# What are the constraints on a matrix that allow it to be “extended” into a unitary?

DaftWulie's answer to Extending a square matrix to a Unitary matrix says that extending a matrix into a unitary cannot be done unless there's constraints on the matrix. What are the constraints?

A necessary and sufficient condition is that, given an $$n\times n$$ matrix $$M$$, you can construct a $$2n\times 2n$$ unitary matrix $$U$$ provided the singular values of $$M$$ are all upper bounded by 1.

# Sufficiency

To see this, express the singular value decomposition of $$M$$ as $$M=RDV$$ where $$D$$ is diagonal and $$R$$, $$V$$ are unitary. Now define $$U=\left(\begin{array}{cc} M & R\sqrt{\mathbb{I}-D^2}V \\ R\sqrt{\mathbb{I}-D^2}V & -M \end{array}\right),$$ which we can only do if the singular values are no larger than 1. Let's verify that it's unitary \begin{align*} UU^\dagger&=\left(\begin{array}{cc} RDV & R\sqrt{\mathbb{I}-D^2}V \\ R\sqrt{\mathbb{I}-D^2}V & -RDV \end{array}\right)\left(\begin{array}{cc} V^\dagger DR^\dagger & V^\dagger\sqrt{\mathbb{I}-D^2}R^\dagger \\ V^\dagger\sqrt{\mathbb{I}-D^2}R^\dagger & -V^\dagger DR^\dagger \end{array}\right) \\ &=\left(\begin{array}{cc} RD^2R^\dagger+R(\mathbb{I}-D^2)R^\dagger & 0 \\ 0 & RD^2R^\dagger+R(\mathbb{I}-D^2)R^\dagger \end{array}\right) \\ &=\mathbb{I}. \end{align*}

# Necessity

Imagine I have a matrix $$M$$ with a singular value $$\lambda>1$$ and corresponding normalised vector $$|\lambda\rangle$$. Assume I construct a unitary $$U=\left(\begin{array}{cc} M & A \\ B & C \end{array}\right).$$ Let's act $$U$$ on the state $$\left(\begin{array}{c} |\lambda\rangle \\ 0 \end{array}\right)$$. We get $$U\left(\begin{array}{c} |\lambda\rangle \\ 0 \end{array}\right)=\left(\begin{array}{c} M|\lambda\rangle \\ B|\lambda\rangle \end{array}\right).$$ This output state must have a norm that is at least the norm of $$M|\lambda\rangle$$, i.e. $$\lambda>1$$. But if $$U$$ is a unitary, the norm must be 1. So it must be impossible to perform such a construction if there exists a singular value $$\lambda>1$$.

• @DaftWulie: This is "a" necessary and sufficient condition. Is it the only one? – Pablo LiManni Jan 10 at 17:54
• You might be able to phrase the condition in another way, but it would be materially equivalent. That’s the point of necessary and sufficient - it is the precise categorisation of what is required. – DaftWullie Jan 10 at 18:24
• "A necessary and sufficient condition is that, given an n×n matrix M, you can construct a 2n×2n unitary matrix U provided the singular values of M are all upper bounded by 1." If I'm reading this correctly (and I am far from sure I am), it seems that this can be rewritten as "Given a matrix $M$, a necessary and sufficient condition for being able to extend $M$ to a 2n×2n unitary matrix is that the singular values of $M$ all be less than or equal to $1$." – Acccumulation Jan 10 at 19:31
• @DaftWullie it is definitely possible to do this with less then doubling the space though. As a trivial example, any matrix obtained by removing one row and column from a unitary matrix can be extended to a unitary matrix by adding a single dimension. Do you have any idea on how one could estimate the minimum number of dimensions that have to be added to a given matrix to make it into a unitary? – glS Jan 11 at 9:42
• @glS Well, I know what I'd do, which is perform a Gram-Schmidt-like procedure, extending one row at a time, ensuring orthonormality with all previous rows. I don't know ho to succinctly write down the dimension number based on properties of $M$ - I've never thought about it. I guess a starting point is by counting the number of singular values equal to 1, and reducing the size of the extension by that much? – DaftWullie Jan 11 at 9:56

$$\newcommand{\bs}{\boldsymbol{#1}}$$Here is a slightly different way to prove what the other excellent answer did.

Note that a matrix $$U$$ is unitary if and only if it sends orthonormal bases into orthonormal bases. This, in particular, means that if $$U$$ is unitary then $$\|U\bs v\|=1$$ for any $$\bs v$$ with $$\|\bs v\|=1$$.

Let us write the SVD of $$M$$ as $$M\bs u_k=s_k\bs v_k$$, where $$s_k\ge0$$ are the singular values of $$M$$.

Note that if $$U$$ is an extension of $$M$$, then $$U\bs u_k=s_k \bs v_k+\bs w_k$$ for some $$\bs w_k$$ orthogonal to $$\bs v_k$$ (and more generally to the whole range of $$M$$).

If follows that if, for any $$k$$, $$s_k>1$$, then $$\|U\bs u_k\|>1$$, and thus $$U$$ is not unitary.

On the other hand, if $$s_k\le1$$ for all $$k$$, let us show how can always construct a unitary $$U$$ that contains $$M$$ as a submatrix. Let us denote with $$\bs v\oplus \bs 0$$ the vectors in the extended $$2n$$-dimensional space that are built by appending zeros to the $$n$$-dimensional vector $$\bs v$$, and with $$\bs 0\oplus\bs v$$ the vectors that are equal to $$\bs v$$ in the last $$n$$ dimensions by zero in the first $$n$$ ones. Being $$\{\bs u_k\}_k$$ a basis for the original space, it follows that $$\{\bs u_k\oplus \bs 0,\bs0\oplus\bs u_k\}_k$$ is a basis for the extended space.

We will define $$U$$ through its action on the vectors $$u_k\oplus \bs 0$$ and $$\bs0\oplus u_k$$ as follows: \begin{align} U(\bs u_k\oplus \bs 0)&=s_k(\bs v_k\oplus\bs 0)+\sqrt{1-s_k^2}(\bs 0\oplus \bs v_k) \\ U(\bs0 \oplus \bs u_k)&=\sqrt{1-s_k^2}(\bs v_k\oplus\bs 0)-s_k(\bs 0\oplus \bs v_k). \end{align}

One can then check that all of these output vectors form an orthonormal system in the extended space, and thus $$U$$ is unitary.