Finding a subspace (possibly intersection of subspace?)

Click For Summary
The discussion revolves around finding a subspace B of the vector space M2x2 such that M2x2 equals the direct sum of subspace A and B. A is defined as a set of matrices of the form [s 2s; 0 t], while the challenge is to identify B such that their intersection is only the zero matrix. Participants express confusion about the problem's wording and the notation used, particularly regarding the real numbers and the nature of the matrices involved. Clarification is sought on the relationships between the elements of A and B, and the requirements for B to be a valid subspace. The conversation highlights the need for a deeper understanding of vector space operations and properties.
Throwback
Messages
10
Reaction score
0

Homework Statement



Let A be the following 2x2 matrix:

s 2s
0 t

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


Homework 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

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}
 
Physics news on Phys.org
Throwback said:

Homework Statement



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?
Throwback said:

Homework 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

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}
 
Throwback said:
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?
 
This should make it easier haha

[PLAIN]http://dl.dropbox.com/u/907375/asd.jpg

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)
 
Last edited by a moderator:
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
\begin{bmatrix}s & 2s \\ 0 & t\end{bmatrix}.
Saying that "A(+)B= M_{22}(R)" means that for any numbers u, x, y, z, there exist numbers a, b, c, d such that
\begin{bmatrix}s & 2s \\ 0 & t\end{bmatrix}+ \begin{bmatrix}a & b \\ c & d\end{bmatrix}= \begin{bmatrix}u & x \\ y & z\end{bmatrix}

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

Now, what relations must a, b, c, and d satisfy?
 
Closure under addition and closure under multiplication?
 
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...
 

Similar threads

Replies
8
Views
2K
Replies
5
Views
5K
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 3 ·
Replies
3
Views
6K