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Finding a subspace (possibly intersection of subspace?)

by Throwback
Tags: intersection, possibly, subspace
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Throwback
#1
Nov5-11, 05:31 PM
P: 13
1. The problem statement, all variables and given/known data

Let A be the following 2x2 matrix:

s 2s
0 t

Find a subspace B of M2x2 where M2x2 = A (+) B


2. Relevant equations

A ∩ B = {0}

if u and v are in M2x2, then u + v is in M2x2
if u is in M2x2, then cu is in M2x2

3. The attempt at a solution

Let B be the following 2x2 matrix:

0 0
r 0

Because they are both subspace, they intersect at the zero vector and thus the set {0}, the zero subspace, is a subspace of M2x2. We then have

M2x2 = A (+) B:

M2x2 = A + B /\ A ∩ B = {0}
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Mark44
#2
Nov5-11, 07:17 PM
Mentor
P: 21,215
Quote Quote by Throwback View Post
1. The problem statement, all variables and given/known data

Let A be the following 2x2 matrix:

s 2s
0 t

Find a subspace B of M2x2 where M2x2 = A (+) B
This doesn't make sense to me. M2x2 is the vector space of 2x2 matrices. It's not a matrix.

It also doesn't make sense to add a matrix - A - and a subspace - B.

What is the exact wording of this problem?
Quote Quote by Throwback View Post


2. Relevant equations

A ∩ B = {0}

if u and v are in M2x2, then u + v is in M2x2
if u is in M2x2, then cu is in M2x2

3. The attempt at a solution

Let B be the following 2x2 matrix:

0 0
r 0

Because they are both subspace, they intersect at the zero vector and thus the set {0}, the zero subspace, is a subspace of M2x2. We then have

M2x2 = A (+) B:

M2x2 = A + B /\ A ∩ B = {0}
Mark44
#3
Nov5-11, 07:24 PM
Mentor
P: 21,215
Quote Quote by Throwback View Post
Find a linear subspace B of M2x2(ℝ) such that M2x2(ℝ) = A (+) B where A is the matrix

s 2s
0 t
That makes more sense, except that I can't read what's in the parentheses in M2x2(ℝ). In my browser it shows up as an empty box. What symbol is that?

This stuff, too.
where {A|s,t in ℝ} ℂ M2x2(ℝ)

s, t in what?

Throwback
#4
Nov5-11, 07:32 PM
P: 13
Finding a subspace (possibly intersection of subspace?)

This should make it easier haha



In case the image isn't showing either, the symbol that isn't showing is the "R" for real numbers, so s,t in R and M(R)
HallsofIvy
#5
Nov5-11, 07:38 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,348
I take it then that you mean B is a subspace of the space of all two by two matrices with real entries. However, you do NOT mean that A is the "matrix" given. Rather, A is the subspace of all two by two matrices, with real entries, of the form
[tex]\begin{bmatrix}s & 2s \\ 0 & t\end{bmatrix}[/tex].
Saying that "[itex]A(+)B= M_{22}(R)[/itex]" means that for any numbers u, x, y, z, there exist numbers a, b, c, d such that
[tex]\begin{bmatrix}s & 2s \\ 0 & t\end{bmatrix}+ \begin{bmatrix}a & b \\ c & d\end{bmatrix}= \begin{bmatrix}u & x \\ y & z\end{bmatrix}[/tex]

Of course, then we must have
[tex]\begin{bmatrix} a & b \\ c & d\end{bmatrix}= \begin{bmatrix}u- s & x- 2s \\ y & z- t\end{bmatrix}[/tex]

Now, what relations must a, b, c, and d satisfy?
Throwback
#6
Nov5-11, 07:49 PM
P: 13
Closure under addition and closure under multiplication?
Throwback
#7
Nov5-11, 08:00 PM
P: 13
I barely see what it's asking...

Given A, I don't have a problem proving that A is a subspace of M22 -- just show there's closure under addition and multiplication. I can find a basis/span, etc.

For this question, I'm somewhat lost. Since A + B = M22, then B = M22 - A, I'm assuming B has to follow the closure requirements, but is that it? It seems like I'm just missing something pretty big here...


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