Why $x^{T}B^TWBx=\|W^{1/2}Bx\|^2_2$ for diagonal matrix W?

Question

I have question reading paper "Graph Sparsification by Effective Resistances" by Daniel A. Spielman and Nikhil Srivastava, on page N4 it says that it is obvious that L is semidefinite positive since $x^{T}B^TWBx=\|W^{1/2}Bx\|^2_2 \geq 0$ how did we get the equality $x^{T}B^TWBx=\|W^{1/2}Bx\|^2_2$?

Answer 1

\begin{align} x^TB^TWBx &= x^TB^TW^{\frac12}W^\frac12Bx \\ &=x^TB^T(W^{\frac12})^TW^\frac12Bx \\ &=(W^\frac12Bx)^T(W^\frac12Bx) \\ &=\|W^\frac12Bx)\|^2 \end{align}

Note that $W$ is a diagonal matrix with positive diagonal entries and hence its square root is also symmetric.

Answer 2

Assuming that $W$ is symmetric and that there exists a matrix $W^{\frac12}$ with $W = W^{\frac12} W^{\frac12},$ we have \begin{align*} x^T B^T WBx &= x^T B^T W^{\frac12} W^{\frac12} Bx \\ &=\langle W^{\frac12} Bx, W^{\frac12} B x \rangle \\ &= \lVert W^{\frac12} Bx \rVert^2_2. \end{align*}

Answer 3

We can define $$\|Ax\|_2^2 = (Ax)^T(Ax) = x^TA^TAx$$

so just set $A = W^{1/2}B$, expand, use the changed order transpose laws and the fact that $W^T=W$

Answer 4

If diagonal elements of $W$ are non-negative then it factorizes as $W^{1/2}W^{1/2}$, where $W^{1/2}$ is a diagonal matrix with square roots.

$$ \|y\|_2^2=y^Ty=(W^{1/2}Bx)^TW^{1/2}Bx=x^TB^T(W^{1/2})^TW^{1/2}Bx=x^TB^TWBx $$