Re: Subspaces with Matrices?

try setting:

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

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).

Re: Subspaces with Matrices?

Quote:

Originally Posted by

**Deveno** try setting:

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

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:

$\displaystyle \left[ {\begin{array}{*{20}{c}} {2y}&y\\ y&y \end{array}} \right] = y\left[ {\begin{array}{*{20}{c}} 2&1\\ 1&1 \end{array}} \right],y \in R$

Therefore S∩T has a dimension of 1?

Re: Subspaces with Matrices?

indeed. in fact, one can see that every element of S has equal entries in the second column, but B_{1} does not, so B_{1} is not in the intersection. but B_{2} clearly is, since B_{2} = 2A_{3} - A_{2}. so the dimension can't be 2, and it is at least 1, which means it IS 1.

Re: Subspaces with Matrices?

Alrighty, so the dimension is 1, and therefore the basis is simply

$\displaystyle \left[ {\begin{array}{*{20}{c}} 2&1\\ 1&1 \end{array}} \right]$?

I have a feeling I'm forgetting something here, but I'm not entirely sure as to what exactly.

Re: Subspaces with Matrices?

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