Description

General Instructions

The following document contains the solutions to the theory-based questions for
Question 2

a) Since A and B are positive semi-definite matrices, they are both symmetric. Thus, for some matrices U and V, we have
A = UTU
B = VTV ATB = (UTU)TVTV = UTUVTV
Now, we have:
tr(ATB) = tr(UTUVTV)
Since:
tr(XY) = tr(YX) We can set X = UT and Y = UVTV
Therefore, we have:
tr(ATB) = tr(UVTVUT) = tr(UVT(UVT)T)
which is of the from tr(MMT), which is always ≥ 0.
□ b) If
tr(ATB) = 0
then,
tr(UVT(UVT)T) = 0
Let M = UVT. Let the entries of M be mij.
tr(MMT) = ∑∑(mij)2 = 0
i j
⟹ mij = 0 ∀i,j
⟹ M = 0
⟹ UVT = 0
AB = UTUVTV = 0
□ Question 3

For A to be a positive semi-definite matrix, xTAx ≥ 0 ∀x ∈ Rn.
We know that,
xTAx = ∑xixjAij
ij
= ∑(xxT)ijAij
ij
Now for any two matrices A and B, the trace of C = ATB is given by
tr(C) = ∑cii
i
We can write cii as the dot product of the ith row of A and the ith column of B
tr AijBji)
= ∑AijBji
ij
Combining the above two equations, we get
xTAx = ∑xixjAij
ij
= ∑Aij(xxT)ij
ij
= tr(ATxxT) ≥ 0
As B is also a positive semi-definite matrix, we can write it as
B = ∑σivivTi
i
Using the above equation we can write,
tr(ATB) = tr ivivTi )
= ∑
i
≥ 0
□