A First Course in Linear Algebra, 2017a

308 Linear Transformations
Since {T (⃗v 1 ),···,T (⃗v r )} is linearly independent, it follows that each c i = 0. Hence ∑ s j=1 a j⃗u j = 0andso,
since the {⃗u 1 ,···,⃗u s } are linearly independent, it follows that each a j = 0 also. Therefore {⃗u 1 ,···,⃗u s ,⃗v 1 ,···,⃗v r }
is a basis for V and so
n = s + r = dim(ker(T )) + dim(im(T ))
♠
The above theorem leads to the next corollary.
Corollary 5.53:
Let T : V → W be a linear transformation where V ,W are subspaces of R n . Suppose the dimension
of V is m. Then
dim(ker(T )) ≤ m
dim(im(T )) ≤ m
This follows directly from the fact that n = dim(ker(T )) + dim(im(T )).
Consider the following example.
Example 5.54:
Let T : R 2 → R 3 be defined by
⎡
T (⃗x)= ⎣
1 0
1 0
0 1
Then im(T )=V is a subspace of R 3 and T is an isomorphism of R 2 and V. Finda2 × 3 matrix A
such that the restriction of multiplication by A to V = im(T ) equals T −1 .
⎤
⎦⃗x
Solution. Since the two columns of the above matrix are linearly independent, we conclude that dim(im(T )) =
2 and therefore dim(ker(T )) = 2 − dim(im(T )) = 2 − 2 = 0 by Theorem 5.52. Then by Theorem 5.51 it
follows that T is one to one.
Thus T is an isomorphism of R 2 and the two dimensional subspace of R 3 which is the span of the
columns of the given matrix. Now in particular,
Thus
Extend T −1 to all of R 3 by defining
⎡
T (⃗e 1 )= ⎣
T −1 ⎡
⎣
1
1
0
1
1
0
⎤
⎡
⎦, T (⃗e 2 )= ⎣
⎤ ⎡
⎦ =⃗e 1 , T −1 ⎣
T −1 ⎡
⎣
0
1
0
⎤
⎦ =⃗e 1
0
0
1
⎤
0
0
1
⎤
⎦
⎦ =⃗e 2