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 2008 Problem 10
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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 2008 Problem 10
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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.

Prerequisites


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 .


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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


ISI MStat PSB 2008 Problem 10
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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 .


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Eigen Values of a Matrix : IIT JAM 2016 Problem Number 13

What are Eigen Values of a Matrix?


The Eigen values of a matrix are the roots of its characteristic equation.

Try the problem from IIT JAM 2016


The largest eigen value of the matrix

$A = \begin{bmatrix}
1 & 4 & 16 \\[0.3em]
4 & 16 & 1 \\[0.3em]
16 & 1 & 4 \\
\end{bmatrix}$ is

$\textbf{(A)}\qquad 16\qquad \textbf{(B)}\qquad 21\qquad \textbf{(C)}\qquad 48\qquad \textbf{(D)}\qquad 64\qquad $

IIT JAM 2016 Problem Number 13

Finding the largest eigen value of a matrix

6 out of 10

Higher Algebra : S K Mapa

Knowledge Graph- Eigen values of a matrix


Eigen value of matrix knowledge graph

Use some hints


In order to find the largest eigen value of the given matrix we have to find all the eigen values of the given matrix, then we can find the largest among them. Now it is very easy to find the eigen values. Give it a try!!!

Now the process of rank starts with finding the characteristics polynomial of the given matrix, i.e., $\textbf{ det }(A-\lambda I)=0$. We have to find the value of this $\lambda $ which is the eigen value. Now it is very easy to find the determinant of $(A-\lambda I)$. Try to cook this up.

So, now let's do some calculation.

$ \textbf{det}(A-\lambda I) = \left| \begin{matrix}
1-\lambda & 4 & 16 \\[0.3em]
4 & 16-\lambda & 1 \\[0.3em]
16 & 1 & 4 -\lambda \\
\end{matrix} \right| =0 $

Now we are half way done just our calculation part remains.

$|A-\lambda I|=(1-\lambda)[64-20 \lambda + \lambda ^2-1]-4[16-4 \lambda -16]+16[4-256+16 \lambda]=0$

$\Rightarrow \quad (1- \lambda )( \lambda ^2-20 \lambda +63)+16 \lambda +256 \lambda -4032=0$

$\Rightarrow \quad \lambda ^2-20 \lambda +63 - \lambda ^3+20 \lambda ^2-63 \lambda +272 \lambda -4032=0$

$\Rightarrow \quad - \lambda ^3+21 \lambda ^2+189 \lambda -3969=0$

$ \Rightarrow \quad \lambda ^3-21 \lambda ^2-189 \lambda +3969=0 $

$\Rightarrow \quad(\lambda-21)(\lambda^2-189)=0$

Therefore $\lambda=21, \quad -13.747727, \quad 13.747727$

So from here we can clearly see

$\lambda=21$ is the largest among all the eigen values.

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Application of eigenvalue in degree 3 polynomial: ISI MMA 2018 Question 14

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]Let A be a 3 x 3 real matrix with all the diagoanl entries equal to 0 . If 1 + i is an eigen value of A , the determinant of A equal ?[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]Sample Questions (MMA ) :2019[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Linear Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Schaum's Outline of Linear Algebra[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]Let A be a 3 x 3 real matrix with trace 0 Now( 1+i) be an eigen value . (Eigenvalues are a special set of scalars associated with a linear system of equations (i.e., a matrix equation) that are sometimes also known as characteristic roots, characteristic values, proper values, or latent roots & Eigen vectors are In linear algebra,) (An eigenvector  or characteristic vector of a linear transformation is a non-zero vector that changes by only a scalar factor when that linear transformation is applied to it. More formally, if T is a linear transformation from a vector space V over a field F into itself and v is a vector in V that is not the zero vector, then v is an eigenvector of T if T(v) is a scalar multiple of v. This condition can be written as the equation ) [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]So ,( 1 + i) be the roots of the characteristic poly of A  Now A is a real matrix, char poly of A  \(\epsilon \mathbb{R}[x]\) [Right!] Therefore ( 1 - i) is also a root of char poly of A[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]deg( char poly of A ) =3 So , it has two imaginary roots & one real root Let real root be r Note tr(A) = 0 => r + 1 +i +1 -i = 0 , => r = -2 Can you play with the determinants now ?[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]We know the determinant of A is the product of all eigen values  (-2)(1+i)(1-i) = detA => det(A) = -2[1 +1] = -4[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="3" _address="0.1.0.3"]

Watch the video

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TIFR 2014 Problem 11 Solution - Nilpotent Matrix Eigenvalues


TIFR 2014 Problem 11 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
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Problem:


Let (A) be an (nxn) matrix with real entries such that (A^k=0) (0-matrix) for some (k\in\mathbb{N}).

Then

A. A has to be the 0-matrix.

B. trace(A) could be non-zero.

C. A is diagonalizable.

D. 0 is the only eigenvalue of A


Discussion:


Let (v) be an eigenvector of (A) with eigenvalue (\lambda).

Then (v \neq 0) and (Av=\lambda v).

Again, (A^2 v=A(Av)=A(\lambda v)=\lambda Av= (\lambda)^2v).

We continue to apply A, applying it k times gives: (A^k v=(\lambda)^k v).

By given information, the left hand side of the above equality is 0.

So (\lambda^k v=0) and remember (v \neq 0).

So (\lambda =0).

Therefore (0) is the only eigenvalue for (A).

So D is true.

We analyse the question a little bit further, to check it satisfies no other options above.

We know (trace(A)=) sum of eigenvalues of A= (\sum 0 =0)

So option B is false.

Take (A=\begin{bmatrix} 0 & 1 // 0 & 0 \end{bmatrix} ).

Then (A^2 =0). But (A) is not the zero matrix.

Also, if (A) were diagonalizable then the corresponding diagonal matrix would be the zero matrix. Which would then imply that (A) is the zero matrix, which in this case it is not. (See  TIFR 2013 Probmem 8 Solution-Diagonalizable Nilpotent Matrix ) So this disproves options A and C.


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TIFR 2013 Problem 38 Solution -Eigenvalue of differentiation

TIFR 2013 Problem 38 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
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Problem Type:True/False?

Let (V) be the vector space of polynomials with real coefficients in variable (t) of degree ( \le 9). Let (D:V\to V) be the linear operator defined by (D(f):=\frac{df}{dt}). Then (0) is an eigenvalue of (D).


Hint:

If 0 were an eigenvalue, what would be its eigenvector?


Discussion:


There are several ways to do this. One possible way is to find out the matrix representation of (D) with respect to standard basis ( {1,t,t^2,...,t^n})( or any other basis) and observe that it is a (strictly) upper triangular matrix with all diagonal entries 0 and therefore the determinant of (D) is 0. This implies that D is not injective, so there is some nonzero vector to which when (D) is applied gives the zero vector. Therefore, (D) has 0 eigenvalue.

Another way to do this is by observing that (D(1)=0(1)), therefore 0 is an eigenvalue of (D) with 1 as an eigenvector.


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