Proof.
Suppose
A
and
B
are similar matrices. By definition, the nullity of
A
is the dimension of its null space. But
N
(
A
) =
E
0
(
A
), the 0eigenspace
of
A
. By the last sentence in the statement of Proposition 4.2,
E
0
(
A
) and
E
0
(
B
) have the same dimension.
Hence
A
and
B
have the same nullity.
Since rank +nullity =
n
,
A
and
B
also have the same rank.
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152
16. SIMILAR MATRICES
4.3. Intrinsic and extrinsic properties of matrices.
It is not un
reasonable to ask whether a matrix that is similar to a symmetric ma
trix is symmetric.
Suppose
A
is symmetric and
S
is invertible.
Then
(
SAS

1
)
T
= (
S

1
)
T
A
T
S
T
= (
S
T
)

1
AS
T
and there is no obvious reason
which this should be the same as
SAS

1
. The explicit example
1
2
0
1
2
3
3
4
1
2
0
1

1
=
8
11
3
4
1

2
0
1
=
8

5
3

2
shows that a matrix similar to a symmetric matrix need not be symmetric.
PAUL  More to say.
5. Diagonalizable matrices
5.1. Diagonal matrices.
A square matrix
D
= (
d
ij
) is
diagonal
if
d
ij
= 0 for all
i
6
=
j
. In other words,
D
=
λ
1
0
0
· · ·
0
0
0
λ
2
0
· · ·
0
0
0
0
λ
3
· · ·
0
0
.
.
.
.
.
.
.
.
.
0
0
0
· · ·
λ
n

1
0
0
0
0
· · ·
0
λ
n
for some
λ
1
, . . . , λ
n
∈
R
. We sometimes use the abbreviation
diag(
λ
1
, . . . , λ
n
) =
λ
1
0
0
· · ·
0
0
0
λ
2
0
· · ·
0
0
0
0
λ
3
· · ·
0
0
.
.
.
.
.
.
.
.
.
0
0
0
· · ·
λ
n

1
0
0
0
0
· · ·
0
λ
n
.
The identity matrix and the zero matrix are diagonal matrices.
Let
D
= diag(
λ
1
, . . . , λ
n
). The following facts are obvious:
(1) the
λ
i
s are the eigenvalues of
D
;
(2) det(
D
) =
λ
1
λ
2
· · ·
λ
n
;
(3) the characteristic polynomial of
D
is (
λ
1

t
)(
λ
2

t
)
· · ·
(
λ
n

t
);
(4)
De
j
=
λ
j
e
j
, i.e., the standard basis vectors
e
j
are eigenvectors for
D
.
Warning:
Although the axis
R
e
j
is contained in
λ
j
eigenspace for
D
its
λ
j
eigenspace might be bigger.
For example, if
λ
i
=
λ
j
, then the
E
λ
i
contains the plane
R
e
i
+
R
e
j
.
5. DIAGONALIZABLE MATRICES
153
5.2. Definition.
An
n
×
n
matrix
A
is
diagonalizable
if it is similar to
a diagonal matrix, i.e., if
S

1
AS
is diagonal for some
S
.
A diagonal matrix is diagonalizable:
if
D
is diagonal, then
I

1
DI
is
diagonal!
Since similar matrices have the same characteristic polynomials, the
characteristic polynomial of a diagonalizable matrix is a product of
n
linear
terms. See “obvious fact” (3) above.
5.3. Example.
If
A
=
5

6
3

4
and
S
=
2
1
1
1
,
then
S

1
AS
=
1

1

1
2
5

6
3

4
2
1
1
1
=
2
0
0

1
so
A
is diagonalizable.
5.4.
The obvious questions are
how do we determine whether
A
is di
agonalizable or not
and if
A
is diagonalizable
how do we find an
S
such
that
S

1
AS
is diagonal.
The next theorem and its corollary answer these
questions.
Theorem
5.1
.
An
n
×
n
matrix
A
is diagonalizable if and only if it has
n
linearly independent eigenvectors.
Proof.
We will use the following fact. If
B
is a
p
×
q
matrix and
C
a
q
×
r
matrix, then the columns of
BC
are obtained by multiplying each column
of
C
by
B
; explicitly,
BC
= [
BC
1
, . . . , BC
r
]
where
C
j
is the
j
th
column of
C
.