Is the square root of a symmetric positive definite matrix also symmetric? The inverse of a SPD matrix is also symmetric.
But what about the square root? Intuitively, I would say yes. But I'm not sure about it.
 A: Each PSD matrix certainly has a symmetric square root.  In fact, PSD matrices have a unique PSD square root.  However, not every square root of a PSD matrix is symmetric.  
For example, the $2 \times 2$ zero matrix is PSD, but
$$
A = \pmatrix{0&1\\0&0}
$$
Is not symmetric, even though $A^2$ is PSD.
EDIT: If, like Pavel (below) you need some positivity in your PSD matrix, you should consider
$$
A = \pmatrix{1&0&0\\0&0&1\\0&0&0}
$$
A: There exist both positive-definite and non-positive-definite square roots for a Symmetric Positive-Definite matrix.
If A is SPD, then A can be represented as UDU* for some unitary matrix U and diagonal matrix D, with the eigenvalues represented in D, and all eigenvalues positive.
If I let R = sqrt(D) then I claim that URU* is a square root of A.  Notice:
(URU*)^2 = URU*URU* = URRU = UDU = A
To find a particular square root, I only need to worry about finding square roots of the eigenvalues of D.  There will be both positive and negative values for the square roots and sometimes a mix of both.
However, a matrix is positive-definite only when all the eigenvalues are positive.  You can indeed find a square root with all positive values, but you can also find some with both positive and negative eigenvalues.
