So, What is the Dimension of the Subspace of Matrices with Trace 0?

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SUMMARY

The dimension of the subspace S of M5(R), consisting of all matrices with trace 0, is determined to be 24. This conclusion arises from the fact that for an n x n matrix, the trace condition provides one linear constraint, reducing the total degrees of freedom. For M5(R), which has 25 entries, the dimension is calculated as 25 - 1 = 24. The confusion in the discussion stemmed from miscalculating the contributions of the diagonal elements and the misunderstanding of the trace condition.

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xenogizmo
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Hey Everyone,
I have this question on my assignment that's confusing me, I think I have the logic right, but for some reason I am getting the wrong answer..
Anyways the question states:

If S is the subspace of M5 (R) consisting of all matrices with trace 0, then dim(S) = ?

Now, I know the trace is the sum of entries in the main diagonal, so I assumed that all the elements on the diagonal are zero..
Which leaves me with 28 non-zero elements above the diagonal and 28 below, which gives a total of 56 elements.. hence a dimension of 56?
However that was wrong.. now I know that for a skew symmetrix matrix of 8x8 we'd have a dimesion of 28 since the ones above are the negative of the ones below.. can anyone please help me out?

I had 6 tries for the question.. and I only have one left! Any help would be greatly appreciated.. I tried 28, 56, and 60, all were wrong..

Thank you very much!
AZH
 
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xenogizmo said:
Hey Everyone,
I have this question on my assignment that's confusing me, I think I have the logic right, but for some reason I am getting the wrong answer..
Anyways the question states:

If S is the subspace of M5 (R) consisting of all matrices with trace 0, then dim(S) = ?

Now, I know the trace is the sum of entries in the main diagonal, so I assumed that all the elements on the diagonal are zero..
Which leaves me with 28 non-zero elements above the diagonal and 28 below, which gives a total of 56 elements.. hence a dimension of 56?
However that was wrong.. now I know that for a skew symmetrix matrix of 8x8 we'd have a dimesion of 28 since the ones above are the negative of the ones below.. can anyone please help me out?

I had 6 tries for the question.. and I only have one left! Any help would be greatly appreciated.. I tried 28, 56, and 60, all were wrong..

Thank you very much!
AZH

First, how did M5 get to be 8 by 8 matrices?

Also:
"Now, I know the trace is the sum of entries in the main diagonal, so I assumed that all the elements on the diagonal are zero.."

No, the sum of the elements on the diagonal must be 0. It certainly is not the case that all elements must be 0. You have one equation connecting n (for n by n matrices). If you knew n-1 of them, you could calculate the remaining one.
 

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