in Linear Algebra edited by
729 views
12 votes
12 votes

Consider two statements below -

  • Statement $1: $ If $A$ is invertible and $\lambda$ is an eigenvalue of $A$, then $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$.
  • Statement $2:$ Let $A$ be a real skew-symmetric $n \times n$ matrix, i.e., $A^T=-A$. If $\lambda$ is an eigenvalue of $A$, then so is $-\lambda$.

Which of the following option is CORRECT?

  1. Statement $1$ is true but Statement $2$ is false
  2. Statement $2$ is true but Statement $1$ is false
  3. Both statements are true
  4. Both statements are false
in Linear Algebra edited by
729 views

1 comment

Proof of statement 2 by :- Vineeth Rambhiya

A is skew symmetric matrix
A'= -A
let x be eigen vector of A with corresponding eigen value λ
Ax = λx
Apply conjugate on both sides
(Ax)* = (λx)*
A*x* = λ*x*
Since A is real matrix
A* = A
Ax* = λ*x* --- eq1

Ax = λx
Left multiply by transpose of x* on both sides
(Right side multiplication of transpose of x*is not possible)

(x*)'Ax = (x*)'λx
(x*)'Ax = λ(x*)'x
(A'x*)'x=λ(x*)'x
(-Ax*)'x=λ(x*)'x
Using eq1
(-λ*x*)'x=λ(x*)'x

Note that λ is a scalar so λ' and λ* are both also scalar

-λ*(x*)'x=λ(x*)'x

-λ*|x|²=λ|x|²

So -λ*=λ

So if λ eigen value if Skew symmetric matrix A then -λ* [negative of conjugate of λ] is also eigen value of matrix A

From this we can also show that Eigen value of skew symmetric matrices come 0 or purely imaginary

Let λ be a+ib
then -λ*= -a+ib

a+ib = -a+ib
So a = 0

So  λ is purely imaginary or 0

The only skew symmetric matrix that satisfies λ = 0 is zero matrix

Since λ is purely imaginary λ= ib. So -λ* = -ib = -λ
Hence the proof

Some corollary/results from this proof

1. The eigen values of skew symmetric matrices come in pair
λ and -λ* and since λ is purely imaginary -λ*= -λ.
So eigen values come in pair in the form of + λ

2. The eigen values of skew symmetric matrices λ and -λ have same Algebraic multiplicities

1
1

3 Answers

8 votes
8 votes
Statement 1:

we know $Ax= \lambda x$ , where $x $ is the eigenvector corresponding to eigenvalue $\lambda $

As $A$ is invertible then $A^{-1}x = \lambda^{-1}x$ so stmt 1 is $True$

Statement 2:

we know that eigenvalues of skew-symmetric matrix(i.e. $A^{T} = – A$) are pure imaginary

$\lambda = \pm ib$ ,where b is some  constant and they are always in pair

so if $\lambda  = ib$ is an eigen value then  $\bar{\lambda} = – ib = -\lambda$ is also be the eigenvalue

Hence stmt 2 is also $True$
4 votes
4 votes

Statement-1

we are given $Ax= \lambda x$ and we have to prove $A^{-1}x= \frac{1}{\lambda}x$

Multiply by $A^{-1}$ on both sides.

$A^{-1}Ax= \lambda A^{-1}x \Rightarrow$ $Ix= \lambda A^{-1}x$

$\mathbf{ \frac{1}{\lambda}x=A^{-1}x}$, which tells us that $\frac{1}{\lambda}$ is eigen value of $A^{-1}$

Therefore Statement-1 is True.

Statement-2

we are given that $A^T=-A$ and $Ax= \lambda x$

Multiply by $x^T$ on both sides

$x^{T}Ax=\lambda x^{T}x \Rightarrow x^{T}Ax=\lambda |x|^{2}$

We know that $a^Tb=b^Ta$

$x^{T}(Ax) \Rightarrow (Ax)^Tx \Rightarrow x^TA^Tx \Rightarrow -x^TAx$ chip in the value of $Ax$ form first line.

we get $x^{T}(Ax) = -x^T\lambda x $

$x^{T}(Ax) = –\lambda x^T x$ 

$x^{T}(Ax) = –\lambda |x|^2$

But $x^{T}(Ax)=\lambda |x|^2$

which means $\lambda |x|^2=-\lambda |x|^2 \Rightarrow \lambda =-\lambda$

source

Therefore Statement-2 is True. 

by

4 Comments

@rhl what is a and b here?

0
0
a and b are column vectors
0
0

ok, my doubt is in 4th line of explanation of statement 2 did you take transpose of $a^Tb$?

0
0
No, this is a property i am using. $a^Tb=b^Ta$ (dot product of 2 vectors.).

Check the next statement in $x^TAx$ , $x^T$ is $a^T$ and $Ax$ is $b$ and then i am applying the above property.
0
0
4 votes
4 votes

The complete proof of statement 2

A is a skew symmetric matrix

$A' = -A$

Let $x$ be an eigen vector of $A$ with corresponding eigen value $\lambda$

$Ax=\lambda x$

Apply conjugate on both sides

$\begin{align*} \overline{Ax} &=\overline{\lambda x} \\ \bar{A}\bar{x} &=\bar{\lambda}\bar{x} \\ A\bar{x} &= \bar{\lambda}\bar{x} --- eq(1) \end{align*}$

$Ax = λx$

Left multiply by the transpose of $\overline{x}$ on both sides

$\begin{align*} (\bar{x})'(Ax) &=(\bar{x})'\lambda x \\ (A'\bar{x})'x &=\lambda (\bar{x})'x \\ (-A\bar{x})'x &=\lambda (\bar{x})'x \end{align*}$

Using $eq1$

$\begin{align*} {(-A\overline{x})}'x &= \lambda {(\overline{x})}'x\\ {(-\overline{\lambda}\overline{x})}'x &= \lambda {(\overline{x})}'x \\ -\overline{\lambda}{(\overline{x})}'x &= \lambda {(\overline{x})}'x \\ -\overline{\lambda}{(\overline{x})}'x - \lambda {(\overline{x})}'x &= 0\\ (-\overline{\lambda}-\lambda) {(\overline{x})}'x &= 0 \end{align*}$

Since by definition Eigen vectors are not zero, 

$\begin{align*} (-\overline{\lambda}-\lambda) &= 0 \end{align*}$

$\begin{align*} \lambda &= -\overline{\lambda} \end{align*}$

Let $λ$ be $a+ib$

then – $\overline{λ} = -a+ib$

$a+ib = -a+ib$

So $a = 0$

So  $λ$ is purely imaginary or $0$

We have proved an important result here that the eigen values of skew-symmetric matrices is 0 or purely imaginary

The only skew-symmetric matrix that satisfies λ = 0 is the zero matrix

For non-zero skew-symmetric matrices, eigen values are purely imaginary

For any matrix A if $\lambda$ is eigen value corresponding to vector $v$, then $\overline{\lambda}$ is eigen value corresponding to vector $\overline{v}$

$\begin{align*} \lambda &= ib\\ \overline{\lambda} &= -ib = -\lambda \end{align*}$

Hence the proof.

Some corollary/results from this proof

  1. The Eigen value of skew symmetric matrices is either $0$ or pure imaginary
     
  2. The eigenvalues of non zero skew-symmetric matrices come in pair of
      $\lambda$ and $-\lambda$ for eigen vectors  $v$ and $\overline{v}$ respectively
     
  3. The eigenvalues of skew-symmetric matrices λ and -λ have the same Algebraic multiplicities – this is extra info
edited by

2 Comments

how come  −$\lambda$ conjugate = -ib  if $\lambda$=ib
 

as if $\lambda$=ib  then -$\lambda$ = -ib  and conjugateOf(-$\lambda$) = ib
1
1
edited by
It is $\lambda = -(\overline{\lambda})$ as eigen values are purely imaginary in case if non-zero skew-symmetric matrices.
Sorry for the incomplete proof
0
0
Answer:

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true