Categories

## ISI MStat PSB 2009 Problem 1 | Nilpotent Matrices

This is a very simple sample problem from ISI MStat PSB 2009 Problem 1. It is based on basic properties of Nilpotent Matrices and Skew-symmetric Matrices. Try it !

## Problem– ISI MStat PSB 2009 Problem 1

(a) Let $A$ be an $n \times n$ matrix such that $(I+A)^4=O$ where $I$ denotes the identity matrix. Show that $A$ is non-singular.

(b) Give an example of a non-zero $2 \times 2$ real matrix $A$ such that $\vec{x’}A \vec{x}=0$ for all real vectors $\vec{x}$.

### Prerequisites

Nilpotent Matrix

Eigenvalues

Skew-symmetric Matrix

## Solution :

The first part of the problem is quite easy,

It is given that for a $n \times n$ matrix $A$, we have $(I+A)^4=O$, so, $I+A$ is a nilpotet matrix, right !

And we know that all the eigenvalues of a nilpotent matrix are $0$. Hence all the eigenvalues of $I+A$ are 0.

Now let $\lambda_1, \lambda_2,……,\lambda_k$ be the eigenvalues of the matrix $A$. So, the eigenvalues of the nilpotent matrix $I+A$ are of form $1+\lambda_k$ where, $k=1,2…..,n$. Now since, $1+\lambda_k=0$ which implies $\lambda_k=-1$, for $k=1,2,…,n$.

Since all the eigenvalues of $A$ are non-zero, infact $|A|=(-1)^n$. Hence our required propositon.

(b) Now this one is quite interesting,

If for any $2\times 2$ matrix, the Quadratic form of that matrix with respect to a vector $\vec{x}=(x_1,x_2)^T$ is of form,

$a{x_1}^2+ bx_1x_2+cx_2x_1+d{x_2}^2$ where $a,b,c$ and $d$ are the elements of the matrix. Now if we equate that with $0$, what condition should it impose on $a, b, c$ and $d$ !! I leave it as an exercise for you to complete it. Also Try to generalize it you will end up with a nice result.

## Food For Thought

Now, extending the first part of the question, $A$ is invertible right !! So, can you prove that we can always get two vectors from $\mathbb{R}^n$, say $\vec{x}$ and $\vec{y}$, such that the necessary and sufficient condition for the invertiblity of the matrix $A+\vec{x}\vec{y’}$ is “ $\vec{y’} A^{-1} \vec{x}$ must be different from $1$” !!

This is a very important result for Statistics Students !! Keep thinking !!

Categories

## ISI MStat PSB 2005 Problem 3 | The Orthogonal Matrix

This is a very subtle sample problem from ISI MStat PSB 2005 Problem 3. Given that one knows the property of orthogonal matrices its just a counting problem. Give it a thought!

## Problem– ISI MStat PSB 2005 Problem 3

Let $A$ be a $n \times n$ orthogonal matrix, where $n$ is even and suppose $|A|=-1$, where $|A|$ denotes the determinant of $A$. Show that $|I-A|=0$, where $I$ denotes the $n \times n$ identity matrix.

### Prerequisites

Orthogonal Matrix

Eigenvalues

Characteristic Polynomial

## Solution :

This is a very simple problem, when you are aware of the basic facts.

We, know that, the eigenvalues of a orthogonal matrix is $-1$ and $1$ .($i$ and $-i$ if its skew-symmetric). But this given matrix $A$ is not skew-symmetric.(Why??).So let for the matrix $A$, the algebraic multiplicity of $-1$ and $1$ be $m$ and $n$, respectively.

So, since $|A|=-1$, hence the algebraic multiplicity of $-1$ is definitely odd, since we know by the property of eigenvalues determinant of a matrix is just the product of its eigenvalues.

Now since, $n$ is even and the algebraic multiplicity of $-1$ i.e. $m$ is odd, hence $n$ is also odd and $n \ge 1$.

Hence, the Characteristic Polynomial of $A$, is $|I\lambda – A |=0$, where $\lambda$ is the eigenvalue of $A$, and in this problem $\lambda=-1$ or $1$.

Hence, putting $\lambda=1$, we conclude that, $|I-A|=0$. Hence we are done !!

## Food For Thought

Now, suppose $M$ is any non-singular matrix, such that $M^2=-I$. What can you say about the column space of $M$ ?

Keep thinking !!

Categories

## ISI MStat PSB 2006 Problem 1 | Inverse of a matrix

This is a very beautiful sample problem from ISI MStat PSB 2006 Problem 1 based on Inverse of a matrix. Let’s give it a try !!

## Problem– ISI MStat PSB 2006 Problem 1

Let A and B be two invertible $n \times n$ real matrices. Assume that $A+B$ is invertible. Show that $A^{-1}+B^{-1}$ is also invertible.

### Prerequisites

Matrix Multiplication

Inverse of a matrix

## Solution :

We are given that A,B,A+B are all invertible real matrices . And in this type of problems every information given is a hint to solve the problem let’s give a try to use them to show that $A^{-1}+B^{-1}$ is also invertible.

Observe that ,$A(A^{-1}+B^{-1})B= (B+A) \Rightarrow |A^{-1}+B^{-1}|=\frac{|A+B|}{|A| |B| }$ taking determinant is both sides as A+B , A and B are invertible so |A+B| , |A| and |B| are non-zero . Hence $A^{-1}+B^{-1}$ is also non-singular .

Again we have , $A(A^{-1}+B^{-1})B= (B+A) \Rightarrow B^{-1} {(A^{-1}+B^{-1})}^{-1} A^{-1} = {(A+B)}^{-1}$ , taking inverse on both sides .

Now as A+B , A and B are invertible so , we have ${(A^{-1}+B^{-1})}^{-1}=B {(A+B)}^{-1} A$ . Hence we are done .

## Food For Thought

If $A \& B$ are non-singular matrices of the same order such that $(A+B)$ and $\left(A+A B^{-1} A\right)$ are also non-singular, then find the value of $(A+B)^{-1}+\left(A+A B^{-1} A\right)^{-1}$.

Categories

## ISI MStat PSB 2007 Problem 2 | Rank of a matrix

This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 2 based on Rank of a matrix. Let’s give it a try !!

## Problem– ISI MStat PSB 2007 Problem 2

Let $A$ and $B$ be $n \times n$ real matrices such that $A^{2}=A$ and $B^{2}=B$
Suppose that $I-(A+B)$ is invertible. Show that rank(A)=rank(B).

### Prerequisites

Matrix Multiplication

Inverse of a matrix

Rank of a matrix

## Solution :

Here it is given that $I-(A+B)$ is invertible which implies it’s a non-singular matrix .

Now observe that ,$A(I-(A+B))=A-A^2-AB= -AB$ as $A^2=A$

Again , $B(I-(A+B))=B-BA-B^2=-BA$ as $B^2=B$ .

Now we know that for non-singular matrix M and another matrix N , $rank(MN)=rank(N)$ . We will use it to get that

$rank(A)=rank(A(I-(A+B)))=rank(-AB)=rank(AB)$ and $rank(B)=rank(B(I-(A+B)))=rank(-BA)=rank(BA)$ .

And it’s also known that $rank(AB)=rank(BA)$ . Hence $rank(A)=rank(B)$ (Proved) .

## Food For Thought

Try to prove the same using inequalities involving rank of a matrix.

Categories

## ISI MStat PSB 2007 Problem 1 | Determinant and Eigenvalues of a matrix

This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 1 based on Determinant and Eigen values and Eigen vectors . Let’s give it a try !!

## Problem– ISI MStat PSB 2007 Problem 1

Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{2}=0 .$ Find the determinant of $I+A$ where I denotes the identity matrix.

Determinant

Eigen Values

Eigen Vectors

## Solution :

Let ${\lambda}_{1} , {\lambda}_{2}$ be two eigen values of A then , ${{\lambda}_{1}}^2 , {{\lambda}_{2}}^2$ .

Now it’s given that $A^2=0$ , so we have ${{\lambda}_{1}}^2=0 , {{\lambda}_{2}}^2 =0$ . You may verify it ! (Hint : use the theorem that $\lambda$ is a eigen value of matrix B and $\vec{x}$ is it’s corresponding eigen value then we can write $Bx=\lambda \vec{x}$ or , use $det(B- \lambda I )=0$ ).

Hence we have ${\lambda}_{1} =0 , {\lambda}_{2}=0$ .

Now , eigen values of Identity matrix I are 1 . So, we can write for eigen value $\vec{x}$ of (A+I) , $(A+I) \vec{x}= Ax+I\vec{x}=0+\vec{x}=\vec{x}$.

Thus we get that both the eigen values of (A+1) are 1 . Again we know that determinant of a matrix is product of it’s eigen values .

So, we have $|A+I|=1$.

Do you think this solution is correct ?

If yes , then you are absolutely wrong . The mistake is in assuming A and I has same eigen vectors $Ax+I\vec{x} \ne \vec{x}$

Correct Solution

We have shown in first part of wrong solution that A has eigen values 0 . Hence the characteristic polynomial of A can be written as , $|A- \lambda I|= {\lambda}^2$ .

Now taking $\lambda =-1$ we get $|A+ I|={(-1)}^2 \implies |A+I|= 1$ .

## Food For Thought

If we are given that $A^{n} = 0$ for positive integer n , instead of $A^2=0$ then find the same .

Categories

## ISI MStat PSB 2005 Problem 1 | The Inductive Matrix

This is a very beautiful sample problem from ISI MStat PSB 2005 Problem 1. It is based on some basic properties of upper triangular matrix and diagonal matrix, only if you use them carefully. Give it a thought!

## Problem– ISI MStat PSB 2005 Problem 1

Let $A$ be a $n \times n$ upper triangular matrix such that $AA^T=A^TA$. Show that $A$ is a diagonal matrix.

### Prerequisites

Upper Triangular Matrix

Diagonal Matrix

Mathematical Induction

## Solution :

This is very beautiful problem, since it deals with some very beautiful aspects of matrix. Lets assume the matrix $A$ to be,

$A=\begin{bmatrix} a_{11} & \vec{{a_1}^T} \\ \vec{0_{n-1}}& A_{n-1} \end{bmatrix}$.

Here, $A_{n-1}$ is a partition matrix of $A$, which is also an upper triangular matrix of order $n-1$, and $\vec{0_{n-1}}$ is a null column vector of order $n-1$.

So, $AA^T= \begin{bmatrix} {a_{11}}^2+\vec{{a_1}^T}\vec{a_1} && \vec{{a_1}^T}{A_{n-1}}^T \\ A_{n-1}\vec{a_1}&& A_{n-1}{A_{n-1}}^T \end{bmatrix}$ .

Again, $A^TA= \begin{bmatrix} {a_{11}}^2&& a_{11}\vec{a_1} \\ a_{11}\vec{a_1}&& \vec{a_1}\vec{{a_1}^T}+{A_{n-1}}^TA_{n-1} \end{bmatrix}$.

Now lets assume that when a $n\times n$ upper triangular matrix $A$, holds $AA^T=A^TA$ , then $A$ is diagonal is true. Then equating the above elements, we have,

$\vec{{a_1}^T}\vec{a_1}=0 \Rightarrow \vec{a_1}=\vec{0}$ , and also $A_{n-1}{A_{n-1}}^T={A_{n-1}}^TA_{n-1}$.

Now observe that, if $A_{n-1}$ (which is actually an $n\times n$ upper triangular matrix), is a diagonal matrix then only $A$ will be also diagonal. So, its the same problem we are trying to prove !! So, just use induction get the proof done !!

## Food For Thought

What if I change the given relation as $AA^*=A^*A$, where $A^*$ is the conjugate matrix of $A$, rest of the conditions remains same. Can you investigate whether $A$ is a diagonal matrix or not ?

Keep thinking !!

Categories

## ISI MStat PSB 2014 Problem 1 | Vector Space & Linear Transformation

This is a very beautiful sample problem from ISI MStat PSB 2014 Problem 1 based on Vector space and Eigen values and Eigen vectors . Let’s give it a try !!

## Problem– ISI MStat PSB 2014 Problem 1

Let $E={1,2, \ldots, n},$ where n is an odd positive integer. Let $V$ be
the vector space of all functions from E to $\mathbb{R}^{3}$, where the vector space
operations are given by $(f+g)(k) =f(k)+g(k)$, for $f, g \in V, k \in E \ (\lambda f)(k) =\lambda f(k),$ for $f \in V, \lambda \in \mathbb{R}, k \in E$
(a) Find the dimension of $V$
(b) Let $T: V \rightarrow V$ be the map given by $T f(k)=\frac{1}{2}(f(k)+f(n+1-k)), \quad k \in E$
Show that T is linear.
(c) Find the dimension of the null space of T.

### Prerequisites

Linear Transformation

Null Space

Dimension

## Solution :

While doing this problem we will use a standard notation for vectors of canonical basis i..e $e_j$ . In $R^{3}$ they are $e_1=(1,0,0) , e_2=(0,1,0)$ and $e_3=(0,0,1)$ .

(a) For $i \in {1 , 2 , \cdots , n}$ and $j \in {1 , 2 , 3}$ , let $f_{ij}$ be the function in $V$ which maps $i \mapsto e_j$ and $k \mapsto (0,0,0)$ where $k \in {1 , 2 , \cdots , n}$ and $k \neq i$. Then ${f_{ij} : i \in {1 , 2 , \cdots , n} , j \in {1 , 2 , 3}}$ is a basis of $V$ .

It looks somewhat like this , $f_{11}(1)={(1,0,0)} ,f_{11}(2)={(0,0,0)} , \cdots , f_{11}(n)={(0,0,0)}$

$f_{12}(1)={(0,1,0)} ,f_{12}(2)={(0,0,0)} , \cdots , f_{12}(n)={(0,0,0)}$ , $\cdots , f_{n3}(1)={(0,0,0)} ,f_{n3}(2)={(0,0,0)} , \cdots , f_{n3}(n)={(0,0,1)}$

Hence , dimension of $V$ is 3n.

(b) To show T is linear we have to show that $T(af(k)+bg(k)) =aT(f(k))+bT(g(k))$ for some scalar a,b .

$T(af(k)+bg(k))=\frac{ af(k)+bg(k)+af(n+1-k)+bg(n+1-k)}{2} = a \frac{f(k)+f(n+1-k)}{2} + b \frac{g(k)+g(n+1-k)}{2} = aT(f(k))+bT(g(k))$.

Hence proved .

(c) $f\in ker T$ gives $f(k)=-f(n+1-k)$ so, the values of $f$ for the last $\frac{n-1}{2}$ points are opposite to first $\frac{n-1}{2}(i.e. f(n)=-f(1) ~\text{etc.})$ so we can freely assign the values of f for first $\frac{n-1}{2}$ to any of $e_j$ .Hence, the null space has dimension $\frac{3(n-1)}{2}.$

## Food For Thought

let $T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ be a non singular linear transformation.Prove that there exists a line passing through the origin that is being mapped to itself.

Prerequisites : eigen values & vectors and Polynomials

Categories

## ISI MStat PSB 2010 Problem 1 | Tricky Linear Algebra Question

This is a very beautiful sample problem from ISI MStat PSB 2010 Problem 1 based on Matrix multiplication and Eigen values and Eigen vectors . Let’s give it a try !!

## Problem– ISI MStat PSB 2010 Problem 1

Let $A$ be a $4 \times 4$ matrix with non-negative entries such that the sum of the entries in each row of $A$ equals 1 . Find the sum of all entries in matrix $A^{5}$ .

### Prerequisites

Matrix Multiplication

Eigen Values

Eigen Vectors

## Solution :

Doing this problem you have to use the hint given in the question . Here the hint is that the sum of the entries in each row of $A$ equals 1 . How can you use that ? Think about it!

Here comes the trick .

Let V be a vector such that $V={[1,1,1,1]}^{T}$ . Now if we multiply A by V then we will get V i.e $AV=V$ .

This is because it is given that the sum of the entries in each row of $A$ equals 1 .

So, from $AV=V$ we can say that 1 is an eigen value of A .

Hence $A^5V=A^4(AV)=A^4V=A^3(AV)= \cdots = V$ . From here we can say the sum of all the entries of each rows of $A^5$ is 1.

Therefore the sum of all the entries of $A^5$ is also 4 .

## Food For Thought

Let $A$ and B be $n \times n$ matrices with real entries satisfying $tr(A A^{T}+B B^{T})=tr(A B+A^{T} B^{T})$ .
Prove that $A=B^{T}$ .

Hint : Use properties of trace that’s the trick here .

Categories

## 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

$A_{nxn}$ = \begin{pmatrix} a & b & \cdots & b \\b& a & \cdots & b \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \cdots & a \end{pmatrix} ;

where $a \neq b$ and $a+(n-1)b= 0$.

Suppose B=A+ $\frac{\vec{1} \vec{1′}}{n}$ where $\vec{1} =(1,1,…..,1)’$ is nx1 vector.

Show that,

(a) B is non-singular .

(b) $A{B}^{-1} A =A$

### 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,

$A=(a-b) I_{n} +b \vec{1} \vec{1′}$.

which reduces, $B = (a-b) I_{n} + (b + \frac{1}{n}) \vec{1}\vec{1′}$.

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 $\vec{u}$ $\vec{v}$ ,

we have $|M+\vec{u}\vec{v’}|=|M|(1+ \vec{v’} M^{-1}\vec{u})$, when $\vec{v’}M^{-1}\vec{u} \neq -1$.

for, this particular problem , $\vec{v}$=$\vec{u}$= $\sqrt{(b+ \frac{1}{n})}\vec{1}$ , $M= (a-b)I_n$ , which is non-singular, and clearly $nb \neq 0$,

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, $|B|= |(a-b)I_{n}+ (b + \frac{1}{n}) \vec{1}\vec{1′}|$= $(a-b)^{n} ( 1+\vec{1′} \frac{I_n}{(a-b)}\vec{1})$ =$(a-b)^{n-1}( a+(n-1)b+1 )$ =$(a-b)^{n-1} \neq 0$

as $a+(n-1)b=0$ and $a \neq b$, 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 $A\vec{1}=\vec{0}$ why ??? , so, $B\vec{1}=\vec{1} \Rightarrow \vec{1}= B^{-1}\vec{1}$ …………….(*)

So, $B=A+ \frac{\vec{1} \vec{1′}}{n} \Rightarrow I_n= B^{-1}A+ \frac{1}{n} B^{-1}\vec{1}\vec{1′}=B^{-1}A+ \frac{1}{n}\vec{1}\vec{1′}$ ….using(*)

now left multiplying A, to the above matrix equation, we have

$A=AB^{-1}A+\frac{1}{n}A\vec{1}\vec{1′}=AB^{-1}A$. hence , we are done !!

## Food For Thought

Suppose, it is given that $Trace(A)=Trace(A^2)=n$ , 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, $\lambda$ which one may find for a square matrix C, such that, for a non-null $\vec{x}$,

$C\vec{x}= \lambda\vec{x}$, for a matrix order n, one will find n such scalars or eigenvalues, say $\lambda_1,…..,\lambda_1$ ,

then $\lambda_1+ …..+ \lambda_n= Trace(C)$, and , for symmetric matrices, all eigenvalues are real, so you don’t need to worry much but you can obviously verify it !! ]

Categories

## ISI MStat PSB 2018 Problem 1 | System of Linear Equations

This is a beautiful sample problem from ISI MStat PSB 2018 Problem 1. This is based on finding the real solution of a system of homogeneous equations. We provide a detailed solution with prerequisites mentioned explicitly.

## Problem– ISI MStat PSB 2018 Problem 1

Find all real solutions ${(x_{1}, x_{2}, x_{3}, \lambda)}$ for the system of equations
$x_{2}-3 x_{3}-x_{1} \lambda =0$
$x_{1}-3 x_{3}-x_{2} \lambda =0$
$x_{1}+x_{2}+x_{3} \lambda =0$

## Solution

We are given the system of homogeneous equations as follows ,

$-x_{1} \lambda + x_{2} – 3 x_{3}=0$
$x_{1}-x_{2} \lambda-3 x_{3} =0$
$x_{1}+x_{2}+x_{3} \lambda =0$

Let , $A= \begin{pmatrix} – \lambda & 1 & -3 \\ 1 & – \lambda & -3 \\ 1 & 1 & \lambda \end{pmatrix}$

As A is a $3 \times 3$ matrix so ,if the Rank of A is <3 then it has infinitely many solution and

if Rank of A is 3 then it has only trivial solution i.e $x_{1}=x_{2}=x_{3}=0$

Let’s try to find the rank of matrix A from it’s row echelon form ,

$A= \begin{pmatrix} – \lambda & 1 & -3 \\ 1 & – \lambda & -3 \\ 1 & 1 & \lambda \end{pmatrix}$ $R_{1} \leftrightarrow R_{2}$ $\begin{pmatrix} 1 & – \lambda & -3 \\ – \lambda & 1 & -3 \\ 1 & 1 & \lambda \end{pmatrix}$ $R_{1} \lambda + R_{2} \to R_{2} ,R_{3}- R_{1} \to R_{3}$ $\begin{pmatrix} 1 & – \lambda & -3 \\ 0 & 1 – {\lambda}^2 & -3-3 \lambda \\ 0 & 1 + \lambda & 3+ \lambda \end{pmatrix}$

Now see when the determinant of A =0 as then the Rank(A) will be <3 and then it has infinitely many solutions

Det(A)= $|A|=0 \Rightarrow (1- {\lambda}^2)(3+ \lambda)+{(1+ \lambda)}^2 3=0 \Rightarrow (1+ \lambda)( \lambda +2)( \lambda -3 )=0$ $\Rightarrow \lambda =-1 ,-2 ,3$

So, if $\lambda=-1,-2,3$ then Rank(A) <3 hence it has infinitely many solutions

Now from here we can say that if $\lambda \ne -1,-2,3$ then Rank (A) =3 then the system of homogeneous equations has only trivial solution i.e $x_{1}=x_{2}=x_{3}=0$

For $\lambda =-1$ system of homogeneous equation is as follows ,

$x_{1} + x_{2} – 3 x_{3}=0$
$x_{1}+x_{2} -3 x_{3} =0$
$x_{1}+x_{2}-x_{3} =0$

Solving this we get $x_{2}=-x_{1}$ and $x_{3}=0$ . Hence solution space is {$x_{1}(1,-1,0)$} , $x_{1} \epsilon \mathbb{R}$ .

Similarly , for $\lambda =-2,3$ we have solution space { $x_1(1,1,0)$ } and { $x_1 (1,1, -2/3 )$ } respectively .

Therefore , real solutions (x1,x2,x3,λ) for the system of equations are $(0,0,0, \lambda )$ , $\lambda \epsilon \mathbb{R}$ and $(x_{1},-x_{1},0,-1)$, $(x_1,x_1,0,-2) ,(x_1,x_1, \frac{-2}{3} x_1)$$x_{1} \epsilon \mathbb{R}$ .