# Eigenvectors for a 2*2 Matrix

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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?

Zulfi.

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kuruman
Homework Helper
Gold Member
The eigenvectors must be normalized, no?

Gold Member
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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Mark44
Mentor

## 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?

Zulfi.

kuruman
Homework Helper
Gold Member
Can you post what it says? I cannot open the file.

Gold Member
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
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]

Zulfi.

Mark44
Mentor
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
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).