Linear Algebra, Theory And Applications, 2012a

7.1. EIGENVALUES AND EIGENVECTORS OF A MATRIX 159
where z ∈ F. You would obtain the same collection of vectors if you replaced z with 4z.
Thus a simpler description for the solutions to this system of equations whose augmented
matrix is in (7.3) is
⎛
5
⎞
z ⎝ −2 ⎠ (7.4)
4
where z ∈ F. Now you need to remember that you can’t take z = 0 because this would
result in the zero vector and
Eigenvectors are never equal to zero!
Other than this value, every other choice of z in (7.4) results in an eigenvector. It is a good
idea to check your work! To do so, I will take the original matrix and multiply by this vector
and see if I get 5 times this vector.
⎛
⎝
5 −10 −5
2 14 2
−4 −8 6
⎞ ⎛
⎠ ⎝
5
−2
4
⎞ ⎛
⎠ = ⎝
25
−10
20
⎞ ⎛
⎠ =5⎝
so it appears this is correct. Always check your work on these problems if you care about
getting the answer right.
The variable, z is called a free variable or sometimes a parameter. The set of vectors in
(7.4) is called the eigenspace and it equals ker (λI − A) . You should observe that in this case
the eigenspace has dimension 1 because there is one vector which spans the eigenspace. In
general, you obtain the solution from the row echelon form and the number of different free
variables gives you the dimension of the eigenspace. Just remember that not every vector
in the eigenspace is an eigenvector. The vector, 0 is not an eigenvector although it is in the
eigenspace because
Eigenvectors are never equal to zero!
Next consider the eigenvectors for λ =10. These vectors are solutions to the equation,
⎛ ⎛
⎝10 ⎝ 1 0 0 ⎞ ⎛
⎞⎞
⎛
5 −10 −5
0 1 0 ⎠ − ⎝ 2 14 2 ⎠⎠
⎝ x ⎞ ⎛
y ⎠ = ⎝ 0 ⎞
0 ⎠
0 0 1 −4 −8 6 z 0
That is you must find the solutions to
⎛
⎝ −2 −4 −2
5 10 5
4 8 4
⎞ ⎛
⎠ ⎝ x ⎞ ⎛
y ⎠ = ⎝ 0 ⎞
0 ⎠
z 0
which reduces to consideration of the augmented matrix
⎛
⎞
−2 −4 −2 0 ⎠
⎝ 5 10 5 0
4 8 4 0
The row reduced echelon form for this matrix is
⎛
⎝ 1 2 1 0
0 0 0 0
0 0 0 0
⎞
⎠
5
−2
4
⎞
⎠