- #1

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^{6x6}consisting of all lower triangular matrices, what is the dimension of S?

Does anyone know the properties about dimension of lower triangular matrices?

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- Thread starter lypaza
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- #1

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Does anyone know the properties about dimension of lower triangular matrices?

- #2

Mark44

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In your subspace, each matrix is guaranteed to have at least how many 0 elements, and where are they in each matrix?

- #3

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Never mind. I think I found it :D

Editing: Yes, it is 21 :D

Editing: Yes, it is 21 :D

- #4

Mark44

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Yessireebob

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If S is the subspace of R

I found out that trace is sum of the diagonal.

And I also found out that symmetric matrices have zero trace.

The problem is I don't know how to find the basis of 5x5 symmetric matrices?

- #6

Mark44

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That's not true. A matrix A is symmetric if A = AOk, I think this one is a little bit trickier

If S is the subspace of R^{5x5}consisting of all matrices with trace 0, then what is dimension of S?

I found out that trace is sum of the diagonal.

And I also found out that symmetric matrices have zero trace.

So let's leave symmetric matrices out of this, since there is no mention of them in your problem statement. What single equation describes every matrix in your subspace S?

The problem is I don't know how to find the basis of 5x5 symmetric matrices?

- #7

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you mean

ax_{1} + bx_{2} + cx_{3} + dx_{4} + ex_{5} = 0 ?

where a, b, c, d, e are elements on diagonal

ax

where a, b, c, d, e are elements on diagonal

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- #8

Mark44

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Here you have one equation with five unknowns. How many degrees of freedom do you have?

- #9

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4? So the dimension is 4?

- #10

Mark44

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Right.

- #11

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But when I submitted the answer, it said incorrect....

http://img192.imageshack.us/img192/6831/49103767.png [Broken]

http://img192.imageshack.us/img192/6831/49103767.png [Broken]

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- #12

Mark44

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- #13

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So we were dealing with R

This is R

- #14

vela

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While the condition for the trace only involves elements on the diagonal, but those aren't the only ones that matter. You still have the ability to arbitrarily set the 20 off-diagonal elements, so you have a total of 20+4 degrees of freedom for the matrix as a whole.

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HallsofIvy

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