Question

Let A∈R^nxn be a symmetric matrix and assume that B∈Rmxn such that A=B^TB.

Show that A is positive Semidefinite

Answer 1

To show that a symmetric matrix A=BTB is positive semidefinite, we prove that the quadratic form xTAx is non-negative for any vector x by expressing it in terms of the dot product of y=Bx with itself, yTy, which is always non-negative.

Step-by-step explanation:

If A is a symmetric matrix in Rnxn and B is a matrix in Rmxn such that A = BTB, to show that A is positive semidefinite, we have to prove that for any vector x in Rn, the quadratic form xTAx is non-negative.

Considering the quadratic form:

• xTAx = xT(BTB)x
• Let y = Bx; then xTAx = (Bx)T(Bx) = yTy
• yTy represents the dot product of y with itself, which is the sum of the squares of the components of y. Since the sum of squares is always non-negative, yTy ≥ 0.

Therefore, xTAx ≥ 0 for any vector x, which means that A is positive semidefinite. This stems from the fact that A can be expressed as a product of BT and B, showing that the eigenvalues of A must be non-negative, as they are the squares of the singular values of B.