Notes 3/MATH 223 & MATH 247/Winter, 2009
1. Matrices of linear mappings
Suppose we have:
vector spaces and ;
dim=, dim=;
basis = () of ; basis = () of ;
and
linear mapping .
We define the matrix of relative to the bases and .The notation for this matrix is , where, in the subscript position, the two bases and are supposed to be indicated.
For short, let us write for this matrix now. Its definition is as follows.
is an matrix ( =dimension of the codomain space ; =dimension of the domain space ; );
;
or, in more detail,
= the the column of = the coordinate vector of relative to the basis (note: is applied to the th basis vector in ) .
Theorem For any vector in , we have
=
(Partly) in words: the coordinate vector of relative to the basis equals the matrix described above times the coordinate vector of relative to the basis .
Proof Let be . Then . Therefore, since is linear, . Thus,
= = = .
Here, we made use of the fact that “taking coordinate vectors is a linear operation”:
;
and
.
Most common special case: when , and = . Now, is a linear operator. We abbreviate as . The columns of are the coordinate vectors , for ; and the equality of the theorem becomes, for arbitrary in :
= .
Example 1. =, ; , ; ==, the standard basis of , ==, the standard basis of ; defined by the formula
Now: = = = ;
= = =
The matrix = is
= [,] = [,] = .
Given, as an example, , we have: = = ;
= = ; = = .
On the other hand:
= = = =
We see that the theorem is verified in this special case (which, of course, is not a proof of the theorem itself). But we see even more: not only that the results of the two calculations are the same, but also, that the intermediate expressions are also, essentially, the same. The theorem itself has a very easy proof, one that is not more complicated than the calculations above.
Example 2. The spaces and , as well as the mapping the same as in Example 1; the bases , consist of vectors defined as follows:
; .
Now: = = = ;
= = =
But, now we need and . If = , then
= = = ;
which means that (!)
=
Denoting the matrix by , and accepting the fact that is invertible, this means that
=
[ is the so-called transition, or change-of-basis, matrix for the change of basis from , the standard basis in , to the “new” basis . Later, there will be more on “change-of-basis”.]
We can calculate
= ; and so
= = = .
Similarly,
= = = .
Thus, the matrix = is
= [,] = .
The last matrix can be used directly to calculate if is given in the form ; the result will be given in the form .
For instance, let . Then = , and thus
= = = ,
which means that
. (*)
If we want to check this result , then we calculate
= = ;
from the definition of directly:
= ;
On the other hand, from the formula (*)
= = ;
unbelievably, the same!
Example 3. = , = ;
= (), the standard basis of :
= (), the standard basis of :
;
defined by , with the fixed matrix .
First of all, it is easy to see, without calculation, that so defined is a linear mapping:
;
and similarly for the other condition.
We have that dim()=6 , dim()=4. thus, the matrix is a matrix.
The columns of are
.
We need to calculate the values at hand.
= = = ,
= = = ,
= = =
= = =
= = =
= = = ;
from which, the matrix is
=
2. Composition of linear mappings
Suppose we have vector spaces and . Let us also suppose that we have linear mappings and as in
.
Then we can form the composite linear mapping
: ,
defined as follows:
.
In other words, is applied to any by first applying to , and then, applying to the result . Since the domain of is supposed to coincide with the codomain of , the space , and is in , can be meaningfully applied to , and the result will be in .
It is easy to see that and being linear implies that is a linear.
Let us now also assume that we have respective selected bases , and of the spaces and . Then we can form the following matrices:
, , and .
(Note carefully the subscripts in these formulas. They are the only ones that are meaningfully applicable in our context.)
Theorem Under the above notation, . In an abbreviated notation:
Remark: the theorem says that “composition of linear mappings corresponds to matrix multiplication.
Proof Let be any vector in . Let . Then = . We have . Therefore,
=
= ;
at the exclamation mark, we used the associativity of matrix multiplication:
,
with . We have shown that
= .
The definition of the matrix says that
= .
Therefore = . Since this holds true for all in , we must have that , as asserted.
2