Page 275 - A Course in Linear Algebra with Applications
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8.1: Basic Theory of Eigenvectors 259
The general solution of this system is £1 = |(—1 + y/—l) and
— d , where d is an arbitrary scalar. Thus the eigenvectors
x 2
of A associated with the eigenvalue C\ are the non-zero vectors
of the form
Notice that these, together with the zero vector, form a 1-
2
dimensional subspace of C . In a similar manner the eigen-
vectors for the eigenvalue 3 — \/—T are found to be the vectors
of the form
where d ^ 0. Again these form with the zero vector a subspace
2
ofC .
It should be clear to the reader that the method used in
this example is in fact a general procedure for finding eigen-
vectors and eigenvalues. This will now be described in detail.
The characteristic equation of a matrix
Let A be an n x n matrix over a field of scalars F, and let
X be a non-zero n-column vector over F. The condition for X
to be an eigenvector of A is AX = cX, or
{A - d n)X = 0,
where c is the corresponding eigenvalue. Hence the eigenvec-
tors associated with c, together with the zero vector, form
the null space of the matrix A — cl n. This subspace is often
referred to as the eigenspace of the eigenvalue c.
Now (A — d n)X = 0 is a linear system of n equations in n
unknowns. By 3.3.2 the condition for there to be a non-trivial
solution of the system is that the coefficient matrix have zero
determinant,
det(A - cl n) = 0.