- #1

- 4

- 0

whose diagonal entries are zero. ?

i know that the dimension is the number of vectors that are the basis for this subspace ,but i cannot figure out what is the basis for this subspace ?

any help will be appreciated ,

thanks in advance

- Thread starter baher
- Start date

- #1

- 4

- 0

whose diagonal entries are zero. ?

i know that the dimension is the number of vectors that are the basis for this subspace ,but i cannot figure out what is the basis for this subspace ?

any help will be appreciated ,

thanks in advance

- #2

Bacle2

Science Advisor

- 1,089

- 10

- #3

- 1,765

- 125

What's the difference between a 2 by 2 matrix and a column vector with 4 entries?

Presumably, you know how to find a basis for all the column vectors with 4 entries.

- #4

- 4

- 0

[0 b]

[c 0] for some b, c.

Since these matrices are generated by

[0 1].[0 0]

[0 0],[1 0], the dimension equals 2. is that a right answer ?

- #5

- 1,765

- 125

Yes, that's it.

- #6

Bacle2

Science Advisor

- 1,089

- 10

do those matrices generate, but that no smaller ( in size/cardinality) set generates the whole space.

- #7

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

[tex]\begin{bmatrix}0 & a \\ b & 0\end{bmatrix}= \begin{bmatrix}0 & a \\ 0 & 0 \end{bmatrix}+ \begin{bmatrix}0 & 0 \\ b & 0 \end{bmatrix}= a\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}+ b\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}[/tex]

which makes it clear what a basis is.

- #8

- 1

- 0

WHAT IS THE RESULT span(spanV)=???

- #9

- 8

- 0

Try doing a generalized row-reduction using Gaussian elimination.

- Last Post

- Replies
- 10

- Views
- 6K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 20K

- Last Post

- Replies
- 2

- Views
- 3K

- Replies
- 1

- Views
- 572

- Replies
- 2

- Views
- 3K

- Last Post

- Replies
- 9

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 5K

- Last Post

- Replies
- 5

- Views
- 7K