Uniqueness of square root of a PD matrix

I have a symmetric positive definite covariance matrix $C$ and I'd like to get its square root $B$ such that $B'B=C$. My question is under what assumption(s) could this decomposition be unique, or unique up to something. Here's what I know, (1)If $B$ is triangular, then Cholesky decomposition could guarantee uniqueness. (2)If $B$ if symmetric positive definite, $B$ is unique as well. (3)Without further assumption, $B$ is generally not unique and is invariant to orthogonal transformation. Let $U$be unitary i.e. $U'U=I$ then any $UB$ satisfies $B'U'UB=B'IB=C$.

(4)If $B$ is symmetric(without PD assumption), $B$ is probably not unique. What do we know about this class of symmetric $B$?

(5)How to get a sparse $B$ (a lot of 0)?

(1) The Cholesky factorization is not unique. E.g., $$B=\pmatrix{1&1\\0&-1}\quad\text{and}\quad B=\pmatrix{1&1\\0&1}$$ are two Cholesky factors of the same matrix. The uniqueness of the Cholesky factorization is guaranteed if, e.g., the diagonal entries of $B$ are positive.

(2) Yes, a positive definite matrix has a unique positive definite square root.

(4) If $B$ is a unique positive definite square root of $C$, then just by switching the signs of (some of) the eigenvalues of $B$ you can obtain another square root which is symmetric but indefinite.

(5) Depends on what is that $B$ supposed to be now. If $C$ is sparse, you can use the sparse Cholesky factorization to obtain (hopefully) sparse Cholesky factors.

• Sorry, I forgot to mention that B does not have negative elements – Mark Wang Feb 10 '15 at 14:03
• very helpful, thanks! – Mark Wang Feb 10 '15 at 14:06
• @MarkWang You're welcome! If your comment refers to (1), you probably mean the diagonal of $B$. The Cholesky factor with positive diagonal of course can have negative off-diagonal entries. – Algebraic Pavel Feb 10 '15 at 14:07
• Yes, I meant what you said! Thanks again for your clarification. – Mark Wang Feb 10 '15 at 14:10

What you know in items 1 and 2 appears to be false. For instance, for the $2 \times 2$ identity, we have $$I^2 = I \\ \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}^2 = I.$$ So even in this most elementary case, there's no unique square root. (The same goes for the $1 \times 1$ identity, which is the square of both $+1$ and $-1$.)

• I forgot to mention that $B$ only has nonnegative elements – Mark Wang Feb 10 '15 at 14:04
• I think (2) is correct, because symmetric positive definite matrix has unique PD square root – Mark Wang Feb 10 '15 at 14:07