Our first decomposition applies only to diagonalizable (Definition DZM)
matrices, and yields a decomposition into a sum of very simple matrices.
Theorem ROD Rank One Decomposition Suppose that A is a
diagonalizable matrix of size n
and rank r. Then
there are r square
matrices {A}_{1},\kern 1.95872pt {A}_{2},\kern 1.95872pt {A}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {A}_{r},
each of size n
and rank 1
such that
Furthermore, if {λ}_{1},\kern 1.95872pt {λ}_{2},\kern 1.95872pt {λ}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {λ}_{r} are the
nonzero eigenvalues of A, then
there are two sets of r linearly
independent vectors from {ℂ}^{n},
such that {A}_{k} = {λ}_{k}{x}_{k}{y}_{k}^{t},
1 ≤ k ≤ r.
□
Proof The proof is constructive. Generally, we will diagonalize
A, creating a nonsingular
matrix S and a
diagonal matrix D.
Then we split up the diagonal matrix into a sum of matrices with a single nonzero
entry (on the diagonal). This fundamentally creates the decomposition in the
statement of the theorem, the remainder is just bookkeeping. The vectors in
X and
Y will result from
the columns of S
and the rows of {S}^{−1}.
Let {λ}_{1},\kern 1.95872pt {λ}_{2},\kern 1.95872pt {λ}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {λ}_{n} be the
eigenvalues of A
(repeated according to their algebraic multiplicity). If
A has rank
r, then
\mathop{ dim}\nolimits \left (N\kern -1.95872pt \left (A\right )\right ) = n − r (Theorem RPNC).
The null space of A is the
eigenspace of the eigenvalue λ = 0
(Theorem EMNS), so it follows that the algebraic multiplicity of
λ = 0 is
n − r,
{α}_{A}\left (0\right ) = n − r.
Presume that the complete list of eigenvalues is ordered so that
{λ}_{k} = 0 for
r + 1 ≤ k ≤ n.
Since A
is hypothesized to be diagonalizable, there exists a diagonal matrix
D and an invertible
matrix S, such that
D = {S}^{−1}AS. We can rearrange
tis equation to read, A = SD{S}^{−1}.
Also, the proof of Theorem DC says that the diagonal elements of
D are the
eigenvalues of A
and we have the flexibility to assume they lie on the diagonal
in the same order as we have specified above. Now, let
{X}^{∗} = \left \{{x}_{
1},\kern 1.95872pt {x}_{2},\kern 1.95872pt {x}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {x}_{n}\right \} be the columns
of S, and let
{Y }^{∗} = \left \{{y}_{
1},\kern 1.95872pt {y}_{2},\kern 1.95872pt {y}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {y}_{n}\right \} be the rows
of {S}^{−1} converted
to column vectors. With little motivation other than the statement of the theorem,
define size n
matrices {A}_{k},
1 ≤ k ≤ n by
{A}_{k} = {λ}_{k}{x}_{k}{y}_{k}^{t}. Finally, let
{D}_{k} be the size
n matrix that is totally
zero, other than having {λ}_{k}
in row k and
column k.
With everything in place, we compute entry-by-entry,
So by Definition ME we have the desired equality of matrices. The careful reader will have
noted that {A}_{k} = O,
r + 1 ≤ k ≤ n, since
{λ}_{k} = 0 in these instances.
To get the sets X
and Y from
{X}^{∗} and
{Y }^{∗}, simply discard the last
n − r vectors. We can safely
ignore (or remove) {A}_{r+1},\kern 1.95872pt {A}_{r+2},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {A}_{n}
from the summation just derived.
One last assertion to check. What is the rank of
{A}_{k},
1 ≤ k ≤ r? Every row
of {A}_{k} is a scalar
multiple of {y}_{k}^{t},
row k of the
nonsingular matrix {S}^{−1}
(Theorem MIMI). As a row of a nonsingular matrix,
{y}_{k}^{t} cannot be all zeros.
In particular, row i
of {A}_{k} is obtained as a
scalar multiple of {y}_{k}^{t}
by the scalar {α}_{k}{\left [{x}_{k}\right ]}_{i}.
We have restricted ourselves to the nonzero eigenvalues of
A, and as
S is nonsingular,
some entry of {x}_{k}
is nonzero. This all implies that some row of
{A}_{k} will be nonzero. Now
consider row-reducing {A}_{k}.
Swap the nonzero row up into row 1. Use scalar multiples of this row to zero out
every other row. This leaves a single nonzero row in the reduced row-echelon form, so
{A}_{k} has rank
one. ■
We record two observations that was not stated in our theorem above. First, the vectors in
X, chosen as columns of
S, are eigenvectors of
A. Second, the product of
two vectors from X and
Y in the opposite order,
by which we mean {y}_{i}^{t}{x}_{
j},
is the entry in row i
and column j of the
matrix product {S}^{−1}S = {I}_{
n}
(Theorem EMP). In particular,
With n = 2
distinct eigenvalues, Theorem DED tells us that
A
is diagonalizable, and with no zero eigenvalues we see that
A has
full rank. Theorem DC says we can construct the nonsingular matrix
S with
eigenvectors of A
as columns, so we have
The algebraic and geometric multiplicities of each eigenvalue are equal, so Theorem DMFE
tells us that A
is diagonalizable. With a single zero eigenvalue we see that
A has
rank 4 − 1 = 3.
Theorem DC says we can construct the nonsingular matrix
S with
eigenvectors of A
as columns, so we have