Question 4563: Let T, U: V to W be linear transformations. Prove that:
a. R(T+U) is a subset of R(T)+R(U).
b. If W is finite-dimensional, then rank(T+U)=< rank(T)+rank(U).
c. Deduct from b that rank(A+B)=< rank(A)+rank(B) for any m x n matrices
A and B.
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! Let T, U: V to W be linear transformations. Prove that:
a. R(T+U) is a subset of R(T)+R(U).
b. If W is finite-dimensional, then rank(T+U)=< rank(T)+rank(U).
c. Deduct from b that rank(A+B)=< rank(A)+rank(B) for any m x n matrices
A and B.
Proof: a) x',y' are in R(T+U)
--> there exist x,y in V such that (T+U)(x) = x', (T+U)(y) = y'
--> x'+y'=(T+U)(x)+(T+U)(y)=(T+U)(x+y)= T(x)+U(x) + T(y)+ U(y)
= T(x+y) + U(x+y)
---> x'+y' is in R(T)+R(U)
x' is in R(T+U) and c is a scalar
--> there exist x,y in V such that (T+U)(x) = x'
---> (T+U)(cx) = c(T+U)(x) = cx' = T(cx)+ U(cx) for some x in V
--> cx' is in R(T) + R(U)
This shows R(T+U) is a subspace of R(T)+R(U).
b) Note T+U: V-->W is a linear transformation,
rank(T) and rank(U) <= dim W (finite)
Hence, rank(T) + rank(U) is finite.
By a) R(T+U) is a subspace of R(T)+R(U),
let W' be the intersection of R(T) and R(U).
we have dim R(T) + dim R(U) = dim [R(T)+R(U)]+ dim(W')
that is, rank(T) + rank(U) = rank(T+U) + dim(W')
Hence, rank(T+U) <= rank(T)+rank(U)
c) Consider
Define two linear transformations T, U from R^n to R^m
(or from F^n to F^m, column vectors of size n or m)
by T(ei) = E Aij fj and U(ei) = E Bij fj,
(where E means sumation){ei+i=1,,..,n} {fj|j=1,2,..m} are
standard basis of R^n & R^m resp.
Since rank(T) = rank(A) and rank(U) = rank(B)
By b), we have rank(T+U) <= rank(T)+rank(U)
Hence , rank(A+B)<= rank(A)+rank(B) for any m x n matrices
A and B.
Kenny
PS. You have to use the theorems in the text tofill some details.
In a) R(T+U) should be a subspace ofR(T)+R(U) ,
as a subset does not make much sense
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