Eigenvalues & Eigenvectors

1. Jun 25, 2009

fuzzyorama

Eigenvalues & Eigenvectors !!!SOLVED!!!

1. The problem statement, all variables and given/known data
Find the eigenvalues and eigenvectors of matrix
$$A = \left( \begin{array}{cc} 2 & 2 \\ 3 & 1 \end{array} \right)$$

2. Relevant equations
$$Ax = \lambda x$$

3. The attempt at a solution
Solving
$$\left\vert \begin{array}{cc} 2 - \lambda & 2 \\ 3 & 1 - \lambda \end{array} \right\vert$$
I get the eigenvalues $$\lambda = -1, 4$$

When $$\lambda = 4$$

$$\left( \begin{array}{cc} -2 & 2 \\ 3 & -2 \end{array} \right)\left( \begin{array}{c} x \\ y \end{array} \right) = 4\left( \begin{array}{c} x \\ y \end{array} \right)$$

Then I get these equations $$-2x + 2y = 4x \mbox{~and~} 3x -3y = 4y$$
From the first equation, $$y = 3x$$. So is

$$x\left( \begin{array}{c} 4 \\ 12 \end{array} \right)$$ the eigenvector?

Last edited: Jun 26, 2009
2. Jun 25, 2009

HallsofIvy

Staff Emeritus
That will certainly work. But remember that the set of all eigenvectors, corresponding to a given eigenvalue, form a subspace- there is no such thing as the eigenvector, there are an infinite number of them. Any multiple of an eigenvector is also an eigenvector. Here you got y= 3x. Any reason for then taking x= 4? The fact that the eigenvalue is 4 (if that's where you got the "4") does not imply that x must be 4. I might be inclined to take x= 1 so that y= 3(1) and the eigenvector is
$$\left(\begin{array}{c}1 \\ 3\end{array}\right)$$
or you might divide that by its length, $\sqrt{10}$ to get a unit length eigenvector,
$$\left(\begin{array}{c}\frac{1}{\sqrt{10}}\\ \frac{3}{\sqrt{10}}\end{array}\right)$$
Any of those are eigenvectors corresponding to eigenvalue 4. Now, what are the eigenvectors corresponding to eigenvalue -1?

3. Jun 25, 2009

fuzzyorama

I thought that just by replacing the y in
$$\left( \begin{array}{cc} -2 & 2 \\ 3 & -2 \end{array} \right)\left( \begin{array}{c} x \\ y \end{array} \right) = 4\left( \begin{array}{c} x \\ y \end{array} \right)$$

with 3x would give me the eigenvector. I then multiplied it with the eigenvalue 4 and took the x out. So the eigenvector can also be
$$\left(\begin{array}{c}2 \\ 6\end{array}\right) \left(\begin{array}{c}3 \\ 9\end{array}\right) \left(\begin{array}{c}5 \\ 15\end{array}\right)$$ ?
When $$\lambda = -1$$, I get
$$\left( \begin{array}{cc} 3 & 2 \\ 3 & 2 \end{array} \right)\left( \begin{array}{c} x \\ y \end{array} \right) = -\left( \begin{array}{c} x \\ y \end{array} \right)$$

and the equations 3x + 2y = -x and 3x +2y = -y. From these two equations I get y = x so the eigenvector is $$-\left(\begin{array}{c}1 \\ 1\end{array}\right)$$ ?

4. Jun 25, 2009

Staff: Mentor

Your work is easy enough to check, and you should do this. If $\lambda$ is an eigenvalue with associated eigenvector x, then Ax better be equal to $\lambda$x. If not, then you have a mistake.

5. Jun 26, 2009

fuzzyorama

OMG I got a typo. a22 should be -3. Apologies.

$$\left( \begin{array}{cc} -2 & 2 \\ 3 & -3 \end{array} \right)\left( \begin{array}{c} x \\ y \end{array} \right) = 4\left( \begin{array}{c} x \\ y \end{array} \right)$$
From what I understand, by that you mean from the equation
$$\left( \begin{array}{cc} -2 & 2 \\ 3 & -3 \end{array} \right)\left( \begin{array}{c} x \\ y \end{array} \right) = 4\left( \begin{array}{c} x \\ y \end{array} \right)$$

where the eigenvector is
$$\left(\begin{array}{c}1 \\ 3\end{array}\right)$$
I can replace the x with 1 and y with 3:
$$\left( \begin{array}{cc} -2 & 2 \\ 3 & -3 \end{array} \right)\left( \begin{array}{c} 1 \\ 3 \end{array} \right) = 4\left( \begin{array}{c} 1 \\ 3 \end{array} \right)$$
Solving the $$Ax$$ part, I get
$$\left( \begin{array}{cc} -2 & 2 \\ 3 & -3 \end{array} \right)\left( \begin{array}{c} 1 \\ 3 \end{array} \right) = \left( \begin{array}{c} -2(1) + 2(3) \\ 3(1) + (-3)(3) \end{array} \right) = \left( \begin{array}{c} 4 \\ -6 \end{array} \right)$$
but it doesn't equal to
$$\lambda x = \left( \begin{array}{c} 4(1) \\ 4(3) \end{array} \right) = \left( \begin{array}{c} 4 \\ 12 \end{array} \right)$$

Does this mean I got it wrong?

6. Jun 26, 2009

Pengwuino

No, you're looking at $$Ax = \lambda x$$ where x is your eigenvector and lambda is your eigenvalue and A is still your original $$A = \left( \begin{array}{cc} 2 & 2 \\ 3 & 1 \end{array} \right)$$. You plug in your eigenvector at this point and verify you get the vector back multiplied by its respective eigenvalue.

The reason you don't have simply 1 eigenvector is that you can start as $$Ax = \lambda x$$ and say, let your new eigenvector be 3x where x is what you call "the eigenvector", that is $$A(3x) = \lambda (3x)$$ which gives you $$3Ax = 3\lambda x$$ which is simply $$Ax = \lambda x$$ still, thus any multiple of the eigenvector you found is another eigenvector.

7. Jun 26, 2009

fuzzyorama

Ahh.. so I don't actually need to multiply the eigenvector with the eigenvalue to see if they are equal?

8. Jun 26, 2009

Pengwuino

Yes you still need to. Remember, the idea here is finding a vector that when acted upon by your matrix A, you return the vector multiplied by the eigenvalue. No matter what constant you multiply your eigenvector by, that is x -> cx where c is a constant, when you multiply from the left with A you still must return that eigenvalue multiplied by cx.

9. Jun 26, 2009

fuzzyorama

Oh wait!
$$\left( \begin{array}{cc} 2 & 2 \\ 3 & 1 \end{array} \right)\left(\begin{array}{c}1 \\ 3\end{array}\right) = 4\left(\begin{array}{c}1 \\ 3\end{array}\right)$$
$$\left(\begin{array}{c}8 \\ 6\end{array}\right) = \left(\begin{array}{c} 4 \\ 12 \end{array}\right)$$

But it doesn't matter because both are multiples of $$\left(\begin{array}{c}1 \\ 3\end{array}\right)$$ ?

10. Jun 26, 2009

Pengwuino

No, they must be exact. You're still using the wrong eigenvectors, I thought you realized that earlier when you mentioned the typo you made. When you found your eigenvector for lambda = 4, you found 2 equations that were inconsistent $$-2x + 2y = 4x \mbox{~and~} 3x -3y = 4y$$ . This should have immediately told you that something was wrong.

Basically, at this point: $$\left( \begin{array}{cc} -2 & 2 \\ 3 & -2 \end{array} \right)\left( \begin{array}{c} x \\ y \end{array} \right) = 4\left( \begin{array}{c} x \\ y \end{array} \right)$$ you should know that somethings wrong since the rows have to be linearly dependent (that is, rows are a multiple of eachother) to solve for eigenvectors.

11. Jun 26, 2009

Matterwave

This is the wrong equation to use. The matrix on the right has to be A, and not the modified one with the subtracted eigenvalues.

If you want to use the modified matrix as you are, you need to set the right hand side equal to 0.

Basically you're making the mistake of this:

y=x Original equation

and then:
y-x=x WRONG

y-x=0 CORRECT

but in matrix form.

I make this mistake myself quite often.

12. Jun 26, 2009

fuzzyorama

Thanks for pointing that out Now I will redo everything.
$$A = \left( \begin{array}{cc} 2 & 2 \\ 3 & 1 \end{array} \right)$$
$$\left\vert \begin{array}{cc} 2 - \lambda & 2 \\ 3 & 1 - \lambda \end{array} \right\vert = 0$$

$$(2 - \lambda )(1 - \lambda) - (3)(2) = 0$$
$$2 - 2\lambda - \lambda + \lambda^2 - 6 = 0$$
$$\lambda^2 -3\lambda - 4 = 0$$

$$\therefore \lambda = -1, 4$$

When $$\lambda = -1$$,
$$\left( \begin{array}{cc} 2 - (-1) & 2 \\ 3 & 1 - (-1) \end{array} \right)\left( \begin{array}{c} x \\ y \end{array} \right) = 0$$
$$\left( \begin{array}{cc} 3 & 2 \\ 3 & 2 \end{array} \right)\left( \begin{array}{c} x \\ y \end{array} \right) = 0$$

$$3x + 2y = 0$$
$$2y = -3x$$
$$\therefore y = -\frac{3}{2}x$$

Now I put this in my original matrix A ($$Ax = \lambda x$$)
$$\left( \begin{array}{cc} 2 & 2 \\ 3 & 1 \end{array} \right)\left( \begin{array}{c} x \\ -\frac{3}{2}x \end{array} \right) = -\left( \begin{array}{c} x \\ -\frac{3}{2}x \end{array} \right)$$
$$\therefore -x\left( \begin{array}{c} 1 \\ -\frac{3}{2} \end{array} \right)$$

Next, when $$\lambda = 4$$,
$$\left( \begin{array}{cc} 2 - 4 & 2 \\ 3 & 1 - 4 \end{array} \right)\left( \begin{array}{c} x \\ y \end{array} \right) = 0$$
$$\left( \begin{array}{cc} -2 & 2 \\ 3 & -3 \end{array} \right)\left( \begin{array}{c} x \\ y \end{array} \right) = 0$$

$$-2x + 2y = 0$$
$$3x - 3y = 0$$
$$\therefore x = y$$

$$\left( \begin{array}{cc} 2 & 2 \\ 3 & 1 \end{array} \right)\left( \begin{array}{c} x \\ x \end{array} \right) = 4 \left( \begin{array}{c} x \\ x \end{array} \right)$$
$$\therefore x\left( \begin{array}{c} 1 \\ 1 \end{array} \right)$$

Dayum this took quite a while heh.

Last edited: Jun 26, 2009
13. Jun 26, 2009

Pengwuino

At $$\therefore y = -\frac{3}{2}x$$, what this is telling you is that the solution to your system of equation is ANY matrix such that the y-entry is -3/2 of the x-entry. For example, you can try $$\left( {\begin{array}{*{20}c} 2 \\ { - 3} \\\end{array}}\right)$$ for example. Any constant multiple of that eigenvector is an eigenvector as well! You can normalize it and use $$\left( {\begin{array}{*{20}c} {\frac{2}{{\sqrt {13} }}} \\ {\frac{{ - 3}}{{\sqrt {13} }}} \\ \end{array}} \right)$$ for example! Plug it into your original equation and you'll see both work.

Also, having the x multiplying your eigenvector is kinda confusing as it look slike you're multiplying your eigenvector by another matrix x.

14. Jun 26, 2009

fuzzyorama

I got it right?

Understood! So it can also be
$$\left( {\begin{array}{*{20}c} 4 \\ { - 6} \\\end{array}}\right)$$ ?

15. Jun 26, 2009

Pengwuino

Yes, it can be any constant multiple. Simply plug it in and see what pops out to make sure you have an eigenvector.

16. Jun 26, 2009

fuzzyorama

$$\left( \begin{array}{cc} 2 & 2 \\ 3 & 1 \end{array} \right)\left( \begin{array}{c} 2 \\ -3 \end{array} \right) = -\left( \begin{array}{c} 2 \\ -3 \end{array} \right)$$

$$2(2) + 2(-3) = -2$$
$$3(2) + 1(-3) = 3$$

I GOT IT RIGHT!!! YOU GUYS ARE THE AWESOMEST! THANKS :!!) :!!) :!!)