This is a very beautiful sample problem from ISI MStat PSB 2010 Problem 1 based on Matrix multiplication and Eigen values and Eigen vectors . Let’s give it a try !!

**Problem**– ISI MStat PSB 2010 Problem 1

Let \(A\) be a \(4 \times 4\) matrix with non-negative entries such that the sum of the entries in each row of \( A\) equals 1 . Find the sum of all entries in matrix \(A^{5}\) .

**Prerequisites**

Matrix Multiplication

Eigen Values

Eigen Vectors

## Solution :

Doing this problem you have to use the hint given in the question . Here the hint is that the sum of the entries in each row of \( A\) equals 1 . How can you use that ? Think about it!

Here comes the trick .

Let V be a vector such that \( V={[1,1,1,1]}^{T} \) . Now if we multiply A by V then we will get V i.e \( AV=V \) .

This is because it is given that the sum of the entries in each row of \( A\) equals 1 .

So, from \( AV=V \) we can say that 1 is an eigen value of A .

Hence \( A^5V=A^4(AV)=A^4V=A^3(AV)= \cdots = V \) . From here we can say the sum of all the entries of each rows of \(A^5 \) is 1.

Therefore the sum of all the entries of \( A^5\) is also 4 .

## Food For Thought

Let \(A\) and B be \( n \times n \) matrices with real entries satisfying \(tr(A A^{T}+B B^{T})=tr(A B+A^{T} B^{T})\) .

Prove that \( A=B^{T}\) .

Hint : Use properties of trace that’s the trick here .

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