## 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 โฅ 0.

โก b) If

tr(ATB) = 0

then,

tr(UVT(UVT)T) = 0

Let M = UVT. Let the entries of M be mij.

tr(MMT) = โโ(mij)2 = 0

i j

โน mij = 0 โi,j

โน M = 0

โน UVT = 0

AB = UTUVTV = 0

โก Question 3

For A to be a positive semi-definite matrix, xTAx โฅ 0 โx โ Rn.

We know that,

xTAx = โxixjAij

ij

= โ(xxT)ijAij

ij

Now for any two matrices A and B, the trace of C = ATB is given by

tr(C) = โcii

i

We can write cii as the dot product of the ith row of A and the ith column of B

tr AijBji)

= โAijBji

ij

Combining the above two equations, we get

xTAx = โxixjAij

ij

= โAij(xxT)ij

ij

= tr(ATxxT) โฅ 0

As B is also a positive semi-definite matrix, we can write it as

B = โฯivivTi

i

Using the above equation we can write,

tr(ATB) = tr ivivTi )

= โ๎

i

โฅ 0

โก

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