Originally Posted by

**Noxide** Hello people.

I just had a midterm today and did really well. There was a bonus question on the test that I had no idea how to do. Could someone please show me?

Prove or give a counter example:

a) If S and T are subspaces of R^n, then S (union) T is also a subspace R^n

is this b/c every element of s and every element of T are elements of R^n, so every element of S U T is also an element of R^n?

This is always false UNLESS one of them is contained in the other one.

For example take S = {(x,y) in R^2; x = y} and T = {(x,0) in R^2}.

Both S, T are subspaces but their union isn't since if you take (1,1) in S and (1,0) in T and you sum them you get (2,1), which isn't neither in S nor in T.

b) If S and T are subspaces of R^n, then S (intersection) T is also a subspace of R^n

is this b/c every element of s and every element of T are elements of R^n, so every element of S (intersection) T is also an element of R^n?

I don't understand what you mean by "is this b/c every element of s and every element of T are elements of R^n, so every element of S (intersection) T is also an element of R^n?"...in fact, I douibt whether you have an accurate idea what this means.

Anyway this is always true and you can easily prove it using the definition of subspace.

Tonio

lol, I guess this is a bonus bonus question, but my prof made this a try it yourself question in his notes and i have no clue how to do it....

using the fact that rank(AB) <= rank(B) prove that if A and B are square matrices satisfying AB = I(sub)n then A^-1 = B and B^-1 = A

Well, if AB = I then rk(B) >= rk(AB) = rk(I) = n, but clearly rk(B) <= n for any nxn square matric ==> rk(B) = n and B is invertible. The same is true for A and thus from AB = I we get A, B are inverses of each other.

Tonio

all i know so far is that rank is the number of pivot columns of a matrix

idk why our prof is introducing if then statements now, it's the last chapter in our textbook!

anyway, help would be much appreciated!