The rank equal the number linearly independent elements
106 Chapter 3 Matrices and Linear Transformations
It should be made clear that applying a linear transformation to a basis does not produce the basis for the new vector space. It only shows where the basis vectors end up in the new vector space — in our case in terms of the standard basis. In fact, a transformed basis may be no longer linearly independent. Take as another example
These two examples illustrate one useful property. If the rank of a linear transformation T equals the number of elements in a transformed basis β, then we can say that β is linearly independent. In fact, the rank is equal to the number of linearly independent elements in β, and those linearly independent elements will span the range of T.
3.3.4 Matrices and Linear Transformations
b0 | ⎡⎢⎢⎢⎣ | a0,0 | a0,1 |
|
|
⎤⎥⎥⎥⎦ | ⎡⎢⎢⎢⎣ | x0 | |||
---|---|---|---|---|---|---|---|---|---|---|---|
b1 | a1,0 | a1,1 | x1 | ||||||||
... | ... | ... | ... | ||||||||
· · · | |||||||||||
bm−1 | am−1,0 | am−1,1 | am−1,n−1 | xn−1 |
will store, in order, each of these transformed basis vectors as the columns of A, or
A =�a0 a1 · · · an−1�
Using our matrix multiplication definition to compute the product of A and a vector x in V, we see that the result for element i in b is
�c0 |
|
� | = | �x0 |
|
� | ⎡⎢⎢⎢⎣ | a0,0 | a0,1 | ⎤ ⎥⎥⎥⎦ |
||
---|---|---|---|---|---|---|---|---|---|---|---|---|
a1,0 | a1,1 | |||||||||||
... | ... | |||||||||||
am−1,0 | am−1,1 | · · · | am−1,n−1 |
The column space and row space are not necessarily the same vector space. As an example, take the matrix
⎡⎣ | 0 | 1 | 0 |
|
---|---|---|---|---|
0 | 0 | 1 | ||
0 | 0 | 0 |