum_{i=1}^na_{1i}\overline{v_{ni}} \\ \sum_{i=1}^na_{2i}\overline{a_{1i}} & \sum_{i=2}^na_{2i}\overline{a_{2i}} & ... & \sum_{i=1}^na_{2i}\overline{v_{ni}} \\ ... & ... & ... & ... \\ \sum_{i=1}^na_{ni}\overline{a_{1i}} & \sum_{i=2}^na_{ni}\overline{a_{2i}} & ... & \sum_{i=1}^na_{ni}\overline{v_{ni}} \end{bmatrix}$

The resulting matrix consists of summations of multiplications of the corresponding elements of the rows of the original matrix. But since we are dealing with a matrix, we know that the number of linear independent columns equals the number of linear independent rows and conclude that the rows of the matrix also form vectors which on the other hand form a basis for $C^n \implies A'(A')^* = I$

Since $A'(A')^* = (A')^*A' = I$ we can take $ U = A' $. The statement follows.

Is my reasoning correct ? If yes, do I need to add some details/be more formal in some places? If no, it would be great if you could give me some hints.

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Seems about right to me. The point is that a matrix is unitary if and only if its columns form an orthonormal basis with respect to the complex Euclidean inner product. In fact, if $U = [v_1 | \cdots | v_n]$ then $(UU^\dagger)_{ij} = \langle v_i, v_j \rangle$ (or $U^\dagger U$ depending on your convention). Thus your question really boils down to: can any orthonormal set be extended to an orthonormal basis (in finite dimensions)? Yes, by Gram-Schmidt, as you indicated.

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