Subspaces with Matrices?

**mathdude2012** · Apr 8th 2012, 06:17 PM

Hi everyone,
I've just got a question today. I've given it a few attempts, but I keep getting stuck, and I'm not sure where I'm going wrong.

Let S and T be subspaces of ${{M}_{2}}(\mathbb{R})$ , where:

S is generated by the matrices $\{ {A_1},{A_2},{A_3}\}$ , where ${A_1} = \left( {\begin{array}{*{20}{c}} 0&1\\ 0&1 \end{array}} \right),{A_2} = \left( {\begin{array}{*{20}{c}} 0&1\\ 1&1 \end{array}} \right),{A_3} = \left( {\begin{array}{*{20}{c}} 1&1\\ 1&1 \end{array}} \right)$
and T is generated by the matrices $\{ {B_1},{B_2}\}$ , where ${B_1} = \left( {\begin{array}{*{20}{c}} 1&1\\ 1&0 \end{array}} \right),{B_2} = \left( {\begin{array}{*{20}{c}} 2&1\\ 1&1 \end{array}} \right)$ .

I've already done the first part of the question, determining dim(S) and dim(T) [Just had to prove the matrices contained are linearly independent.]

The part I need help with is determining the basis and dimension of $S \cap T$ .

I've tried creating a spanning set, and then proving that they're linearly independent, and it indicates that they are. Which doesn't make sense, because then it implies that the dimension theorem is wrong.
I figure I'll throw it out there, I can't find any examples on the internet for anything similar, and I'm only getting more confused.

**Deveno** · Apr 8th 2012, 07:18 PM

try setting:

$\begin{bmatrix}c&a+b+c\\b+c&a+b+c \end{bmatrix} = \begin{bmatrix}x+2y&x+y\\x+y&y \end{bmatrix}$

that's a system of 4 linear equations in 5 unknowns. find the solution space.

(hint # 2: dim(S∩T) ≤ dim(S), dim(T), so it can be 2 at the most. if an element for your basis of T is not in S, then dim(S∩T) = 0, or 1.

if there is any non-zero matrix in S∩T, the dimension cannot be 0).

**mathdude2012** · Apr 8th 2012, 07:43 PM

Originally Posted by Deveno

try setting:

$\begin{bmatrix}c&a+b+c\\b+c&a+b+c \end{bmatrix} = \begin{bmatrix}x+2y&x+y\\x+y&y \end{bmatrix}$

that's a system of 4 linear equations in 5 unknowns. find the solution space.

(hint # 2: dim(S∩T) ≤ dim(S), dim(T), so it can be 2 at the most. if an element for your basis of T is not in S, then dim(S∩T) = 0, or 1.

if there is any non-zero matrix in S∩T, the dimension cannot be 0).

So in this case, the solution set would be:
$\left[ {\begin{array}{*{20}{c}} {2y}&y\\ y&y \end{array}} \right] = y\left[ {\begin{array}{*{20}{c}} 2&1\\ 1&1 \end{array}} \right],y \in R$

Therefore S∩T has a dimension of 1?

**Deveno** · Apr 8th 2012, 10:17 PM

indeed. in fact, one can see that every element of S has equal entries in the second column, but B₁ does not, so B₁ is not in the intersection. but B₂ clearly is, since B₂ = 2A₃ - A₂. so the dimension can't be 2, and it is at least 1, which means it IS 1.

**mathdude2012** · Apr 8th 2012, 10:20 PM

Alrighty, so the dimension is 1, and therefore the basis is simply
$\left[ {\begin{array}{*{20}{c}} 2&1\\ 1&1 \end{array}} \right]$ ?
I have a feeling I'm forgetting something here, but I'm not entirely sure as to what exactly.

**Deveno** · Apr 9th 2012, 12:09 AM

that's "a" basis. remember, bases are not unique. dimension, however, is invariant.

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