Finding a subspace (possibly intersection of subspace?)

  • Thread starter Throwback
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  • #1
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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}
 

Answers and Replies

  • #2
34,667
6,380

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?

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}
 
  • #3
34,667
6,380
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?
 
  • #4
13
0
This should make it easier haha

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

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:
  • #5
HallsofIvy
Science Advisor
Homework Helper
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961
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?
 
  • #6
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Closure under addition and closure under multiplication?
 
  • #7
13
0
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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