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} $.

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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.

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Rank:IIT JAM 2018 PROBLEM 9

[et_pb_section fb_built="1" _builder_version="4.1"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][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_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Let's Try A Warm Up MCQ

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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 $latex \mathbb{R} $ of the polynomial functions of degree less than or equal to 3 defined on $latex \mathbb{R} $. Let $latex 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"]

Start with hints

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[/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 $latex \beta= \{e_1,....,e_n\} $ of the vector space $latex V $. Step 2: Write down the matrix $latex [T]_{\beta}^{\beta} $ Step 3: Calculate the rank of the matrix $latex [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 $latex V $ is $latex \{1,x,x^{2},x^{3}\} = \beta$ $latex (Tf) (x) =f(x) - xf^{'}(x)$ $latex (T1) (x) = 1 - 0 = 1$; $latex (Tx) (x) = x - x = 0$; $latex (T x^{2}) (x)= x^{2} - 2x^{2} = -x^{2}$ ; $latex (T x^{3}) (x) = -2x^{3} $ So, $latex [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 $latex 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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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

What is this topic: Linear Algebra
What are some of the associated concept: Trace,Rank-Nullity Theorem
Book Suggestions: Linear Algebra done Right by Sheldon Axler

ISI MStat PSB 2006 Problem 1 | Inverse of a matrix

Problem- ISI MStat PSB 2006 Problem 1

Prerequisites

Solution :

Food For Thought

Similar Problems and Solutions

Outstanding Statistics Program with Applications

ISI MStat PSB 2007 Problem 2 | Rank of a matrix

Problem- ISI MStat PSB 2007 Problem 2

Prerequisites

Solution :

Food For Thought

Similar Problems and Solutions

Outstanding Statistics Program with Applications

Rank:IIT JAM 2018 PROBLEM 9

Let's Try A Warm Up MCQ

Understand the problem

Start with hints

Watch the video

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Similar Problems

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

Problem Type:True/False?

Hint:

Discussion:

HELPDESK

Problem- ISI MStat PSB 2006 Problem 1

Prerequisites

Solution :

Food For Thought

Similar Problems and Solutions

Related Program

Outstanding Statistics Program with Applications

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Problem- ISI MStat PSB 2007 Problem 2

Prerequisites

Solution :

Food For Thought

Similar Problems and Solutions

Related Program

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube

Let's Try A Warm Up MCQ

Understand the problem

Start with hints

Watch the video

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Similar Problems

Problem Type:True/False?

Hint:

Discussion:

HELPDESK