Possibility of constructing an orthogonal matrix

Question

I have a $n \times n$ matrix $A$, and for each column $j$ of $A$, $\sum_{i}(A[i,j])^2 = 1$ holds. (i.e. sums of squares of each column adds to $1$.)

I want to build a new $2n \times 2n$ matrix that needs to be orthogonal.

$$ M = \left[\begin{array}{c|c} A & C\\ \hline B & D\end{array}\right] $$

Moreover, $C$ must be a "$0$" matrix (i.e., all entities are $0$).

Size being $2n$ is not strictly necessary. Building a larger $M$ like $(3n+3) \times (3n+3)$ etc. is also acceptable, as long as all entities on the right side of $A$ is zero.

What I am trying to achive is, to show that such an $M$ exists so I can use it in the remaining part of my proof.

I can build $M$'s for specific cases but I stuck at figuring a generic algorithm or proving such an $M$ exists for generic case.

Any help appreciated. Thank you

EDIT: $A$ is not necessarily orthogonal. What I am trying to achieve is to construct an orthogonal matrix only by changing the values of $B$ and $D$.

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Motivation

I am writing a computer science paper. For the paper, I am trying to convert a probabilistic finite automaton (PFA) to a quantum finite automaton (QFA). One of the steps of this process requires that I construct the matrix $M$ with the aforementioned properties.

Answer 1

Let $D=I, C = \mathbb{0}$. Then every column vector has norm $1$. Use dot product of column vectors and $B = C =0$ to show that any two column vectors $v_i, v_j (1\le i\le n, n+1 \le j \le 2n)$ are orthogonal.

Answer 2

Too long for a comment

If $A$ is orthogonal the trivial solution by user sz applies: $D=I,B=C=0$. When $A$ is not orthogonal and you require that $C$ must be zero I have doubts that $D$ exists.

• The $n$ columns of $D$ must be orthogonal (since $C=0$).

• Also: Each of the $n$ columns of $D$ must be orthogonal to every column of $B$. Since the columns of $D$ span $\mathbb R^n$ every column of $B$ must be zero.

This implies that $A$ must be orthgonal but we assumed it is not.

What exactly do you mean by "a larger $M$ like $(3n+3)\times (3n+3)$ is also accepetable" ?