Eigenvectors!?

Question

Find the eigenvalues & corresponding eigenvectors of the matrix A =
[ 1 2 4 ]
[ 1 2 4 ]
[ 1 2 4 ]

ok so i've found the eigenvalues to be 7 and 0 but i can't find the eigenvectors. help plssssss~~~

Answer 1

Remember that v is an eigenvector with eigenvalue λ iff (A-λI)v = 0. So, plug in the eigenvalues you've found, and look at the kernel of those matrices. First, let's look at the eigenvalue 0. A-0I = A, so we need vectors that are in the kernel of this matrix. By inspection, we immediately find the vectors (2, -1, 0) and (4, 0, -1) are in the kernel, and since these are linearly independent and the nullity of A is at most 2 (since for it to be 3, A would have to be the zero matrix), we know that we've found a basis for E₀ (which is the space of eigenvectors with eigenvalue 0).

To find the basis of E₇, first we find A-7I, which is:

[-6, 2, 4]
[1, -5, 4]
[1, 2, -3]

Here, it's harder to find eigenvectors by inspection (although not too hard, if you think about it), but to make it easier, we can reduce this to rref (this does not change the kernel). First, let's swap the first and third rows so we don't have to do any division:

[1, 2, -3]
[1, -5, 4]
[-6, 2, 4]

Subtract the first row from row two, and add 6 times row 1 to row 2:

[1, 2, -3]
[0, -7, 7]
[0, 14, -14]

Divide the second row by -2:

[1, 2, -3]
[0, 1, -1]
[0, 14, -14]

And now subtract 14 times row 2 from row 3 and twice row 2 from row 1:

[1, 0, -1]
[0, 1, -1]
[0, 0, 0]

And now the eigenvector is obvious -- it is (1, 1, 1). So from this and our earlier result a basis of eigenvectors is:

(2, -1, 0), (4, 0, -1), and (1, 1, 1).

Answer 2

get the matrix

L*(I3) - A ..... (I3) is the identity matrix... L is the eigenvalue
and reduce ...
the solution space is the eigenspace...


so for L = 0
L*(I3) - A =
[-1 -2 -4]
[-1 -2 -4]
[-1 -2 -4]

after reduction it becomes...
[1 2 4]
[0 0 0]
[0 0 0]

x + 2y + 4z = 0
x = -2y - 4z

so...
x = -2s - 4t
y = s
z = t

the basis for the eigenspace is:
{ [-2 1 0]T , [-4 0 1]T} .. these are your eigenvectors for L=0.
There is a T behind because they need to be transposed... the answers are column vectors... §

Now you can do the same on L = 7.