# Eigenvalue/vector and similar matrices

1. Mar 15, 2009

### A_lilah

1. The problem statement, all variables and given/known data
Find a matrix B such that B^2 = A

A = 3x3 =

9 -5 3
0 4 3
0 0 1

2. Relevant equations

B^2 = A

A = XDX^(-1) (similar matrices rule)

also used to find eigenvectors: A - λI

3. The attempt at a solution

Thoughts: If A = XDX^(-1), then B^2 = XDX(-1), and B = X * D^(1/2) * X^-1
So, if I could find D and X for A, I could find B. D = diagonal matrix where the diagonal elements are the eigenvalues of A, and A is lower triangular, so it's eigenvalues are:

λ1 = 9, λ2 = 4, λ3 = 1,

so D = 3x3 =

9 0 0
0 4 0
0 0 1

To find X, I need the eigenvectors:

A-9I =
0 -5 3
0 -5 3
0 0 -8
so (-5)v2 + (3)v3 = 0, or v2 = (3/5)v3
and the eigenvector (representative) = [0 1 3/5]^T

I suspect there is something wrong here because my bottom row in A-9I isn't all 0 and I think it is supposed to be... I don't know what I did wrong though.

For the other eigenvalues, I got eigenvectors that were:

λ2 = 4,
[1 1 0]^T

λ3 = 1
[1 1 -1]

X = 3x3 =

0 1 1
1 1 1
(3/5) 0 -1

X^-1 =
-1 1 0
(8/5) (-3/5) 1
(-3/5) (3/5) -1

To get D^(1/2) I square-rooted the diagonal elements:
3 0 0
0 2 0
0 0 1

Then find X * D^(1/2) * X^-1, which = B, but when I square B, it doesn't = A...
Any input would be great! Thanks

Last edited: Mar 16, 2009
2. Mar 16, 2009

### tiny-tim

Hi A_lilah!

Hint: is the square of a triangular matrix triangular?

if so, what happens to the diagonal elements?

3. Mar 16, 2009

### HallsofIvy

Staff Emeritus
You've completely ignored the third row: -8v3= 0 so v3= 0. Yes, the second row then gives v2= (3/5)v3 and since v3= 0, v2= 0. Now the important point: the fact that the first column is 0 does NOT mean that v1= 0! It says there is NO CONDITION on v1. With v2= v3= 0, 0v1-5v2+ 3v3= 0, 0v1- 5v2+ 3v3= 0, and 0v1+ 0v2- 8v3= 0 are true for ALL VALUES of v3: [1, 0, 0]^T is an eigenvector.

yes, this is correct.

Did you check it?
$$\left[\begin{array}{ccc}9 & -5 & 3 \\ 0 & 4 & 3 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{c}1 \\ 1 \\ 1\end{array}\right]= \left[\begin{array}{c} 7 \\ 7 \\ 1\end{array}\right]$$
No, that is not [1 1 -1]^T again.

For eigenvalue 1, you get the three equations 8v1- 5v2+ 3v3= 0, 3v2+ 3v3= 0, and 0= 0. The last equations says that v3 can be anything. From the second equation, 3v2= -3v3 so v2= -v3 and, putting that into the first equation, 8v1+ 5v3+ 3v3= 0, v1= -v3. You can write down the eigenvector corresponding to eigenvalue 1 from that.

Last edited: Mar 16, 2009
4. Mar 16, 2009

### A_lilah

Thank you both so much! Of course I should have looked at the other rows ~ silly of me. Thanks again :)