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Evaluating a wrong value of Eigenvector

  • Thread starter zak100
  • Start date
  • #1
zak100
Gold Member
439
10

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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Answers and Replies

  • #2
12,692
9,239

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
zak100
Gold Member
439
10
Hi,
Thanks for your response.
For evaluating, eigen vectors, I used the equation
Ax = λx

Is it right or not?

Zulfi.
 
  • #4
12,692
9,239
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
zak100
Gold Member
439
10
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
12,692
9,239
[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
zak100
Gold Member
439
10
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 cant 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
12,692
9,239
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 cant 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
33,310
5,004
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 cant 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
zak100
Gold Member
439
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
12,692
9,239
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.
 

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