Get inspired by the success stories of our students in IIT JAM MS, ISI  MStat, CMI MSc DS.  Learn More

# ISI MStat PSB 2011 Problem 1 | Linear Algebra

This is a quite pleasant sample problem from ISI MStat PSB 2011 Problem 1. It is mainly about, patterns in matrices and determinants and using a special kind of determinant decomposition, which is widely used in Statistics . Give it a try !

## Problem- ISI MStat PSB 2011 Problem 1

Let A be a nxn matrix, given as

$$A_{nxn}$$ = \begin{pmatrix} a & b & \cdots & b \\b& a & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \cdots & a \end{pmatrix} ;

where $$a \neq b$$ and $$a+(n-1)b= 0$$.

Suppose B=A+ $$\frac{\vec{1} \vec{1'}}{n}$$ where $$\vec{1} =(1,1,.....,1)'$$ is nx1 vector.

Show that,

(a) B is non-singular .

(b) $$A{B}^{-1} A =A$$

### Prerequisites

Basic Matrix multiplication

Determinants.

Matrix decomposition .

## Solution :

While attacking a problem related to matrices, the primary approach which I find very helpful is, realizing if there is any subtle pattern hidden. Similarly, here also a pattern is very much prominent exhibited in the matrix A, hence we will break A as,

$$A=(a-b) I_{n} +b \vec{1} \vec{1'}$$.

which reduces, $$B = (a-b) I_{n} + (b + \frac{1}{n}) \vec{1}\vec{1'}$$.

now, we can find Determinant of B, by using a known decomposition , which follows from determinants of partitioned matrices.

we have the decomposition as for some non-singular M and column vectors $$\vec{u}$$ $$\vec{v}$$ ,

we have $$|M+\vec{u}\vec{v'}|=|M|(1+ \vec{v'} M^{-1}\vec{u})$$, when $$\vec{v'}M^{-1}\vec{u} \neq -1$$.

for, this particular problem , $$\vec{v}$$=$$\vec{u}$$= $$\sqrt{(b+ \frac{1}{n})}\vec{1}$$ , $$M= (a-b)I_n$$ , which is non-singular, and clearly $$nb \neq 0$$,

If you are not familiar with these, then become friendly with this decomposition, as it is has very important and frequent applications in Statistics. Best to derive it !

So, $$|B|= |(a-b)I_{n}+ (b + \frac{1}{n}) \vec{1}\vec{1'}|$$= $$(a-b)^{n} ( 1+\vec{1'} \frac{I_n}{(a-b)}\vec{1})$$ =$$(a-b)^{n-1}( a+(n-1)b+1 )$$ =$$(a-b)^{n-1} \neq 0$$

as $$a+(n-1)b=0$$ and $$a \neq b$$, So, hence B is non-singular. Also, B is invertible, we will need this to do the next part

for the second part (b), observe that $$A\vec{1}=\vec{0}$$ why ??? , so, $$B\vec{1}=\vec{1} \Rightarrow \vec{1}= B^{-1}\vec{1}$$ ................(*)

So, $$B=A+ \frac{\vec{1} \vec{1'}}{n} \Rightarrow I_n= B^{-1}A+ \frac{1}{n} B^{-1}\vec{1}\vec{1'}=B^{-1}A+ \frac{1}{n}\vec{1}\vec{1'}$$ ....using(*)

now left multiplying A, to the above matrix equation, we have

$$A=AB^{-1}A+\frac{1}{n}A\vec{1}\vec{1'}=AB^{-1}A$$. hence , we are done !!

## Food For Thought

Suppose, it is given that $$Trace(A)=Trace(A^2)=n$$ , Can You show that all the eigenvalues of A are equal to 1 ? Is it true for any symmetric matrix,following the given condition? Give it a thought !

[ In case, you don't know what eigenvalues are, its a scalar, $$\lambda$$ which one may find for a square matrix C, such that, for a non-null $$\vec{x}$$,

$$C\vec{x}= \lambda\vec{x}$$, for a matrix order n, one will find n such scalars or eigenvalues, say $$\lambda_1,.....,\lambda_1$$ ,

then $$\lambda_1+ .....+ \lambda_n= Trace(C)$$, and , for symmetric matrices, all eigenvalues are real, so you don't need to worry much but you can obviously verify it !! ]

## Subscribe to Cheenta at Youtube

This site uses Akismet to reduce spam. Learn how your comment data is processed.