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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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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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