A First Course in Linear Algebra, 2017a

300 Linear Transformations
You can exhibit an isomorphism of these two spaces as follows.
⎡ ⎤ ⎡ ⎤
1
0
T (⃗e 1 )= ⎢ 2
⎥
⎣ 1 ⎦ ,T (⃗e 2)= ⎢ 1
⎥
⎣ 0 ⎦ ,T (⃗e 3)=
1
1
and extend linearly. Recall that the matrix of this linear transformation is just the matrix having these
vectors as columns. Thus the matrix of this isomorphism is
⎡ ⎤
1 0 1
⎢ 2 1 1
⎥
⎣ 1 0 2 ⎦
1 1 0
You should check that multiplication on the left by this matrix does reproduce the claimed effect resulting
from an application by T .
♠
Consider the following example.
Example 5.46: Finding the Matrix of an Isomorphism
⎡
⎢
⎣
1
1
2
0
⎤
⎥
⎦
Let V = R 3 and let W denote
⎧⎡
⎪⎨
span ⎢
⎣
⎪⎩
1
2
1
1
⎤
⎡
⎥
⎦ , ⎢
⎣
0
1
0
1
⎤
⎡
⎥
⎦ , ⎢
⎣
1
1
2
0
⎤⎫
⎪⎬
⎥
⎦
⎪⎭
Let T : V ↦→ W be defined as follows.
⎡ ⎤
⎡ ⎤ 1
1
T ⎣ 1 ⎦ = ⎢ 2
⎣ 1
0
1
Find the matrix of this isomorphism T .
⎡
⎥
⎦ ,T ⎣
0
1
1
⎤
⎦ =
⎡
⎢
⎣
0
1
0
1
⎤
⎡
⎥
⎦ ,T ⎣
1
1
1
⎤
⎦ =
⎡
⎢
⎣
1
1
2
0
⎤
⎥
⎦
Solution. First note that the vectors
⎡
⎣
1
1
0
⎤
⎡
⎦, ⎣
0
1
1
⎤
⎡
⎦, ⎣
are indeed a basis for R 3 as can be seen by making them the columns of a matrix and using the reduced
row-echelon form.
Now recall the matrix of T is a 4 × 3matrixA which gives the same effect as T . Thus, from the way
we multiply matrices,
⎡
A⎣
1 0 1
1 1 1
0 1 1
⎤
⎦ =
⎡
⎢
⎣
1
1
1
⎤
⎦
1 0 1
2 1 1
1 0 2
1 1 0
⎤
⎥
⎦