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?

Pablo LiManni

Posted 2019-01-10T12:22:28.853

Reputation: 181



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.


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*}


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 $\lambda>1$.


Posted 2019-01-10T12:22:28.853

Reputation: 35 722

1@DaftWulie: This is "a" necessary and sufficient condition. Is it the only one? – Pablo LiManni – 2019-01-10T17:54:44.920

1You 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 – 2019-01-10T18:24:16.927

"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 – 2019-01-10T19:31:23.763

1@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 – 2019-01-11T09:42:49.837

@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 – 2019-01-11T09:56:15.147


$\newcommand{\bs}[1]{\boldsymbol{#1}}$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.


Posted 2019-01-10T12:22:28.853

Reputation: 12 247