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Basis and dimension

Yankel

Active member
Jan 27, 2012
398
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 !!
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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