# Another reminder on finding eigenvectors

1. May 3, 2012

### eherrtelle59

Another question with respect to finding eigenvectors.

If I remember correctly, I should be able to look at certain 2 by 2 matrices and practically write down the eigenvalues and eigenvectors.

For example, I have a diagonal matrix, I know immediately what the eigenvalues and eigenvectors are.

E.g. M = \begin{bmatrix}
1 &0 \\[0.3em]
0 & x \\[0.3em]

\end{bmatrix}

Well, I know immediately λ_1 =1, λ_2 = x and that the eigenvectors are e_1 = (1 0) and e_2 = (0 1).

Now, what about an upper diagonal matrix?

Take M = \begin{bmatrix}
-1 & -1 \\[0.3em]
0 & x-(1/4) \\[0.3em]

\end{bmatrix}

I can see λ_1=1 and that e_1 = (1 0)

How do you find the second eigenvalue and eigenvector?

2. May 3, 2012

### HallsofIvy

In a "triangular matrix", just as in a diagonal matrix, the numbers on the diagonal are the eigenvalues. In your example they are -1, NOT 1, and x- (1/4).

More generally, if your matrix is
$$\begin{bmatrix} a & 0 \\ -1 & b\end{bmatrix}$$
then a and b are the eigenvalues.

The eigenvector corresponding to eigenvalue a must satisfy
$$\begin{bmatrix} a & -1 \\ 0 & b\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}ax & ay\end{bmatrix}$$
so that we have equations ax- y= ax and by= ay. The first equation, on subtracting ax from both sides, gives -y= 0 so y= 0. The second equation, which is equivalent to (b- a)y= 0 also gives y= 0. An eigenvector corresponding to eigenvalue a is <1, 0> as you say.

The eigenvector corresponding to eigenvalue b must satisfy
$$\begin{bmatrix} a & 1 \\ 0 & b\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}bx \\ by\end{bmatrix}$$
which gives the equations ax- y= bx and by= by. The first equation is the same as y= (b- a)x while the second equation is satisfied by any y. Any eigenvector is of the form <x, y>= <x, (b-a)x>= x<1, b-a>.

In your case, with a= 1 and b= x- 1/4, 1 is an eigenvalue with corresponding eigenvector <1, 0> (or any multiple) and x- 1/4 is an eigenvalue with corresponding eigenvector <1, x- 5/4>.

3. May 3, 2012

### eherrtelle59

This makes sense, but according to what I have here, the eigenvector should be
λ_2 = <-1 x+ (3/4)>

This is assuming x-(1/4) > 0. Would that make a difference or is what I have a typo?

4. May 3, 2012

### eherrtelle59

i.e. b-a is 1-(-1/4)=3/4, right?

5. May 4, 2012

### HallsofIvy

No, 1- (-1/4)= 1+ 1/4 = 5/4, not 3/4.

6. May 4, 2012

### eherrtelle59

Typo on my part.

we get for the "b value" an equation -v_1-v_2 = (x-1/4)*v_1

This is -v_2 =(x+3/4)*v_1

7. May 4, 2012

### eherrtelle59

That is, using your equation ax- y= bx

a=-1, b=(x-1/4)

8. Jul 20, 2012

### Katherine Mos

perhaps you should credit the author for the quote at the bottom of your post -- Edna St. Vincent Millay