Page 264 - Advanced engineering mathematics

244 CHAPTER 7 Matrices and Linear Systems

yields z = 0. But then y = 0, so x = 0 also. Then, from the third component, x − w =−w = 0
implies that w = 0. We conclude that
< x, y, z,w >=< 0,0,0,0 >.
The only vector T maps to the zero vector is the zero vector, so T is one-to-one. Clearly T
is not onto, since T does not map any vector to a 7-vector with a nonzero sixth or seventh
component.

m
n
Every linear transformation T : R → R can be associated with a matrix A T that carries all
n
of the information about the transformation. Recall that the standard basis for R consists of the
n orthogonal unit vectors
e 1 =< 1,0,··· ,0 >,e 2 =< 0,1,0,··· ,0 >,
··· ,e n =< 0,0,··· ,0,1 >
m
with a similar basis (with m components) for R .Now let A T be the matrix whose columns are
m
of the images in R of T (e 1 ), T (e 2 ),··· , T (e n ) with coordinates written in terms of the standard
m
basis in R .The A T is an m × n matrix that represents T in the sense that
⎛ ⎞
x 1
x 2
⎜ ⎟
T (x 1 , x 2 ,··· , x n ) = A T ⎜ . ⎟.
⎜ ⎟
.
⎝ . ⎠
x n
Thus we can compute T (X) as the matrix product of A T with the column matrix of the compo-
nents of X. Note that A T is m × n, and X (written as a column matrix) is n × 1, so A T X is m × 1.
m
Hence, it is a vector in R .
EXAMPLE 7.37
Let T (x, y) =< x − y,0,0 >, as in Example 7.34. Then
T (1,0) =< 1,0,0 > and T (0,1) =< −1,0,0 >
so
⎛ ⎞
1 −1
A T = 0 0 ⎠ .
⎝
0 0
Now observe that
⎛ ⎞
x − y

x
A T = ⎝ 0 ⎠ ,
y
0
3
giving the coordinates of T (x, y) with respect to the standard basis for R . We can therefore read
the coordinates of T (x, y) as a matrix product.
EXAMPLE 7.38
In Example 7.36 we had
T (x, y, z,w) =< x − y + 2z + 8w, y − z, x − w, y + 4w,5x + 5y − z,0,0 >.
For the matrix of T , compute
T (1,0,0,0) =< 1,0,1,0,5,0,0 >, T (0,1,0,0) =< −1,1,0,1,5,0,0 >
T (0,0,1,0) =< 2,−1,0,0,−1,0,0 >, T (0,0,0,1) =< 8,0,−1,4,0,0,0 >.

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October 14, 2010 14:23 THM/NEIL Page-244 27410_07_ch07_p187-246