# Thread: [SOLVED] Bonus Question Linear Algebra

1. ## [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!

2. 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!
.................................................. ...

3. 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