Evaluating a wrong value of Eigenvector

In summary: Then you should also agree with this:A\vec{x}=\lambda \vec{x}but it doesnt seem like you do.Zulfi.Yes and yes. Yes, the equation itself ##A\vec{x}=\lambda \vec{x}## is right, and yes, that's not what you used.This equation is actually two equations.1. row: ##0\cdot x + 1 \cdot y = (-2) \cdot x##2. row: ##-2 \cdot x +(-3)\cdot y = (-2) \cdot y##
  • #1
462
11

Homework Statement


I think I am evaluating a wrong value for eigen vector. Wrong in the sense that first row is giving different value and next row is giving a different value.
The matrix is:
[0 1]
[-2 -3]
My eigen values are λ=-1 & λ=-2. I get the fine vectors for λ=-1 but λ=-2 x vanishes.

Homework Equations


A-λI = 0
& Ax = λx

The Attempt at a Solution


[0 1] * [x] = -2 * [x]
[-2 -3] [y] [y]

First row gives: y = -2x -2y
2x = 3y
Now x = 2/3 y

μ =[2, 3] (instead of column, i am showing row vector)

But the second row gives me: -2x -3y = -2x -2y (therfore x vanishes)

they have evaluated
μ=[1, -2] (instead of column, i am showing row vector)

Some body please guide me.

Zulfi.
 

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  • #2
zak100 said:

Homework Statement


I think I am evaluating a wrong value for eigen vector. Wrong in the sense that first row is giving different value and next row is giving a different value.
The matrix is:
[0 1]
[-2 -3]
My eigen values are λ=-1 & λ=-2. I get the fine vectors for λ=-1 but λ=-2 x vanishes.

Homework Equations


A-λI = 0
& Ax = λx

The Attempt at a Solution


[0 1] * [x] = -2 * [x]
[-2 -3] [y] [y]

First row gives: y = -2x -2y
2x = 3y
Now x = 2/3 y

μ =[2, 3] (instead of column, i am showing row vector)

But the second row gives me: -2x -3y = -2x -2y (therfore x vanishes)

they have evaluated
μ=[1, -2] (instead of column, i am showing row vector)

Some body please guide me.

Zulfi.
Both eigenvalues are correct, but I don't follow your solution. It seems you first deal with ##\lambda = -2##. For this we have
## \left( \begin{bmatrix}0&1\\-2&-3\end{bmatrix} - (-2) \cdot \begin{bmatrix}1&0\\0&1\end{bmatrix} \right) \cdot \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}##
which results in ##2x+y=0## and ##-2x-y=0## which is different from what you wrote.
 
  • #3
Hi,
Thanks for your response.
For evaluating, eigen vectors, I used the equation
Ax = λx

Is it right or not?

Zulfi.
 
  • #4
zak100 said:
Hi,
Thanks for your response.
For evaluating, eigen vectors, I used the equation
Ax = λx

Is it right or not?

Zulfi.
Yes and yes. Yes, the equation itself ##A\vec{x}=\lambda \vec{x}## is right, and yes, that's not what you used.

This equation is actually two equations.
1. row: ##0\cdot x + 1 \cdot y = (-2) \cdot x##
2. row: ##-2 \cdot x +(-3)\cdot y = (-2) \cdot y##
 
  • #5
Hi,
I have shown my calculations. But if you are not following and i can do it on a paper and post it as a file. Is it fine?

May be you are right because I am getting the same prob in another question also.

Zulfi.
 
  • #6
zak100 said:
[0 1] * [x] = -2 * [x]
[-2 -3] [y] [y]
This does not make sense.
First row gives: y = -2x -2y
2x = 3y
Now x = 2/3 y
This is wrong.
 
  • #7
Hi,

RHS of row1 is not correct. I am getting : -2x -2y
because λ * x is:
-2 * [ x]
[y]
So its : -2x -2y.
In your case its just -2 *x. I can't understand this because if you agree with this:
A* x = λ *x

RHS = λ * x
& λ = -2
& X = [x]
[y]

Please guide me. Thanks for your time.

Zulfi.
 
  • #8
zak100 said:
Hi,

RHS of row1 is not correct. I am getting : -2x -2y
because λ * x is:
-2 * [ x]
[y]
So its : -2x -2y.
In your case its just -2 *x. I can't understand this because if you agree with this:
A* x = λ *x

RHS = λ * x
& λ = -2
& X = [x]
[y]

Please guide me. Thanks for your time.

Zulfi.
You have to understand how matrix multiplication goes. The equation you have is actually matrix (A) times vector (x,y in column) = eigenvalue (-2) [times identity matrix (I), which is not written] times vector (x,y in column). Now matrix multiplication is always row times column for short. This means it is
$$
\begin{bmatrix}a&b\\c&d\end{bmatrix} \cdot \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}a\cdot x + b \cdot y \\ c\cdot x + d \cdot y \end{bmatrix}
$$
This means for our case that we have
$$
\begin{bmatrix}0&1\\-2&-3\end{bmatrix} \cdot \begin{bmatrix}x\\y\end{bmatrix}= \begin{bmatrix}0\cdot x + 1\cdot y\\ (-2)\cdot x+ (-3)\cdot y\end{bmatrix} = (-2)\cdot \begin{bmatrix}1&0\\0&1\end{bmatrix} \cdot \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}-2 & 0 \\0 &-2 \end{bmatrix} \cdot \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}(-2)\cdot x + 0\cdot y \\ 0\cdot x + (-2)\cdot y\end{bmatrix}
$$
and now you can compare
$$
\begin{bmatrix}0\cdot x + 1\cdot y\\ (-2)\cdot x+ (-3)\cdot y\end{bmatrix} =\begin{bmatrix}y\\-2x-3y\end{bmatrix} = \begin{bmatrix}(-2)\cdot x + 0\cdot y \\ 0\cdot x + (-2)\cdot y\end{bmatrix}= \begin{bmatrix}-2x\\-2y\end{bmatrix}
$$
row by row.
 
  • #9
zak100 said:
Hi,

RHS of row1 is not correct. I am getting : -2x -2y
because λ * x is:
-2 * [ x]
[y]
So its : -2x -2y.
In your case its just -2 *x. I can't understand this because if you agree with this:
A* x = λ *x

RHS = λ * x
& λ = -2
& X = [x]
[y]

Please guide me. Thanks for your time.

Zulfi.
You're making this much harder than it really is.
Instead of working directly with ##Ax = \lambda x##, use ##(A - \lambda I)x = 0##
Here, ##A = \begin{bmatrix}0 & 1 \\ -2 & -3 \end{bmatrix}##, and ##\lambda = -2##, so ##A - \lambda I = \begin{bmatrix}2 & 1 \\ -2 & -1 \end{bmatrix}##.
The last matrix is equivalent to ## \begin{bmatrix}2 & 1 \\ 0 & 0 \end{bmatrix}## (which can be obtained by very simple row reductionn), so the matrix equation reduces to 2x + y = 0. We can arbitrarily set y = 1, so x = -1/2. That is, ##\begin{bmatrix} -1/2 \\ 1 \end{bmatrix}##.

All eigenvectors for ##\lambda = -2## for matrix A are some multiple of this vector.
 
  • #10
Hi,
Thanks I would try by using Identity matrix & redo all my examples.

This means I would use:
Ax =λ*I * x.

This would correct my multiplication.

Zulfi.
 
  • #11
zak100 said:
Hi,
Thanks I would try by using Identity matrix & redo all my examples.

This means I would use:
Ax =λ*I * x.

This would correct my multiplication.

Zulfi.
You can as well simply write ##(-2)\cdot \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-2x\\-2y\end{bmatrix}## which is the same.
 

1. What is an eigenvector?

An eigenvector is a vector that remains in the same direction after being multiplied by a square matrix. In other words, it is a special vector that is only scaled by a scalar value when multiplied by a matrix.

2. How is an eigenvector calculated?

Eigenvectors are calculated by finding the non-zero solutions to the equation (A - λI)v = 0, where A is the square matrix, λ is the eigenvalue, and v is the eigenvector. This equation is known as the characteristic equation of the matrix.

3. What does it mean when a wrong value is obtained for an eigenvector?

Obtaining a wrong value for an eigenvector means that the calculated eigenvector does not satisfy the characteristic equation, which can happen due to calculation errors or incorrect methods used to find the eigenvector.

4. How can a wrong value of an eigenvector affect the overall evaluation of a system?

In many cases, a wrong value of an eigenvector can result in incorrect interpretations and predictions of a system. Eigenvectors are often used to understand the behavior and stability of a system, so a wrong value can lead to inaccurate conclusions and decisions.

5. What are some potential solutions for evaluating a wrong value of an eigenvector?

If a wrong value is obtained for an eigenvector, it is important to check the calculation process and ensure that the correct methods are being used. Additionally, it may be helpful to use software or other tools to double-check the results. In some cases, recalculating the eigenvector using a different method may also help to obtain a more accurate value.

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