# Test of Mathematics Solution Subjective 62 - System of Equations

This is a Test of Mathematics Solution Subjective 62 (from ISI Entrance).

The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

## Problem

Consider the system of equations x + y = 2, ax + y = b. Find conditions on a and b under which
(i) the system has exactly one solution;
(ii) the system has no solution;
(iii) the system has more than one solution.

## Solution

### Key Idea

Solution to the linear equations

$$a_{11} x + a_{12} y = b_1 \\ a_{21} x + a_{22} y = b_2$$

$$a_{11} \times a_{22} - a_{12} \times a_{21} \neq 0$$ implies there is a unique solution.

$$a_{11} \times a_{22} - a_{12} \times a_{21} = 0$$ implies there is either no solution or infinitely many solution.

No solution if: $$\frac{a_{11}}{a_{21} } = \frac{a_{12}}{a_{22}} \neq = \frac{b_1}{b_2}$$; infinitely many solutions other wise.

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# 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 !!

# 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 !!

# 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 !!

# 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 !! ]

# 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 $$(x_{1},-x_{1},0,-1)$$, $$(x_1,x_1,0,-2) ,(x_1,x_1, \frac{-2}{3} x_1)$$$$x_{1} \epsilon \mathbb{R}$$ .

# Data, Determinant and Simplex

This is a beautiful problem connecting linear algebra, geometry and data. Go ahead and delve into the glorious connection.

## Problem

Given a matrix $$\begin{bmatrix}a & b \\c & d \end{bmatrix}$$ with the constraint $$1 \geq a, b, c, d \geq 0; a + b + c + d = 1$$, find the matrix with the largest determinant.

Is there any statistical significance behind this result?

## Solution ( Geometrical )

#### Step 1

Take two vectors $$v = (a,c) and w = (b,d)$$ such that their addition lies on $$v +w lies on x + y = 1$$ line. Now, we need to find a pair of vectors {$$v, w$$}such that the area formed by these two vectors is maximum.

#### Step 2

Rotate the parallelogram such that CF lies on the X - axis.

Now, observe that this new parallelogram has an area same as the initial one. Can you give a new parallelogram with a larger area?

#### Step 3

Just extend the vertices to the end of the simplex OAB. Observe that the new parallelogram has a larger area than the initial parallelogram. Is there any thing larger?

#### Step 4

Now, extend it to a rectangle. Voila! It has a larger area. Now therefore, given any non rectangular parallelogram we can find a rectangle with a larger area than the parallelogram. So, let's search in the region of rectangles. What do you guess is the answer?

#### Step 5

A Square!

Let the rectangle has length $$x, y$$ and area $$xy$$. Now, observe that $$xy$$ is maximized with respect to $$x+y = 1$$ when $$x = y = \frac{1}{2}$$. [Use AM - GM Inequality].

So, $$v = (0,\frac{1}{2})$$ and $$w = (\frac{1}{2},0)$$ maximizes the determinant.

## Challenge 1

Prove it using algebraic methods borrowed from this geometrical thinking. Your solution will be put upon here.

## Challenge 2

Can you generalize this result for $$n \times n$$ matrices? If, yes prove it. Just algebrify the steps.

## Statistical Significance

Lung Cancer and Smoker Data

Observe that that if, we divide every thing by 1000, we get a matrix.

So, the question is about association of Smoking and Lung Cancer. Given these 1000 individuals let's see how the distribution of the numbers result in what odd ratio?

For the categorical table data $$\begin{bmatrix}a & b \\c & d \end{bmatrix}$$ the odd's ratio is defined as $$\frac{ad}{bc} = \frac{det(\begin{bmatrix}a & b \\c & d \end{bmatrix})}{bc} + 1$$

The log odd's ratio is defined as $$log(ad) - log(bc)$$.

Observe the above data, observe that Log Odd's Ratio is almost behaving like the determinant. When $$X = 1$$ and $$X = 0$$ depend on Y uniformly, no information of dependence is released. Hence, Log Odd's Ratio is 0 and so is the Determinant.

Try to understand, why the Log Odd's ratio is behaving same as Odd's Ratio?

$$log(x)$$ is increasing and so is $$x$$ hence, $$log(ad) - log(bc)$$ must have the same nature as $$ad -bc$$.

Stay Tuned! Stay Blessed!

# Eigen Value of a matrix | IIT JAM 2017 | Problem 58

Try this problem from IIT JAM 2017 exam (Problem 58) and know how to evaluate Eigen value of a Matrix.

## Eigen value of a Matrix | IIT JAM 2017 | Problem 58

Let $\alpha, \beta, \gamma, \delta$ be the eigenvalues of the matrix
$$\begin{bmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{bmatrix}$$
Then find the value of $\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}$

### Key Concepts

Linear Algebra

Matrix

Eigen Values

Answer: $6$

IIT JAM 2017 , Problem 58

## Try with Hints

Characteristic Equation of a matrix $A$ of order $n$ is defined by $|A-xI|=0$, where $x$ is scalar and $I$ is the identity matrix of order $n$.

The roots of the characteristic equation are called the Eigen Values of that matrix.

Now it is easy. Give it a try.

Let

$$A=\begin{bmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{bmatrix}$$

Then its characteristic equation is $|A-xI|=0$ for some scalar $x$ and $I$ the identity matrix of order $4$

i.e., $\begin{vmatrix} -x & 0 & 0 & 0 \\ 1 & -x & 0 & -2 \\ 0 & 1 & -x & 1 \\ 0 & 0 & 1 & 2-x \end{vmatrix} =0$

$\Rightarrow -x\begin{vmatrix} -x & 0 & -2 \\ 1 & -x & 1 \\ 0 & 1 & 2-x \end{vmatrix} =0$

$\Rightarrow -x[-x\{(-x)(2-x)\}-1]-2(1-0)]=0$

$\Rightarrow x^{4}-2 x^{3}-x^{2}+2 x=0$

$\Rightarrow x(x^{3}-2 x^{2}-x+2)=0$

$\Rightarrow x(x-1)(x-2)(x+1)=0$

Then the eigen values are : $0,1,-1,2$

Then, $\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2} =0^2+1^2+(-1)^2+2^2=6$ [ANS]

# Minimal Polynomial of a Matrix | TIFR GS-2018 (Part B)

Try this beautiful problem from TIFR GS 2018 (Part B) based on Minimal Polynomial of a Matrix. This problem requires knowledge of linear algebra.

## Minimal Polynomial of a matrix - TIFR GS- Part B (Problem 10)

The minimal polynomial of $\begin{pmatrix} 2 &1& 0&0 \\ 0 &2&0&0 \\ 0&0&2&0\\ 0&0&1&5\end{pmatrix}$ is

• $(x-2)(x-5)$
• $(x-2)^2(x-5)$
• $(x-2)^3(x-5)$
• None of these

### Key Concepts

Linear Algebra

Matrix / Vector Space

Characteristic Polynomial

Answer: $(x-2)^2(x-5)$

TIFR GS -2018 (Part- B) | Problem No 10

Graduate Texts in Mathematics : Springer-Verlag

## Try with Hints

Some Definitions and Results Needed :

1. Monic Polynomial : A polynomial is said to be monic if the coefficient of the highest degree term is 1.
2. Characteristic Polynomial of a matrix : Let $A$ be a square matrix of order $n$ then the polynomial $|A-\lambda I_n|$ is called its characteristic polynomial and $|A-\lambda I_n|=0$ is called the characteristic equation. [$I_n$ is the identity matrix of order $n$]
3. Cayley - Hamilton Theorem : If $p(\lambda)$ is the characteristic polynomial of an $n\times n$ matrix $A$ over a field $F$, then the matrix $A$ satisfies the equation $p(x)=0$, i.e., $p(A)=0$. In other words, every square matrix satisfies its own characteristic equation.
4. Minimal Polynomial : The monic polynomial of lowest degree satisfied by a square matrix $A$ is called its minimal polynomial
5. Let $p(\lambda)$ and $m(\lambda)$ be the characteristic and minimal polynomials of a square matrix $A$ of order $n$ respectively. Then either both $p(\lambda)$ and $m(\lambda)$ are of degree $n$ or $m(\lambda)$ is a factor of $p(\lambda)$ .
6. Minimal polynomial is unique.

So all the ingredients you need to cook the problem are given... Can you make it delicious ?

To find the Characteristic equation of the given matrix :

Let $A= \begin{pmatrix} 2 &1& 0&0 \\ 0 &2&0&0 \\ 0&0&2&0\\ 0&0&1&5\end{pmatrix}$

Then, $|A-\lambda I_4|= \begin{vmatrix} 2-\lambda &1& 0&0 \\ 0 &2-\lambda &0&0 \\ 0&0&2-\lambda &0\\ 0&0&1&5-\lambda \end{vmatrix}$

$\quad = (2-\lambda)^3(5-\lambda) = p(\lambda)$ [say]

then, the characteristic equation of $A$ is $p(x)=(x-2)^3(x-5)=0$

Then By Cayley Hamilton Theorem $(A-2I_4)^2(A-5I_4)=O_{4 \times 4}$ [$O_{4 \times 4}$ is the NULL MATRIX of order $4$]

As minimal polynomial is unique then if minimal polynomial is a polynomial of degree $4$ it is same as the characteristic polynomial by Property 5 in the first hint and if minimal polynomial is less than degree $4$ then it is a factor of characteristic polynomial.

all the factors of characteristic polynomial are :

$p_1(x)=(x-2)(x-5),\quad p_2(x)=(x-2)^2(x-5),\\ \text{ and } p(x)=(x-2)^3(x-5)$

Lets Find $p_1(A)$ i.e., $(A-2I_4)(A-5I_4)$ :

$(A-2I_4)(A-5I_4)=\begin{pmatrix} 0 &1& 0&0 \\ 0 &0&0&0 \\ 0&0&0&0\\ 0&0&1&3\end{pmatrix} \times \begin{pmatrix} -3 &1& 0&0 \\ 0 &-3&0&0 \\ 0&0&-3&0\\ 0&0&1&0\end{pmatrix}$

$\quad\quad= \begin{pmatrix} 0 &-3& 0&0 \\ 0 &0&0&0 \\ 0&0&0&0\\ 0&0&0&0\end{pmatrix} \ne O_{4 \times 4}$

Now, $p_2(A)$ i.e., $(A-2I_4)^2(A-5I_4)$ :

$(A-2I_4)^2(A-5I)=\begin{pmatrix} 2 &1& 0&0 \\ 0 &2&0&0 \\ 0&0&2&0\\ 0&0&1&5\end{pmatrix}^2 \times \begin{pmatrix} -3 &1& 0&0 \\ 0 &-3&0&0 \\ 0&0&-3&0\\ 0&0&1&0\end{pmatrix}$

$\quad\quad= \begin{pmatrix} 0 &0& 0&0 \\ 0 &0&0&0 \\ 0&0&0&0\\ 0&0&3&9\end{pmatrix} \times \begin{pmatrix} -3 &1& 0&0 \\ 0 &-3&0&0 \\ 0&0&-3&0\\ 0&0&1&0\end{pmatrix} = O_{4 \times 4}$

Therefore the lowest degree monic polynomial satisfied by $A$ is $(x-2)^2(x-5)$.

Hence the minimal polynomial is $(x-2)^2(x-5)$

# Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.1" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Consider the vector space V over $\mathbb{R}$ of the polynomial functions of degree less than or equal to 3 defined on $\mathbb{R}$. Let $T : V \longrightarrow V$ defined by \$latex (Tf)(x) = f(x)-xf'(x). Then the rank of T is  (a) 1  (b) 2 (c) 3 (d) 4 [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.1" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.1"]IIT JAM 2018 Problem 9[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.1" open="off"]Vector Space [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.1" open="off"]Abstract Algebra By S.K Mapa[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.1" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0"][et_pb_tab title="Hint 0" _builder_version="4.1" hover_enabled="0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.1" hover_enabled="0"]Rank(T) = dim(Range(T)) There is one easy way to calculate rank of every linear transformation. Step 1:  Take by basis  $\beta= \{e_1,....,e_n\}$ of the vector space $V$. Step 2: Write down the matrix $[T]_{\beta}^{\beta}$ Step 3: Calculate the rank of the matrix  $[T]_{\beta}^{\beta}$ Now can you follow these steps to get the answer?  [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1" hover_enabled="0"]Standard Basis of $V$ is $\{1,x,x^{2},x^{3}\} = \beta$ $(Tf) (x) =f(x) - xf^{'}(x)$ $(T1) (x) = 1 - 0 = 1$; $(Tx) (x) = x - x = 0$; $(T x^{2}) (x)= x^{2} - 2x^{2} = -x^{2}$ ; $(T x^{3}) (x) = -2x^{3}$ So, $[T]_{\beta}^{\beta} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -2 \\ \end{pmatrix}$ Hence the rank is $3$[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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