Basis & Dimension of 2x2 Matrix Subspaces: W1 & W2

In summary, the conversation discusses two subspaces, W1 and W2, in the vector space of 2x2 matrices. W1 is the subspace of upper triangular matrices, and W2 is the subspace spanned by three given matrices. The conversation then poses two questions: the dimension of the intersection of W1 and W2, and whether a specific matrix belongs to the intersection. The answer is that the intersection has a dimension of 2, and the given matrix does not belong to the intersection.
  • #1
Yankel
395
0
Hello

I have this problem, I find it difficult, any hints will be appreciated...

Two subspaces are given (W1 and W2) from the vector space of matrices from order 2x2.

W1 is the subspace of upper triangular matrices

W2 is the subspace spanned by:

\[\left(\begin{matrix}1&0\\1&2\\\end{matrix}\right)\]

\[\left(\begin{matrix}1&1\\2&1\\\end{matrix}\right)\]

\[\left(\begin{matrix}2&1\\0&0\\\end{matrix}\right)\]

a. what is the dimension of the intersection of W1 and W2 ?

b. does:

\[\left(\begin{matrix}6&1\\0&4\\\end{matrix}\right)\]

belong to the intersection ?

thanks !
 
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  • #2
Yankel said:
Hello

I have this problem, I find it difficult, any hints will be appreciated...

Two subspaces are given (W1 and W2) from the vector space of matrices from order 2x2.

W1 is the subspace of upper triangular matrices

W2 is the subspace spanned by:

\[\left(\begin{matrix}1&0\\1&2\\\end{matrix}\right)\]

\[\left(\begin{matrix}1&1\\2&1\\\end{matrix}\right)\]

\[\left(\begin{matrix}2&1\\0&0\\\end{matrix}\right)\]

a. what is the dimension of the intersection of W1 and W2 ?

b. does:

\[\left(\begin{matrix}6&1\\0&4\\\end{matrix}\right)\]

belong to the intersection ?

thanks !
Let:
\(b_1=\left(\begin{matrix}1&0\\1&2\\\end{matrix} \right) \)

\(b_2=\left(\begin{matrix}1&1\\2&1\\\end{matrix} \right) \)

\(b_3=\left(\begin{matrix}2&1\\0&0\\\end{matrix} \right) \)

The it is obvious (and by that I mean an exercises left to the reader) to show that a linear combination of \(b_1, b_2\) and \(b_3\) which is upper triangular is or the form:

\(u(\alpha,\beta)=2\alpha b_1+\alpha b_2 + \beta b_3\)

Thus the upper triangular matrices in \(W1\) form a 2D subspace of 2x2 matrices, which tells us that the intersection of \(W1\) and \(W2\) is 2 dimensional.

The second part just requires that you determine if

\(u(\alpha,\beta) = \left(\begin{matrix}6&1 \\ 0&4 \\ \end{matrix} \right) \)

has a solution.

CB
 

FAQ: Basis & Dimension of 2x2 Matrix Subspaces: W1 & W2

What is the basis of a 2x2 matrix subspace?

The basis of a 2x2 matrix subspace is a set of linearly independent vectors that span the subspace. It is the smallest set of vectors that can be used to represent all other vectors in the subspace through linear combinations.

How is the dimension of a 2x2 matrix subspace determined?

The dimension of a 2x2 matrix subspace is determined by the number of linearly independent vectors in its basis. In other words, it is the number of vectors needed to span the subspace.

Can a 2x2 matrix subspace have more than one basis?

Yes, a 2x2 matrix subspace can have infinitely many bases. This is because there are infinitely many ways to choose linearly independent vectors that can span the subspace.

How does the intersection of two 2x2 matrix subspaces affect their dimensions?

If the intersection of two 2x2 matrix subspaces is non-empty, then the dimension of the intersection will be less than or equal to the minimum of the dimensions of the two subspaces. This is because the intersection can only contain vectors that are in both subspaces.

Can the sum of two 2x2 matrix subspaces have a dimension greater than the sum of their individual dimensions?

No, the dimension of the sum of two 2x2 matrix subspaces will always be less than or equal to the sum of their individual dimensions. This is because the sum can only contain vectors that are in at least one of the subspaces.

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