Find eigenvalue for matrix B= {[3,4,12],[4,-12,3],[12,3,-4]}

[LosTacos]

Homework Statement

Find eigenvalue for matrix B= {[3,4,12],[4,-12,3],[12,3,-4]}

Homework Equations

The Attempt at a Solution

I set up the charactersitic polynomial and got the equation:
Pa(x) = (x-3)(x+12)(x+4) = x³ + 13² - 144 + 144 = x³ + 13²

So I have 3 eigenvalues: 0, 13, -13. Is this correct?

 

[LosTacos]

Actually I forgot to calculate teh determinants so I got:

Pa(x) = (x-3)(x+12)(x+4) + 2197 = x³ + 13x² + 2053

 

[hsetennis]

Those are the correct eigenvalues (if it is conventional to include 0 in your class). It may be good to note multiplicity of eigenvalues.

 

[LosTacos]

From the characteristic polynomial x³ + 13x² + 2053, how do I get the correct eigenvalues?

 

[Mark44]

Homework Statement

Find eigenvalue for matrix B= {[3,4,12],[4,-12,3],[12,3,-4]}

Homework Equations

The Attempt at a Solution

I set up the charactersitic polynomial and got the equation:
Pa(x) = (x-3)(x+12)(x+4) = x³ + 13² - 144 + 144 = x³ + 13²

So I have 3 eigenvalues: 0, 13, -13. Is this correct?

Two of them are correct, but 0 is not an eigenvalue. It should be -13 (a repeated eigenvalue).

From the characteristic polynomial x³ + 13x² + 2053, how do I get the correct eigenvalues?

You can't, since this isn't the correct characteristic polynomial. The eigenvalues are the roots of the characteristic polynomial.

 

[LosTacos]

Okay I understand. So my eigenvalues are 13 and -13. If I was asked to find the basis for both of these, how do I go about doing that. I tried to solve the equation [13I₂ - A I 0] however if ran into a wall. I row reduced it to get the matrix = {[2, 2/5, 6/5], [0,1,3], [0,0,1]}. But I wasnt sure where to go from there

 

[Mark44]

Okay I understand. So my eigenvalues are 13 and -13. If I was asked to find the basis for both of these, how do I go about doing that. I tried to solve the equation [13I₂ - A I 0]

You should be solving the matrix equation (13I₃ - A)x = 0, or equivalently, (A - 13I)x = 0.
This would be the work to find the eigenvector for λ = 13.

however if ran into a wall. I row reduced it to get the matrix = {[2, 2/5, 6/5], [0,1,3], [0,0,1]}.

This is incorrect. Show me the matrix you started with, and a step or two of your work.

But I wasnt sure where to go from there

 

[LosTacos]

(A - 13I)x = 0:

{[10, 4, 12, 0],[4, 1, 3, 0],[12, 3, 17]} = {[1, 2/5, 6/5],[4, 1, 3, 0],[12, 3, 17, 0]} = {[1, 2/5, 6/5, 0], [0, -3/5, -9/5, 0],[0, -9/5, 13/5, 0]} = {[1, 2/5, 6/5, 0],[0, 1, 3, 0], [0, 0, 8, 0]}

 

[Mark44]

(A - 13I)x = 0:

{[10, 4, 12, 0],[4, 1, 3, 0],[12, 3, 17]}

You started off with an incorrect matrix.

Here is A:
$$ \begin{bmatrix} 3 & 4 & 12\\ 4 & -12 & 3 \\ 12 & 3 &-4 \end{bmatrix}$$
To get A - 13I, subtract 13 from the entries on the main diagonal.

= {[1, 2/5, 6/5],[4, 1, 3, 0],[12, 3, 17, 0]} = {[1, 2/5, 6/5, 0], [0, -3/5, -9/5, 0],[0, -9/5, 13/5, 0]} = {[1, 2/5, 6/5, 0],[0, 1, 3, 0], [0, 0, 8, 0]}