A First Course in Linear Algebra, 2017a

270 Linear Transformations
In this case, A will be a 2 × 3 matrix, so we need to find T (⃗e 1 ),T (⃗e 2 ),andT (⃗e 3 ). Luckily, we have
been given these values so we can fill in A as needed, using these vectors as the columns of A. Hence,
[ ]
1 9 1
A =
2 −3 1
In this example, we were given the resulting vectors of T (⃗e 1 ),T (⃗e 2 ),andT (⃗e 3 ). Constructing the
matrix A was simple, as we could simply use these vectors as the columns of A. The next example shows
how to find A when we are not given the T (⃗e i ) so clearly.
Example 5.9: The Matrix of Linear Transformation: Inconveniently
Defined
Suppose T is a linear transformation, T : R 2 → R 2 and
[ ] [ ] [ ] [ ]
1 1 0 3
T = , T =
1 2 −1 2
Find the matrix A of T such that T (⃗x)=A⃗x for all⃗x.
♠
Solution. By Theorem 5.6 to find this matrix, we need to determine the action of T on ⃗e 1 and ⃗e 2 . In
Example 9.91, we were given these resulting vectors. However, in this example, we have been given T
of two different vectors. How can we find out the action of T on ⃗e 1 and ⃗e 2 ? In particular for ⃗e 1 , suppose
there exist x and y such that [ ] [ ] [ ]
1 1 0
= x + y
(5.2)
0 1 −1
Then, since T is linear,
[ 1
T
0
] [ 1
= xT
1
] [
+ yT
0
−1
]
Substituting in values, this sum becomes
[ ] 1
T
0
[ 1
= x
2
] [ 3
+ y
2
]
(5.3)
Therefore, if we know the values of x and y which satisfy 5.2, we can substitute these into equation
5.3. By doing so, we find T (⃗e 1 ) which is the first column of the matrix A.
We proceed to find x and y. We do so by solving 5.2, which can be done by solving the system
x = 1
x − y = 0
We see that x = 1andy = 1 is the solution to this system. Substituting these values into equation 5.3,
we have
[ ] [ ] [ ] [ ] [ ] [ ]
1 1 3 1 3 4
T = 1 + 1 = + =
0 2 2 2 2 4