in Linear Algebra edited by
5,824 views
25 votes
Let A be the $ 2 × 2 $ matrix with elements $a_{11} = a_{12} = a_{21} = +1 $ and $ a_{22} = −1 $ . Then the eigenvalues of the matrix $A^{19}$ are

  1. $1024$ and $−1024$
  2. $1024\sqrt{2}$ and $−1024 \sqrt{2}$
  3. $4 \sqrt{2}$ and $−4 \sqrt{2}$
  4. $512 \sqrt{2}$ and $−512 \sqrt{2}$
in Linear Algebra edited by
by
5.8k views

Subscribe to GO Classes for GATE CSE 2022

4 Answers

52 votes
 
Best answer

Characteristic Equation is $ |A  -\mu I| = 0$

$ \begin{vmatrix} 1-\mu & 1\\1 &-1-\mu\end{vmatrix} = 0$

$\implies (1-\mu)(-1-\mu)-1=0$

$\implies -1-\mu+\mu+\mu^{2}-1=0$

$\implies \mu^{2}-2=0$

$\implies \mu=+\sqrt 2 $ and $-\sqrt 2$

According to properties of Eigen values,

eigen values of $A^{19}=$ (eigen value of A)$^{19}$

$=(\sqrt 2)^{19}$ and $(-\sqrt 2)^{19}$

Hence, Ans is option (D).

edited by

2 Comments

Even though I got the answer but if we try to convert the above matrix into an upper triangular matrix, then we get the eigenvalues as -2, 1. So how is that possible??
0
Because Elementary row operations change eigenvalues of a given matrix.
2
6 votes

Eigenvalues Of Matrix Powers

Suppose A is a square matrix,λ is an eigenvalue of A, and s≥0 is an integer. Then λs is an eigenvalue of As.

http://linear.ups.edu/html/section-PEE.html

1 vote

At the time of solving I just forgot this property & get two method, one method is working fine but other is not,but can't understand why the other method is not working.

1st Method:

By using Cayley-hamilton Theorem from the characteristic eqn I get $\lambda ^{2} - 2 = 0$ which is $A^{2} - 2 = 0$

$A^{2} = 2I \Rightarrow ({A^{2}})^9 = 2^{9}I$

i.e $A^{18}=512I$,   Now post-multiplied by A on both side $A^{18}.A=512I.A$.  which becomes $A^{19}=\begin{bmatrix} 512 &512 \\ 512&-512 \end{bmatrix}$

computing eigen values here we'll get $512\sqrt2$ & $-512\sqrt2$  

This method is ok & it's working.

 

2nd method:

finding a pattern to compute A^19 which 1st compute A^2 by A*A , then A^3 by A^2 * A, then A^4 by A^3 * A

by this we'll get a pattern like,

$A^{n} =\begin{bmatrix} n &0 \\ 0 &n \end{bmatrix}$  when n is even 

$A^{n} =\begin{bmatrix} n-1 &n-1 \\ n-1 &1-n \end{bmatrix}$  when n is odd

likewise $A^{19} =\begin{bmatrix} 18 &18 \\ 18 &-18 \end{bmatrix}$

but computing eigen values on $A^{19}$ will give $-18\sqrt2$  &  $18\sqrt2$

 

don't know why this 2nd method is not working,Anyone please give a reason why this is not working.

2 Comments

$2^{nd}$ method is not working because your pattern is not correct. Just calculate manually, $A^6$ ,  $A^{10}$ , , $A^{15}$ etc. your pattern is working for small $n$. if you assume something is correct then check whether it is actually correct or not for every 'n' using induction. Here, using inductive hypothesis, you assume, it is true for 'n', then you have to prove that it will follow the same pattern for n+1, n+2,... then it will be correct.
1
1st one is preferable
0
0 votes
Its based on the property of eigen values . If A is a matrix and has eigen values x1,x2,,,,,,,xn then eigen values of A^n will x1^n,x2^n......xn^n

So for given question eigen value of A is 1.414 so we have to find (1.414)^19 which gives option d.

eNJOY~!
Answer:

Related questions

Ask
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