Eigenvectors for a 2*2 Matrix

  • Thread starter zak100
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  • #1
zak100
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Homework Statement


Consider the following Matrix:

Row1 = 2 2

Row2 = 5 -1
Find its Eigen Vectors



Homework Equations


Ax = λx & det(A − λI)= 0.

The Attempt at a Solution


First find the det(A − λI)= 0. which gives a quadratic eq.
roots are
λ1 = -3 and λ2 = 4 (Eigen values)

Then using λ1, I got:

x1= -2/5 x2

Now how can i find out the vectors?

Some body please guide me.

Zulfi.
 
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Answers and Replies

  • #2
kuruman
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The eigenvectors must be normalized, no?
 
  • #3
zak100
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Hi,
I dont know about that. I am attaching the example which I have followed. It has not discussed about Normalization.

Zulfi
 

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  • #4
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Homework Statement


Consider the following Matrix:

Row1 = 2 2

Row2 = 5 -1
Find its Eigen Vectors



Homework Equations


Ax = λx & det(A − λI)= 0.

The Attempt at a Solution


First find the det(A − λI)= 0. which gives a quadratic eq.
roots are
λ1 = -3 and λ2 = 4 (Eigen values)

Then using λ1, I got:

x1= -2/5 x2
And x2 = x2 (meaning that x2 is arbitrary).

I haven't checked your work, but if it's correct, any eigenvector for ##\lambda = 3## is a scalar multiple of <-2/5, 1>.
zak100 said:
Now how can i find out the vectors?

Some body please guide me.

Zulfi.
 
  • #5
kuruman
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Can you post what it says? I cannot open the file.
 
  • #6
zak100
462
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Hi,
You are not correct. It says: -2 5. I dont know what would be the vector if:
x1 = -2x2.
Will it be : -2 1
It says about my question:
This means that, while there are infinitely many nonzero solutions (solution vectors) of the
equation Ax = −3x, they all satisfy the condition that the first entry x1 is −2/5 times the
second entry x2.
[ 2t ] = t [2 ]
−5t -5

(Note my square bracket is small. It should enclose 2t & -5t and the other one should enclose 2 -5)
,

where t is any real number. The nonzero vectors x that satisfy Ax = −3x are called
eigenvectors associated with the eigenvalue = −3. One such eigenvector is
u1 =[ 2 −5]

Some body please guide .

Zulfi.
 
  • #7
36,216
8,199
Hi,
You are not correct.
Who are you replying to? Use the Quote feature to copy what someone else is saying.
zak100 said:
It says: -2 5. I dont know what would be the vector if:
x1 = -2x2.
Will it be : -2 1
It says about my question:
This means that, while there are infinitely many nonzero solutions (solution vectors) of the
equation Ax = −3x, they all satisfy the condition that the first entry x1 is −2/5 times the
second entry x2.
[ 2t ] = t [2 ]
−5t -5

(Note my square bracket is small. It should enclose 2t & -5t and the other one should enclose 2 -5)
,

where t is any real number. The nonzero vectors x that satisfy Ax = −3x are called
eigenvectors associated with the eigenvalue = −3. One such eigenvector is
u1 =[ 2 −5]
That looks good. To summarize what you have, if one eigenvalue is ##\lambda = -3##, its associated eigenvector is ##\vec x = \begin{bmatrix} 2 \\ -5\end{bmatrix}##. You can and should verify this fact by confirming that ##A\begin{bmatrix} 2 \\ -5\end{bmatrix} = -3 \begin{bmatrix} 2 \\ -5\end{bmatrix}##, using the matrix you wrote in post #1.

Now do the same kind of work to find the other eigenvalue and associated eigenvector, and you're done (but you should check this one as well).
 

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