eigenvector. This means that our original supposition that a 1 , a 2 , Á Á Á, and a n are
linearly dependent was wrong. Thus, the eigenvectors a 1 , a 2 , Á Á Á, and a n should
be linearly independent.
(ii) Case II: α n ¼ 0.
Suppose again that a 1 , a 2 , Á Á Á, and a n are linearly dependent. Since as before
c n 6 ¼ 0, we get (12.40) and (12.41) once again. Putting α n ¼ 0 in (12.41) we have
0 ¼ À
c 1
c n
α 1 a 1 þ
c 2
c n
α 2 a 2 þ Á Á Á þ
c nÀ1
c n
α nÀ1 a nÀ1
:
ð12:44Þ
Since eigenvalues are different, we should have α 1 6 ¼ 0, α 2 6 ¼ 0, Á Á Á, and
α n À 1 6 ¼ 0. Then, considering that a 1 , a 2 , Á Á Á, and a n À 1 are linearly independent, for (12.44) to hold we must have c 1 ¼ c 2 ¼ Á Á Á ¼ c n À 1 ¼ 0. But, from
(12.40) we have a n ¼ 0, again in contradiction to that a n is an eigenvector. Thus,
the eigenvectors a 1 , a 2 , Á Á Á, and a n should be linearly independent. These
complete the proof.
∎
12.2 Eigenspaces and Invariant Subspaces
Equations (12.21), (12.30), and (12.31) show that if we adopt an eigenvector as one
of basis vectors, the matrix representation of the linear transformation A in reference
to such basis vectors is obtained so that the leftmost column is zero except for the left
top corner on which an eigenvalue is positioned. (Note that if the said eigenvector is
zero, the leftmost column is a zero vector.) Meanwhile, neither Theorem 12.1 nor
Theorem 12.2 tells about multiplicity of eigenvalues. If the eigenvalues have
multiplicity, we have to think about different aspects. This is a major issue of this
section.
Let us start with a discussion of invariant subspaces. Let A be a (n, n) square
matrix. If a subspace W in V
n is characterized by
x 2 W⟹Ax 2 W,
W is said to be invariant with respect to A (or simply A-invariant) or an invariant
subspace in V
n
¼ Span{a 1 , a 2 , Á Á Á, a n }. Suppose that x is an eigenvector of A and that
its corresponding eigenvalue is α. Then, Span{x} is an invariant subspace of V
n . It is
because A(cx) = cAx = cαx ¼ α(cx) and cx is again an eigenvector belonging to α.
Suppose that dimW ¼ m (m
n) and that W ¼ Span{a 1 , a 2 , Á Á Á, a m }. If W is Ainvariant, A causes a linear transformation within W. In that case, expressing A in
reference to a 1 , a 2 , Á Á Á, a m , a m + 1 , Á Á Á, a n , we have
468
12 Canonical Forms of Matrices
linearly dependent was wrong. Thus, the eigenvectors a 1 , a 2 , Á Á Á, and a n should
be linearly independent.
(ii) Case II: α n ¼ 0.
Suppose again that a 1 , a 2 , Á Á Á, and a n are linearly dependent. Since as before
c n 6 ¼ 0, we get (12.40) and (12.41) once again. Putting α n ¼ 0 in (12.41) we have
0 ¼ À
c 1
c n
α 1 a 1 þ
c 2
c n
α 2 a 2 þ Á Á Á þ
c nÀ1
c n
α nÀ1 a nÀ1
:
ð12:44Þ
Since eigenvalues are different, we should have α 1 6 ¼ 0, α 2 6 ¼ 0, Á Á Á, and
α n À 1 6 ¼ 0. Then, considering that a 1 , a 2 , Á Á Á, and a n À 1 are linearly independent, for (12.44) to hold we must have c 1 ¼ c 2 ¼ Á Á Á ¼ c n À 1 ¼ 0. But, from
(12.40) we have a n ¼ 0, again in contradiction to that a n is an eigenvector. Thus,
the eigenvectors a 1 , a 2 , Á Á Á, and a n should be linearly independent. These
complete the proof.
∎
12.2 Eigenspaces and Invariant Subspaces
Equations (12.21), (12.30), and (12.31) show that if we adopt an eigenvector as one
of basis vectors, the matrix representation of the linear transformation A in reference
to such basis vectors is obtained so that the leftmost column is zero except for the left
top corner on which an eigenvalue is positioned. (Note that if the said eigenvector is
zero, the leftmost column is a zero vector.) Meanwhile, neither Theorem 12.1 nor
Theorem 12.2 tells about multiplicity of eigenvalues. If the eigenvalues have
multiplicity, we have to think about different aspects. This is a major issue of this
section.
Let us start with a discussion of invariant subspaces. Let A be a (n, n) square
matrix. If a subspace W in V
n is characterized by
x 2 W⟹Ax 2 W,
W is said to be invariant with respect to A (or simply A-invariant) or an invariant
subspace in V
n
¼ Span{a 1 , a 2 , Á Á Á, a n }. Suppose that x is an eigenvector of A and that
its corresponding eigenvalue is α. Then, Span{x} is an invariant subspace of V
n . It is
because A(cx) = cAx = cαx ¼ α(cx) and cx is again an eigenvector belonging to α.
Suppose that dimW ¼ m (m
n) and that W ¼ Span{a 1 , a 2 , Á Á Á, a m }. If W is Ainvariant, A causes a linear transformation within W. In that case, expressing A in
reference to a 1 , a 2 , Á Á Á, a m , a m + 1 , Á Á Á, a n , we have
468
12 Canonical Forms of Matrices