Answer to Question #196685 in Linear Algebra for Herman

Question #196685

Suppose U1,U2,..,Um are finite-dimensional subspace of V.

Prove that :

U1+U2+...+Um is finite dementional and

dim(U1+U2+...+Um)≤ dimU1 + dimU2 + ..... + dimUm


1
Expert's answer
2021-05-24T12:02:15-0400

Suppose "u_1,u_2,...,u_m" are finite-dimensional subspaces of V.


Thus, Each "u_j" has a finite basis.


Concatenate these lists to get a spanning list of length "dim(u_1) + \u00b7 \u00b7 \u00b7 + dim(u_m) \\text{ for } u_1 + \u00b7 \u00b7 \u00b7 + u_m."


This shows that "u_1+\u00b7 \u00b7 \u00b7+u_m" is finite dimensional and since any spanning list can be reduced


to a basis then "dim(u_1 + \u00b7 \u00b7 \u00b7 + u_m) \u2264 dim(u_1) + \u00b7 \u00b7 \u00b7 + dim(u_m)."


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