Problem with calculating eigen vector for 2*2 Matrix

  • Thread starter zak100
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  • #1
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Homework Statement


r1= 2 7
r2=-1 -6

Homework Equations


A-lambda*I=0
(A-lambda*I)*x=0

The Attempt at a Solution


I have got following eigen values:
lambda1 = -5 and lambda2=1

A-lambdaI matrix is:
r1 = 7 7
r2 = -1 -1

and x matrix is:
r1 =x
r2 =y

I cant understand why we have to use augmented matrix in this case.
Some body please guide me.

Zulfi.
 

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Answers and Replies

  • #2
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I cant understand why we have to use augmented matrix in this case.
Probably because it is how it is taught by your teacher. You will have to solve ##\begin{bmatrix}2&7\\-1&-6\end{bmatrix}\cdot \begin{bmatrix}x\\y \end{bmatrix}=\lambda \cdot \begin{bmatrix}x\\y\end{bmatrix}## for ##\lambda \in \{\,-5,1\,\}##. How you manage to do this is your choice. Just make sure, that your procedure will do good work for ##5 \times 5## matrices as well.
 
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  • #3
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6,396

Homework Statement


r1= 2 7
r2=-1 -6

Homework Equations


A-lambda*I=0
(A-lambda*I)*x=0

The Attempt at a Solution


I have got following eigen values:
lambda1 = -5 and lambda2=1

A-lambdaI matrix is:
r1 = 7 7
r2 = -1 -1
Here ##\lambda = -5##.
zak100 said:
and x matrix is:
r1 =x
r2 =y

I cant understand why we have to use augmented matrix in this case.
You don't have to use an augmented matrix here, since the right side of your matrix equation is the zero vector.
In a nutshell, you have ##(A - \lambda I)\vec x = \vec 0##
In matrix form, with ##\lambda = -5##, this looks like
##\begin{bmatrix} 7 & 7 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}##

You could do this using an augmented matrix, but since the right side of this equation is the zero vector, it won't change. Just row-reduce the matrix on the left side to find the eigenvector that is associated with ##\lambda = -5##. Do the same kind of work for the other eigenvalue and its eigenvector.
zak100 said:
Some body please guide me.

Zulfi.
 
  • #4
462
11
Hi,
Thanks. I dont have any clear directives. But I would use one which is simple.

lambda1 = -5 and lambda2=1
For λ1 =-5
AX = -5[x ]
y
Now,

[2 7 ] * [x] = [-5x ]
[ -1 -6] [y] [-5y]
Now 2x + 7y = -5x-------(1)
-x -6y = -5y-------(2)

From (1) 2x+5x = -7y => 7x = -7y There μ1 = [-1 ]
1



From (2) also, -x -6y = -5y => -x = y There again μ= [-1 ]
1


For λ2=1

[2 7] * [x] = 1* [x ]
[ -1 -6 ] [y] [ y]


Now 2x +7y = x +> x = -7y, Therefore μ2[=[-7]
[1]

Again, -x -6y =y Therefore μ2 =[-7]
[1]



Verify:
A* μ2 = λ2[x]
[y]

[2 7] * [-7] = 1 * [-7]
[-1 -6] [1] [1]

-14 + 7 = -7 (true)
7-6 = 1(true)


And for μ1:

A * μ1 = λ1 * μ1

[2 7] * [-1] = -5 * [ -1]
[ -1 -6] [1] [1 ]

-2+7 = 5(ture)

1 -6 = -5(true)

Thanks for teaching me.

Zulfi.
 
  • #5
462
11
Hi Mark44,
You are right. My answer is same even if I dont use the augmented matrix. I think for 2*2, we should not be using augmented Matrix. Is it correct?
We will use the augmented matrix if (A - lambda* I)x (where x is a vector) = b (where b is a vector), but we may not get such a case in 2*2?

Zulfi.
 
  • #6
34,692
6,396
Hi Mark44,
You are right. My answer is same even if I dont use the augmented matrix. I think for 2*2, we should not be using augmented Matrix. Is it correct?
You don't need an augmented matrix for 2 x 2, 3 x 3, etc.
zak100 said:
We will use the augmented matrix if (A - lambda* I)x (where x is a vector) = b (where b is a vector), but we may not get such a case in 2*2?
When you're finding eigenvectors, you don't need an augmented matrix.

From the definition of eigenvalue/eigenvector, if ##\lambda_1## is an eigenvalue with associated eigenvector ##\vec{x_1}##, then ##A\vec{x_1} = \lambda_1 \vec{x_1}##, or equivalently, ##(A\vec{x_1} - \lambda_1 I)\vec{x_1} = \vec 0##.

##A\vec{x_1} - \lambda_1 I## is the matrix you're working with, and since the right side of the equation is the zero vector, you don't need an augmented matrix. You would only need an augmented matrix if the right side happened to be a nonzero vector of constants, which it never is when you're finding an eigenvector.

Also, your work in finding the eigenvectors agrees with what I found.
 

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