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# ISI MStat PSB 2011 Problem 1 | Linear Algebra

This is ISI MStat PSB 2011 Problem 1, based on patterns in matrices and determinants, and using a special kind of determinant decomposition. Try this out!

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 !! ]

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