## Description

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

To prove:

A2 = tr(A)A − det(A)I

where tr(A) is the trace of a matrix and det(A) is the determinant of a matrix, the simplest method would be the brute force comparison. Consider A as follows:

[

a

A =11 a12]

a21 a22

Now consider the RHS of the equation

tr(A) = a11 + a22 det(A) = a11a22 − a12a21

a11 tr(A)A − det(A)I = (a11 + a22)[

a21

a2 + a

= [ 11 11a22

a11a21 + a22a21 a2 + a

= [ 11 12a21

a11a21 + a22a21 a12 1 0

] − (a11a22 − a12a21)[ ]

a22 0 1

a11a12 + a22a12 a11a22 − a12a21

2 ] − [ 0

a11a22 + a22

a11a12 + a22a12 a12a21 + a222 ] 0 ]

a11a22 − a12a21

=a11a11 + a12a21 a11a12 + a12a22] a21a11 + a22a21 a21a12 + a22a22

=a11 a12][a11 a12] a21 a22 a21 a22

= AA

= A2

□

Using the result obtained in Question 2, we have:

(AB − BA)2 = (AB − BA)tr(AB − BA) − det(AB − BA)I = −det(AB − BA)I

where I is the identity matrix. Taking trace on both sides we get:

tr((AB − BA)2) = tr(−det(AB − BA)I)

= −2det(AB − BA)

Now,

tr((AB − BA)2) = tr((AB − BA)(AB − BA))

= tr(ABAB − AB2A − BA2B + BABA)

Now, tr(ABAB) = tr(BABA) from the property tr(XY) = tr(YX) and similarly using the same property we can say

tr(ABBA) = tr(ABAB) = tr(AABB)

Using the above equations in the original equation we get

tr((AB − BA)2) = 2tr(ABAB − AABB)

= −2det(AB − BA)

⟹ tr(ABAB − AABB) = −det(AB − BA)

As both terms are equal we can say that

P {tr(ABAB) > tr(A2B2)} = P {det(AB − BA) < 0}

□

We want to prove that the determinant of an upper traingular matrix is the product of its diagonal entries.

We know that in an upper triangular matrix, all the entries below the diagonal are zero.

Thus, for a n × n upper triangular matrix, it will be of the form:

⎡a11 a12 ⋯ a1n ⎤

0 a22 ⋯ ⋮

A =

⋮ ⋮ ⋱ a(n−1)n

⎣ 0 0 ⋯ ann ⎦

As all the terms in the first column are zero except the first one, the determinant of the matrix can be written as:

det(A) = a11 × det(B)

⎡a22 a23 ⋯ a2n ⎤

where B =

⋮ ⋮ ⋱ (n−1)n

nn ⎦

Similarly, we can write det(B) as:

det(B) = a22 × det(C)

⎡a33 a34 … a3n ⎤

0 a44 ⋯ ⋮

where C =

⋮ ⋮ ⋱ a(n−1)n

⎣ 0 0 ⋯ ann ⎦

We can continue this process until we arrive at a matrix of dimensions 1 × 1, whose determinant would be the value of the entry i.e. ann

Thus, we can see that:

det(A)= a11 × a22 ⋯ × ann

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