# [SOLVED] Bonus Question Linear Algebra

• Oct 6th 2009, 09:48 PM
Noxide
[SOLVED] Bonus Question Linear Algebra
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?
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?

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

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!
• Oct 7th 2009, 04:51 AM
tonio
Quote:

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!

.................................................. ...
• Oct 7th 2009, 05:56 AM
viko
given subspaces $\displaystyle V, W \ \in R^2$

Thinking the intersection in R^2, it could be only two things: a line (when both subespaces were the same line) or a point (0,0), when both were different.
(Remember that the (0,0) must be in any subspace).

in both cases you get a subspace.

The union could be: the same line (when both were the same line) or two different lines.

In the first case $\displaystyle \ V\ \cup \ W$ is, of course a subspace.

In this case $\displaystyle \ V \ \cup \ W$ is not a subspace since you can add two element on it and get an element from the outside.

viko
• Oct 7th 2009, 03:37 PM
Noxide