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

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

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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