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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 , $$dim ( c_3 )=1$$

## Previous MStat Posts:

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 , $$dim ( c_3 )=1$$

## Previous MStat Posts:

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