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# ISI MStat PSB 2015 Problem 2 | Vector Space & its Dimension

This is a beautiful problem from ISI MStat PSB 2015 Problem 2. We provide detailed solution with prerequisite mentioned explicitly.

## Problem- ISI MStat PSB 2015 Problem 2

For any $n \times n$ matrix $A=\left(\left(a_{i j}\right)\right),$ consider the following three proper-
ties:

1. $a_{i j}$ is real valued for all $i, j$ and $A$ is upper triangular.
2. $\sum_{j=1}^{n} a_{i j}=0,$ for all $1 \leq i \leq n$
3. $\sum_{i=1}^{n} a_{i j}=0,$ for all $1 \leq j \leq n$
Define the following set of matrices:
$c_n$ = {A: A is $n \times n$ and satisfies (1),(2) and (3) above }

(a) Show that $c_n$ is a vector space for any $n \geq 1$ .

(b) Find the dimension of , $c_n$ when n = 2 and n = 3.

## Prerequisites

• Upper triangular matrix
• Subspace of a vector space
• Dimension of a vector space

## Solution

(a) To show that $c_n$ is a vector space for any $n \geq 1$

So, here if we can show that $c_n$ is a subspace of the vector space of $n\times n$ real matrices with usual matrix addition and scalar multiplication then we are done!

Let's try to show this ,

Putting $a_{i j} =0$ for all i,j then $A= \left(\left(a_{i j}\right)\right),$ satisfies all the properties (1),(2) & (3) .

So, $\begin{pmatrix} 0 & 0 &... & 0 \\ 0 & 0 &... & 0 \\ \vdots & \vdots & \vdots \\ 0 & 0 &... & 0 \end{pmatrix}$ $\epsilon$ $c_n$

Shall show that (i) for all $A , B$ $\epsilon$ $c_n$ , $A + B \epsilon c_n$ and

(ii) for all $A$ $\epsilon$ $c_n$ for all $p_1 \epsilon$ {$\mathbb{R}$ }-{0} , $p_1 A \epsilon c_n$

For (i) Take any $A=((a_{i j})) , B=(( b_{i j}))$ $\epsilon$ $c_n$

Let , D=$A + B$ and if $D=(( d_{i j}))$ then $d_{ij}= a_{i j} + b_{i j}$

Now we will see whether D satisfies all the three properties (1),(2) and (3)

$d_{ij} =0$ when $a_{i j}=0$ and $b_{i j} =0$

Hence as A and B are upper triangular matrix , D is also an upper triangular matrix .

So it satisfies property (1)

Again , $\sum_{j=1}^{n} a_{i j}=0,$ for all $1 \leq i \leq n$ and $\sum_{j=1}^{n} b_{i j}=0,$ for all $1 \leq i \leq n$ ,

then $\sum_{j=1}^{n} d_{i j}=0,$ for all $1 \leq i \leq n$ as $d_{ij}=a_{i j} + b_{i j}$

Hence it satisfies property (2) .

Now we have $\sum_{i=1}^{n} a_{i j}=0,$ for all $1 \leq j \leq n$ and $\sum_{i=1}^{n} b_{i j}=0,$ for all $1 \leq j \leq n$ ,then $\sum_{i=1}^{n} d_{i j}=0,$ for all $1 \leq j \leq n$ as $d_{ij}=a_{i j} + b_{i j}$

Hence it satisfies the properties (3)

For (ii) Take any $A=((a_{i j}))$ $\epsilon$ $c_n$

take any $p_1 \epsilon$ {$\mathbb{R}$ }-{0}

Let, $K=p_1 A$ and if $K=(( k_{i j}))$ then $d_{ij}= p_1 a_{i j}$

Then , $k_{ij} =0$ when $a_{i j}=0$

Hence as A is an upper triangular matrix , K is also an upper triangular matrix .

So it satisfies property (1)

Again , $\sum_{j=1}^{n} a_{i j}=0,$ for all $1 \leq i \leq n$ then $\sum_{j=1}^{n} k_{i j}=0,$ for all $1 \leq i \leq n$ as $k_{ij}=p_1 a_{i j}$

Hence it satisfies property (2) .

Now we have $\sum_{i=1}^{n} a_{i j}=0,$ for all $1 \leq j \leq n$ ,then $\sum_{i=1}^{n} k_{i j}=0,$ for all $1 \leq j \leq n$ as $k_{ij}=p_1 a_{i j}$

Hence it satisfies the properties (3)

So, $c_n$ is closed under vector addition and scalar multiplication.

Therefore , $c_n$ is a subspace of the vector space of $n \times n$ real matrices with usual matrix addition and scalar multiplication . Hence we are done !

(b) n=2 ,

$A=((a_{i j}))$ $\epsilon$ $c_2$ then , $A= \begin{pmatrix} a_{11} & a_{12} \\ 0 & a_{22} \end{pmatrix}$by property (1) , $a_{11}+a_{12}=0 , a_{22}=0$---(I) by property (2) and $a_{11}=0 , a_{12}+a_{22}=0$---(II) by property (3) .

Now solving (I) and (II) we get $A= \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$

Giving , $c_2$ = { $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ } hence $dim(c_2)=0$

n=3

$A=((a_{i j}))$ $\epsilon$ $c_3$ then , $A= \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a_{22}& a_{23} \\ 0 & 0& a_{33} \end{pmatrix}$ by property (1) , $a_{11}+a_{12}+ a_{13}=0 , a_{22}+a_{23}=0 , a_{33}=0$---(I) by property (2) and $a_{11}=0 , a_{12}+a_{22}=0 a_{13}+a_{23}+a_{33}=0$---(II) by property (3) .

Now solving (I) and (II) we get $a_{11}=0 , a_{33}=0$ $a_{13}=-a_{12}=a_{22}=-a_{23}=-a_{13}=t$ (say) then ,

$A= t \begin{pmatrix} 0 & -1 & -1 \\ 0 & 1 & -1 \\ 0 & 0& 0\end{pmatrix}$ , $t \epsilon R$

Giving , $c_3$= {t $\begin{pmatrix} 0 & -1 & -1 \\ 0 & 1 & -1 \\ 0 & 0& 0\end{pmatrix}$} ,$t \epsilon R$ .

Hence , $dim ( c_3 )=1$

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