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 .


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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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 2008 Problem 10
Outstanding Statistics Program with Applications

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!

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?

Prerequisites

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.

Triagles and vectors

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?

Triangles and Parallelograms

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?

Triangle and rectangle

Step 5

A Square!

Triangle and 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

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 \( log(ad) - log(bc)\).

Data

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 \(ad -bc\).

Share your ideas here. I will write in more details about this phenemenon.

Stay Tuned! Stay Blessed!

Definite Integral & Expansion of a Determinant |ISI QMS 2019 |QMB Problem 7(a)

Try this beautiful problem from ISI QMS 2019 exam. This problem requires knowledge of expansion of determinant and definite integral.

Definite Integral and Expansion of Determinant - ISI QMS 2019 (QMB Problem 7a)


Let $f(x)= \begin{vmatrix}\sec x & \cos x & \sec^2 x+\cot x\textrm{cosec}^2 x\\ \cos^2x & \cos^2 x & \textrm{cosec}^2x \\ 1 & \cos^2 x & \textrm{cosec}^2 x \end{vmatrix}$. Then find $\displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(x) \mathrm d x$

Key Concepts


Expansion of Determinant

Evaluation of Definite Integral

Check the Answer


Answer: $1-\frac{1}{\sqrt{2}}-\frac{\pi}{8}-\frac12\log 2$

ISI QMS 2019 (QMB Problem 7a)

Secrets in Inequalities.

Try with Hints


Way to proceed with this problem is obtaining a simpler form of $f(x)$ by expanding the given determinant and then integrate it within the given limit.

You can do this ... Give it a try !!!!

Expanding the given determinant :

$f(x)= \begin{vmatrix}\sec x & \cos x & \sec^2 x+\cot x\textrm{cosec}^2 x\\ \cos^2x & \cos^2 x & \textrm{cosec}^2x \\ 1 & \cos^2 x & \textrm{cosec}^2 x \end{vmatrix}$

$=\textrm{sec} x[\cos^2 x \textrm{coesc}^2 x - \cos^x \textrm{coesc}^2 x ]-\\ \quad \cos x [\cos^2 x-\textrm{cosec}^2 x- \textrm{coesc}^2 x]+ \\ \quad (\textrm{sec}^2 x+\textrm{cot} x \textrm{coesc}^2 x)[\cos^4 x-cos^2 x]$

$=0+\cos x[ \textrm{coesc}^2 x (1-cos^2 x)]- \\ \quad (\textrm{sec}^2 x+\textrm{cot} x \textrm{coesc}^2 x)[\cos^2 x(1-\cos^2 x)] $

$=\cos x[\textrm{cosec}^2 x . \sin^2 x]- \\ \quad (\textrm{sec}^2 x+\textrm{cot} x \textrm{coesc}^2 x)[\cos^2x . \sin^2 x] $

$=\cos x-[\sin^2 x+\textrm{cot}x\cos^2 x]$

Therefore $f(x)= \cos x-\sin^2 x-\textrm{cot}x\cos^2 x $

Now integrating $f(x)$ within the given limits we will get our answer.

Integrating $f(x)$ within the given limits :

$I=\displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(x)\quad \mathrm d x\\= \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \quad\mathrm d x - \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin^2 x \quad\mathrm d x - \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \textrm{cot}x\cos^2 x \quad\mathrm d x \\= I_1-I_2-I_3 \text{[say]}$

Evaluating $I_1$ :

Now $I_1 = \displaystyle\int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \quad \mathrm d x = \sin x \bigg|_{{\pi}/{4}}^{{\pi}/{2}}= [\sin (\frac{\pi}{2})-\sin (\frac{\pi}{4})]=[1-\frac{1}{\sqrt 2}] $

Now can you find the values of $I_2 \text{and} I_3$ ?

Evaluating $I_2$ :

$I_2= \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin^2 x \quad \mathrm d x$

$\quad = \frac12\displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} 2\sin^2 x \quad \mathrm d x $

$\quad = \frac12\displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}[1-\cos 2x]\quad\mathrm d x $

$\quad= \frac12\bigg[x-\frac {\sin2x}{2}\bigg]_{ \frac{\pi}{4}}^{\frac{\pi}{2}}$

$\quad =\frac12[\frac{\pi}{2}-\frac{\sin \pi}{2}-\frac{\pi}{4}+\frac{\sin \frac{\pi}{2}}{2}] $

$\quad= \frac{\pi}{8}+\frac14$

Evaluating $I_3$ :

$I_3= \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \textrm{cot} x\cos^2 x \mathrm d x$

$\quad= \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos x(1-\sin^2 x)}{\sin x} \mathrm d x$

$\quad = \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos x}{\sin x} \mathrm d x \quad- \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin x . \cos x \quad \mathrm d x $

$\quad = \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\mathrm d (\sin x)}{\sin x}\quad- \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin x\quad \mathrm d (\sin x) $

$\quad = \bigg[\log (\sin x)\quad-\quad \frac{\sin^2 x}{2}\bigg]_{\frac{\pi}{2}}^{\frac{\pi}{2}}$

$\quad= \bigg[\log (\sin \frac{\pi}{2})-\frac{\sin^2 \frac{\pi}{2}}{2}-\log (\sin \frac{\pi}{4}) + \frac{\sin^2 \frac{\pi}{4}}{2}\bigg]$

$\quad =\bigg[0-\frac12-\log \frac{1}{\sqrt 2}+\frac14\bigg]$

$\quad = \bigg[-\frac12 -\log (1) + \log (\sqrt 2) +\frac14 \bigg]$

$\quad = \bigg[\frac12\log 2- \frac14 \bigg]$

Therefore,

$\quad I= I_1-I_2-I_3$

$\Rightarrow I = 1-\frac{1}{\sqrt 2}-\frac{\pi}{8}-\frac14-\frac12\log(2) +\frac14$

$\Rightarrow I = 1 -\frac{\pi}{8}-\frac{1}{\sqrt 2}-\frac12\log(2)$ [ANS]

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