For time-dependent systems, we define an operator
The question of controllability reduces to: is 
?
Proof:
is a symmetric matrix called the controllability Gramian. Consider
is 
, 
is 
. Then
If
s.t.
For 
as defined above. So
That is, every vector in the range of 
is in the
attainable set. Vice versa, we show that every vector
is also in 
, the
range of the Gramian. Let . Then there is
a control 
such that
Let 
. Since is a real symmetric matrix, the whole
space 
can be decomposed into
Therefore, 
, and 
with 
and 
. Now
Since , 
and 
, or
where
Hence
and
Hence 
. Consequently, 
and the result is
proved.
The result above may be useful when trying to determine the attainable set for time-dependent systems. Similarly as in time-invariant systems
Let 
exists. Suppose we are given 
, 
. Find
s.t. the solution starting from 
will reach 
.
Try
Then
so will be a solution.
If 
and 
(i.e., constants) then
In general
is symmetric.
are 
eigenvalues of 
, 
,
. If 
, then
, 
.
If 
.
In constant sytems
choose 
, so
If
For time invariant systems
is necessary and sufficient for controllability.
Example 
.
If
(Matlab >> ctrb(A,B))
If
then not controllable.
If
still controllable. If
called partial controllability.
That is, there exists a non-singular matrix 
, s.t.
.
and
are the same.
Proof:
(
) We prove that if 
for some
, then the rank of the controllability matrix is
deficient. Let be such that .
Then there is a nonzero vector 
such that
or
Multiply 
by 
and
use the identities
hence
and
This show that the 
rows of the controllability matrix are
linearly dependent. Hence the rank of the matrix is less than .
(
) If there is a such that (1.15)
holds, then
Consider the sequence 
. Let
be the minimal polynomial corresponding to 
and
, i.e., the monic polynimial of lowest degree such that
Let 
be a root of that polynomial. The root exists because
the degree of is at least 
. (If the degree
were 
, then it would mean that 
, and
, a contradiction).
and
Let
Then
hence
Hence a nonzero 
has been found wich violates the Hautus condition.
Example 
.
What is
The columns of 
are linearly independent. To prove controllability,
we just need to show that whatever the value of , there
always is a column of 
which completes the columns
of to a linearly independent set of 
columns.
If 
,
the second column of is 
, with 
. This column and
the two columns of form a basis for 
.
If 
, the second column of is null.
However, the last columns of 
are
which has rank . Hence, the Hautus condition holds when
and when , that is for all
. Therefore, the system is controllable.
Example 
.
We can use the same matrix 
to determine all columns 
such that
the pair 
is not controllable, i.e.,
we find the set of all linearly independent 
, (
vector),
such that the system 
is not controllable.
Form the matrix
Then
The system is not controllable if 
, or 
, regardless of the value of 
. The vectors
are linearly independent vectors in such that
, 
is not controllable.
Let us interpret the vectors .
So, 
is an eigenvector of with associated eigenvalue 
,
is a generalized eigenvector, and 
is an eigenvector
of with associated eigenvalue . Consider now the pairwise
linear combinations of these vectors, with coefficients 
.
The set of noncontrollable directions for consists of planes
in the three dimensional space. Any vector not on one of these two planes
Figure 1.2: There are directions of noncontrollability.
yeilds a such that is controllable. One can see that the
set of noncontrollable directions for is ``thin'', i.e., almost
any perturbation of entries of will result in a controllable system.
In general if 
eigenvector, then
is a matrix of rank . Therefore, eigenvectors do not provide controllable
directions for .
Example 
.
In general, if is a Jordan block, a vector with a
non-zero entry in the bottom row ,
will give a controllable direction.
We can take a transformation that reduces to a diagonalizable form or a
Jordan form.
Let's assume that is diagonalizable.
What is the controllability characteristic of this system?
When is 
non-singular? First assume 
's are distinct.
A necessary condition for controllability is that all 
. But it
is also sufficient because
Vandermonde determinant 
. If 
's are not distinct,
then no controllability.
Example 
.
Eventually obtain
See Figure 1.3.
Example 
.
If two circuits are identical then rank of matrix is . So 
, and 
, 
, i.e., not controllable.
See Figure 1.4.
