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# 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 = \begin{pmatrix} a & b & \cdots & b \\b& a & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \cdots & a \end{pmatrix} ;

where and .

Suppose B=A+ where is nx1 vector.

Show that,

(a) B is non-singular .

(b) ### 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, .

which reduces, .

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  ,

we have , when .

for, this particular problem , = = , , which is non-singular, and clearly ,

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, = = = as and , 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 why ??? , so, ................(*)

So, ....using(*)

now left multiplying A, to the above matrix equation, we have . hence , we are done !!

## Food For Thought

Suppose, it is given that , 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, which one may find for a square matrix C, such that, for a non-null , , for a matrix order n, one will find n such scalars or eigenvalues, say ,

then , 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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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 = \begin{pmatrix} a & b & \cdots & b \\b& a & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \cdots & a \end{pmatrix} ;

where and .

Suppose B=A+ where is nx1 vector.

Show that,

(a) B is non-singular .

(b) ### 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, .

which reduces, .

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  ,

we have , when .

for, this particular problem , = = , , which is non-singular, and clearly ,

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, = = = as and , 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 why ??? , so, ................(*)

So, ....using(*)

now left multiplying A, to the above matrix equation, we have . hence , we are done !!

## Food For Thought

Suppose, it is given that , 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, which one may find for a square matrix C, such that, for a non-null , , for a matrix order n, one will find n such scalars or eigenvalues, say ,

then , 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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