I am trying to figure out what would be the dimenstion of the space of the traceless $n \times n$ matrices. Please help me with this question.

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    $\begingroup$ This question does not show any research effort $\endgroup$ Sep 23 '11 at 4:45

I will not answer your question directly for you have not put any effort into it. However here are some questions that you can ponder about that may or may not help.

The space of $n\times n$ matrices has dimension $n^2$. Now we know that the trace of a matrix is the sum of the elements on its diagonal. So elements off diagonal have no effect on a matrix's trace. Hence are we free to choose such elements?

There are $n$ elements on the diagonal and say we choose $n-1$ of them and stop at the entry $a_{nn}$. Is there a constraint on how I choose $a_{nn}$?

  • 6
    $\begingroup$ Alternatively, note that $\operatorname{tr} : M_n(k) \to k$ is a linear operator on the space of $n \times n$ matrices over the field $k$... $\endgroup$
    – Zhen Lin
    Sep 23 '11 at 7:21
  • $\begingroup$ What if a_nn=0 though? That means the rest of the n-1 entries must also add up to 0 if tr(A)=0 $\endgroup$
    – Mr X
    Mar 26 at 14:39

Let $M_{n\times n}$ be the vector space of n by n square matrices. And so $\forall A\in M_{n\times n}, \ A = \left \{ a_{ij} | (i,j)\in [1,n] \right \}$.

Now every vector space has a basis(not going to prove this here but you can find proofs of it elsewhere on this site) so let's construct one:

Let $E_{ij}$ be the collection of n by n matrices with only 1 nonzero entry. That is, for some $i,j\in [1,n]$,

$E_{ij}= \left \{ e_{kl}\ | e_{kl}=1 \leftrightarrow kl=ij,\ e_{kl}=0\leftrightarrow kl\neq ij \ (1\leq k,l\leq n) \right \}$

This is the standard basis for $M_{n\times n}$. We now can write

$A = \sum_{i,j=1}^{n} a_{ij}E_{ij} \ | \ \forall A\in M_{n\times n}$ and this vector space has dimension $n^{2}$. Now let $K\subset M_{n\times n}$ be subspace of n by n matrices whose trace is equal to zero. Bear in mind that $ (\forall A \in M_{n\times n} ) \ tr(A)=0 \ \leftrightarrow \ \sum_{i=1}^{n}a_{ii}=0$.

So for some $1\leq i\leq n$, $trace(A)-a_{ii}$ = $0-a_{ii} =-a_{ii} = \sum_{j=1}^{i-1}a_{jj} +\sum_{j=i+1}^{n}a_{jj}$. And so you have a linear dependence relation. We can see this by expanding our result for the i^th entry and writing it out as $a_{ii} = -(\sum_{j=1}^{i-1}a_{jj} +\sum_{j=i+1}^{n}a_{jj})$.

So if the trace of a square matrix A = 0, then one of the diagonal entries is constrained by the values of the other (n-1) diagonal entries as it is the negative sum of the other values. Now there are $n^{2}-n$ non-diagonal entries that do not affect the trace so they are independent of tr(A). So the Dim(K)= $(n^{2}-n)+(n-1)=n^{2}-n+n-1= n^{2}-1$.


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