Square Root of a matrix: transpose or inverse of eigen vectors?

Question

Here is described that the square root of a matrix is defined as

K^1/2 = V*D^1/2*V^-1

At the end of scetion 4 of this paper we can see W = K^-1/2es

In the matlab code in createHashTable.m at line 24 we have:

K_half = V_K*diag(d_k)* V_K';

As you can see the last term is V_K' and not V_K^-1. Why?

Answer 1

If K is symmetric then there exists an ortho-normal basis, and $V^{-1} = V^T$