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 .

Similar Problems and Solutions

Related Program

ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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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

Similar Problems and Solutions

Related Program

ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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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 .

Similar Problems and Solutions

Related Program

ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

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Trace & Determinant | ISI MStat 2017 Problem 1 | PSB

This is a beautiful problem from ISI MStat 2017 PSB Problem 1 based on matrices. We provide a detailed solution with the prerequisites mentioned explicitly.

Problem - ISI MStat 2017 Problem 1

Let \(a\) and \(b\) be real numbers. Show that there exists a unique \(2 \times 2\) real symmetric matrix \(A\) with \({trace}(A)=a\) and \( Det(A)=b\) if and only if \(a^{2}=4 b\) .

Prerequisites

Solution

Let , \( A = \begin{bmatrix} x & z \\ z & y \end{bmatrix} \) be a unique \(2 \times 2\) real symmetric matrix , with x,y,z belongs to real .

Given , trace(A)=\( a \Rightarrow (x+y)= a\) ----(1)and

Det(A)=\( b \Rightarrow (xy-z^2)=b \) ----(2)

1st solution : \( x=a-y \) from (1) putting this in (2) we get

\( y(a-y)-z^2=b \Rightarrow ay-y^2-z^2=b \Rightarrow y^2-ay+(b+z^2)=0 \)

Now using , Sridhar Acharya Formula we get ,

\( y= \frac{a \pm \sqrt{a^2-4(b+z^2)}}{2} \)

Now as given , A is unique so y can't take two different values hence this part \(\sqrt{a^2-4(b+z^2)}\) must be zero

i.e \( \sqrt{a^2-4(b+z^2)}=0 \) \( \Rightarrow z= \pm { \sqrt{\frac{a^2-4b}{4}}} \)

Again , as A is unique matrix z can't take two different values . Hence z must be equal to zero

i.e \( z=0 \Rightarrow \sqrt{\frac{a^2-4b}{4}}=0 \Rightarrow a^2=4b \) ( Hence proved)

2nd solution : See if we interchange x by y then all the properties of A remains same i.e trace and determinant . But it can't be possible as we assume that A is unique hence x and y must be equal i.e \( x=y \) .

From (1) we get \( x=y=\frac{a}{2} \)

Again if we interchange z by -z then all the properties of A remains same i.e trace and determinant . But it can't be possible as we assume that A is unique hence z and -z must be equal i.e \( z=-z \Rightarrow z= \frac{1}{2} \) .

From (2) we get \( (xy-z^2)=b \Rightarrow \frac{a^2}{4}-\frac{1}{4} = b \Rightarrow a^2=4b \) (Hence proved )

Now , we will assume that \( a^2=4b\) , where a and b are real and show that the matrix is unique .

Let , \( A = \begin{bmatrix} x & z \\ z & y \end{bmatrix} \) be a \(2 \times 2\) real symmetric matrix with x,y,z belongs to real

Given trace(A)=\( a \Rightarrow (x+y)=a \)---(3)

and Det(A)=\( b \Rightarrow xy-z^2=b\) ---(4)

Another thing is given that \( a^2=4b \)

So using (3) and (4) in (5) we get ,

\( {(x+y)}^2=4(xy-z^2) \Rightarrow x^2+2xy+y^2-4xy+4z^2=0 \Rightarrow {(x-y)}^2+{(2z)}^2=0 \)

i.e sum of two squares is equal to 0 which implies individual squares are equal to zero .

Hence , \( {(x-y)}^2 =0 \) and \( {(2z)}^2=0 \) \( \Rightarrow x=y \) and \( z=0 \)

which give \( x=y=\frac{a}{2} \) and \(z=0\)

Hence , \( A= \begin{bmatrix} \frac{a}{2} & 0 \\ 0 &\frac{a}{2} \end{bmatrix} \) is unique ( proved )

we have proved both if and only if part . Hence we are done!

TIFR 2013 Problem 39 Solution - Rank and Trace of Idempotent matrix

TIFR 2013 Problem 39 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.
The image is a front cover of a book named Linear Algebra Done Right by Sheldon Axler. This book is very useful for the preparation of TIFR Entrance.

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Problem Type:True/False?

If (A) is a complex nxn matrix with (A^2=A), then rank(A)=trace(A).

Hint:

What are the eigenvalues of (A)? What is trace in terms of eigenvalues?

Discussion:

If (v) is an eigenvector of (A) with eigenvalue (\lambda) then (Av=\lambda v), therefore (\lambda v=Av=A^2v=\lambda Av =\lambda^2 v). Therefore, since any eigenvector is non-zero, (\lambda =0 or 1 ).

Sum of eigenvalues is trace of the matrix. So, trace(A)= number of non-zero eigenvalues= total number of eigenvalues - number of 0 eigenvalues

Since (A) satisfies the polynomial (x^2-x), the minimal polynomial is either (x) or (x-1) or (x(x-1)). This means the minimal polynomial breaks into distinct linear factors, so (A) is diagonalizable. Therefore, the algebraic multiplicity of an eigenvalue is same as its geometric multiplicity.

In total there are n eigenvalues (for A is nxn) and the number of 0-eigenvalues is the algebraic multiplicity of 0, which is same as the geometric multiplicity of 0, i.e, the dimension of the kernel of A.

Therefore, trace(A)(=n-)nullity(A).

By the rank-nullity theorem, the right hand side of the above equation is rank(A).

HELPDESK