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# How to Pursue Mathematics after High School?

For Students who are passionate for Mathematics and want to pursue it for higher studies in India and abroad.

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 !! ]

## What to do to shape your Career in Mathematics after 12th?

From the video below, let's learn from Dr. Ashani Dasgupta (a Ph.D. in Mathematics from the University of Milwaukee-Wisconsin and Founder-Faculty of Cheenta) how you can shape your career in Mathematics and pursue it after 12th in India and Abroad. These are some of the key questions that we are discussing here:

• What are some of the best colleges for Mathematics that you can aim to apply for after high school?
• How can you strategically opt for less known colleges and prepare yourself for the best universities in India or Abroad for your Masters or Ph.D. Programs?
• What are the best universities for MS, MMath, and Ph.D. Programs in India?
• What topics in Mathematics are really needed to crack some great Masters or Ph.D. level entrances?
• How can you pursue a Ph.D. in Mathematics outside India?
• What are the 5 ways Cheenta can help you to pursue Higher Mathematics in India and abroad?

## Want to Explore Advanced Mathematics at Cheenta?

Cheenta has taken an initiative of helping College and High School Passout Students with its "Open Seminars" and "Open for all Math Camps". These events are extremely useful for students who are really passionate for Mathematic and want to pursue their career in it.

To Explore and Experience Advanced Mathematics at Cheenta

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