Proving Existence of Positive Eigenvector in 2x2 Matrix with Positive Elements

[Wildcat]

Homework Statement

Let A = matrix [a b] row 1 [c d] row 2 (2x2 matrix) with a>0, b>0,c>0,d>0. Show that A has an eigenvector [x,y] (2x1) with x>0, y>0

Homework Equations

The Attempt at a Solution

I've tried finding the characteristic polynomial by using det((lambda)I - A) =0, then using the quadratic equation to find the roots (thats ugly) so I'm trying multipying the column vector x,y by some eigenvalue to get each column of the original matrix.
if c/y is an eigenvalue that corresponds to the eigenvector [ay/c, y] and d/y is an eigenvalue that corresponds to the eigenvector [ by/d , y] a,b,c,d are all >0 therefore the eigenvector values will by >0

Will this work??

 

[Mark44]

Homework Statement

Let A = matrix [a b] row 1 [c d] row 2 (2x2 matrix) with a>0, b>0,c>0,d>0. Show that A has an eigenvector [x,y] (2x1) with x>0, y>0

Homework Equations

The Attempt at a Solution

I've tried finding the characteristic polynomial by using det((lambda)I - A) =0, then using the quadratic equation to find the roots (thats ugly)

It's not that bad - don't give up so easily. The Quadratic Formula shows that there are two real eigenvalues. You should be able to use one of these eigenvalues to find an eigenvector with positive entries.

so I'm trying multipying the column vector x,y by some eigenvalue to get each column of the original matrix.
if c/y is an eigenvalue that corresponds to the eigenvector [ay/c, y] and d/y is an eigenvalue that corresponds to the eigenvector [ by/d , y] a,b,c,d are all >0 therefore the eigenvector values will by >0

Will this work??

 

[Wildcat]

It's not that bad - don't give up so easily. The Quadratic Formula shows that there are two real eigenvalues. You should be able to use one of these eigenvalues to find an eigenvector with positive entries.

OK, I'll go back and do that. So I'm assuming by your response what I did is incorrect?? If so, can you tell me why?

 

[Wildcat]

OK, I'll go back and do that. So I'm assuming by your response what I did is incorrect?? If so, can you tell me why?

I must be doing something wrong trying to use the quadratic formula. I'm getting my
b=(a+d) a=1 and c=(ad-bc) then when I use the quadratic formula with those values and try to put it back into the (lambda)I - A matrix and row reduce to get to row-echelon form
OH MY! Am I still giving up too soon??

 

[Wildcat]

OK, I'll go back and do that. So I'm assuming by your response what I did is incorrect?? If so, can you tell me why?

nevermind on telling me why its wrong, I figured out what I was doing wrong, I should have known it wasn't that easy!

 

[Mark44]

For the eigenvalues I got lambda = [a + d +/- sqrt((a - d)^2 + 4bc)]/2.

Then work with (A - lambdaI)x = 0 to find the eigenvector associated with each eigenvalue.

 

[Wildcat]

For the eigenvalues I got lambda = [a + d +/- sqrt((a - d)^2 + 4bc)]/2.

Then work with (A - lambdaI)x = 0 to find the eigenvector associated with each eigenvalue.

Thats what I got for the eigenvalues but when I put it in (A-lambdaI)x=0 and try to row reduce it gets really messy. I do need to row reduce right??

 

[Mark44]

Yes.

 

[Wildcat]

Yes.

So using the eigenvalue [(a+d) +√((a-d)^2 +4bc)]/2

I end up with the eigenvector of [1, and some incredibly complex expression] should it be a really complex expression??

 

[Mark44]

It probably shouldn't, but then I haven't worked this through. In any case, all you need to do is show that both coordinates of the eigenvector are positive, so maybe there's a shortcut you can take.

 

[Wildcat]

It probably shouldn't, but then I haven't worked this through. In any case, all you need to do is show that both coordinates of the eigenvector are positive, so maybe there's a shortcut you can take.

Ok thanks for your help. I need to look at it some more.

 

[Wildcat]

So using the eigenvalue [(a+d) +√((a-d)^2 +4bc)]/2

I end up with the eigenvector of [1, and some incredibly complex expression] should it be a really complex expression??

I worked on this some more.
when I put the eigenvalue into the matrix lambdaI -A
I get ([(a+d) +√((a-d)^2 +4bc)]/2 - a)x - by =0 if y=1
then x = 2b/(-a+d+√(a-d)²+4bc) is that correct??

I've tried every scenario I can think of and the denominator is always positive which makes the whole expression positive since b is positive ??

 

[Mark44]

For lambda1, I get [a + d + √((a - d)^2 + 4bc)]/2
For lambda2, I get [a + d - √((a - d)^2 + 4bc)]/2


I think these are what you get.

To make things a little easier, let's define D = √((a - d)^2 + 4bc) and deal with it later.

After substituting lambda1 into the matrix to get lambda1*I - A, I got
$$\left[ \begin{array}{cc} \frac{-a+d + D}{2} & -b\\-c & \frac{a-d-D}{2}\end{array}\right]$$

After a couple of row-reduction steps, I got
$$\left[ \begin{array}{cc} 1 & \frac{2b}{a-d -D}\\0 & 0\end{array}\right]$$

I had to combine two fractions in the row-2, col-2 entry, but everything dropped out, as it should.

The top row of this matrix says that 1x + 2b/(a - d - D) * y = 0, which is the same as saying
x = -2b/(a - d - D) * y

Without loss of generality you can take y = 1, which makes x equal to the coefficient -2b/(a - d - D).

Recall from before that D = √((a - d)^2 + 4bc).

(a - d)^2 < (a - d)^2 + 4bc, since b and c are positive.
So |a - d| = √((a - d)^2) < √((a - d)^2 + 4bc)

This means that the expression a - d - D is negative, and so is -2b, so when y = 1, x > 0.

 

[Wildcat]

For lambda1, I get [a + d + √((a - d)^2 + 4bc)]/2
For lambda2, I get [a + d - √((a - d)^2 + 4bc)]/2


I think these are what you get.

To make things a little easier, let's define D = √((a - d)^2 + 4bc) and deal with it later.

After substituting lambda1 into the matrix to get lambda1*I - A, I got
$$\left[ \begin{array}{cc} \frac{-a+d + D}{2} & -b\\-c & \frac{a-d-D}{2}\end{array}\right]$$

After a couple of row-reduction steps, I got
$$\left[ \begin{array}{cc} 1 & \frac{2b}{a-d -D}\\0 & 0\end{array}\right]$$

I had to combine two fractions in the row-2, col-2 entry, but everything dropped out, as it should.

The top row of this matrix says that 1x + 2b/(a - d - D) * y = 0, which is the same as saying
x = -2b/(a - d - D) * y

Without loss of generality you can take y = 1, which makes x equal to the coefficient -2b/(a - d - D).

Recall from before that D = √((a - d)^2 + 4bc).

(a - d)^2 < (a - d)^2 + 4bc, since b and c are positive.
So |a - d| = √((a - d)^2) < √((a - d)^2 + 4bc)

This means that the expression a - d - D is negative, and so is -2b, so when y = 1, x > 0.

My expression for x is technically the same, but yours would be correct because of the explanation as to why the expression is positive?? I'm just trying to get the reasoning correct.

 

[Mark44]

I don't understand your question. The goal of this problem is to show that there is an eigenvector with both coordinates positive. That's what my explanation is doing. Is there some of it you don't understand?

 

[Wildcat]

I don't understand your question. The goal of this problem is to show that there is an eigenvector with both coordinates positive. That's what my explanation is doing. Is there some of it you don't understand?

Sorry, I do completely understand. I'm just trying to overkill it in my mind so my confidence will increase. Thank you soooo much for your help!