Page 67 - Linear Algebra Done Right
P. 67
Proof: Let
...
a 1,1
.
.
Invertibility a 1,n 53
3.15 M(T) = . . . . .
a m,1 ... a m,n
This means, we recall, that
m
3.16 Tv k = a j,k w j
j=1
for each k. Let v be an arbitrary vector in V, which we can write in the
form 3.12. Thus M(v) is given by 3.13. Now
Tv = b 1 Tv 1 +· · ·+ b n Tv n
m m
= b 1 a j,1 w j +· · ·+ b n a j,n w j
j=1 j=1
m
= (a j,1 b 1 + ··· + a j,n b n )w j ,
j=1
where the first equality comes from 3.12 and the second equality comes
from 3.16. The last equation shows that M(Tv), the m-by-1 matrix of
the vector Tv with respect to the basis (w 1 ,...,w m ), is given by the
equation
a 1,1 b 1 +· · ·+ a 1,n b n
.
M(Tv) = . . .
a m,1 b 1 +· · ·+ a m,n b n
This formula, along with the formulas 3.15 and 3.13 and the definition
of matrix multiplication, shows that M(Tv) =M(T)M(v).
Invertibility
A linear map T ∈L(V, W) is called invertible if there exists a linear
map S ∈L(W, V) such that ST equals the identity map on V and TS
equals the identity map on W. A linear map S ∈L(W, V) satisfying
ST = I and TS = I is called an inverse of T (note that the first I is the
identity map on V and the second I is the identity map on W).
If S and S are inverses of T, then