Extending a square matrix to a unitary matrix

5

Suppose we have a square matrix $M$ of size $n\times n$. It is given that any element $M_{ij}$ of $M$ is a real number and satisfies $0 \leq M_{ij} \leq 1$, $\forall$ $i,j$. No other property for $M$ is known. Is it possible to create a new matrix $U$, $s.t.$:

  1. $U$ is a square matrix of size $2n\times 2n$,
  2. $U$ is of the form $\begin{bmatrix}M&A\\B&C\end{bmatrix}$,
  3. $A,B,C$ are all of size $n\times n$ and all of $A,B,C$ are unique linear transformations of $M$,
  4. The elements of $A,B,C$ can take complex values,
  5. And that $U$ is unitary, $i.e.$, $UU^\dagger = U^\dagger U = I$, (where $I$ is the identitiy matrix, and $U^\dagger$ is symbol for complex conjugate of $U$)?

Thank you

new2quantum

Posted 2019-01-10T10:26:05.743

Reputation: 51

To clarify for 3, do you mean along the lines of "there exist P and Q such that PMQ=A"? Similarly for B,C? – AHusain – 2019-01-10T10:34:06.937

actually more relaxed. what i mean is that $A,B,C$ may be obtained by some transformation (perhaps not linear, not sure) of $M$. for instance $A,B,C$ may be $-M$. What would also work is that $A,B,C$ may be for instance $\frac{1}{\sqrt{n}}$ times the Identity and so on. Basically, we are given $M$ and need to create $2n \times 2n$ unitary matrix from it. – new2quantum – 2019-01-10T10:59:48.547

Answers

4

No. The rows and columns of a unitary $U$ must have a sum-mod-square of 1. $$ \sum_{i}|U_{ij}|^2=\sum_{j}|U_{ij}|^2=1 $$ Your $M$, as specified, could have a 1 element along a whole row so the sum-mod-square of the corresponding row in $U$ would be $n$. So, unless $n=1$, it's impossible without further constraints on $M$.

DaftWullie

Posted 2019-01-10T10:26:05.743

Reputation: 35 722

i understand, so lets assume that $M$ can be normalized to any bound, to satisfy the condition you mention. – new2quantum – 2019-01-10T10:53:11.830

2

See this answer.

– DaftWullie – 2019-01-10T14:21:46.447