The set of $n \times n$ matrices having trace equal to zero is a subspace of $M_{n \times n} \left(F\right)$

I understand that the trace of a matrix is the sum of the diagonal entries. I also understand that an $n \times n$ matrix is defined as a square matrix. However, I do not understand why this means that the aforementioned matrix must therefore be subspace of a square matrix, $\mathrm M_{n \times n} \left(F\right)$.

I would greatly appreciate it if someone could please take the time to make this concept clear to me.

  • It is non-empty since the null matrix $(0)_{n,n}\in E$, where $E$ is the subspace you are looking for.

  • If $M_1,M_2\in E$, $\lambda \in F$, then

$$\mathrm{tr}(M_1+\lambda M_2)=\mathrm{tr}(M_1)+\lambda \mathrm{tr}(M_2)=0$$

because $M_1,M_2\in E$.

So $E$ is indeed a sub-vector space of $M_{n,n}(F)$.


Let $E$ be a vector space. A (non-empty) subset $U$ of $V$ is a subspace (of $V$) if $U$ is closed by addition and product by a scalar :

  • $\forall x,y \in U, x+y \in U$
  • $\forall \lambda \in F, \forall x\in U, \lambda x \in U$

So, here you need to prove that :

  • the sum of two matrix with trace zero has trace zero
  • if you multiply a matrix with trace zero, the result has trace zero

Those two properties are really easy to show


So let $A = \{a_{ij}\}$ and $B = \{b_{ij}\}$. If $A$ and $B$ both have trace $0$, it means that $0 = \sum_{i=1}^n a_{ii} = \sum_{i=1}^n b_{ii}$. Now, $C = A+B$ has diagonal entries $c_{ii} = a_{ii} + b_{ii}$, so that $\sum_{i=1}^n c_{ii} = \sum_{i=1}^n a_{ii} + b_{ii} = \sum_{i=1}^n a_{ii} + \sum_{i=1}^n b_{ii} = 0$, so that $C$ has trace $0$ as well. Similarly, if $A$ has trace $0$, $dA$ will as well, for any $d\in\mathbb{R}$, since $\sum_{i=1}^n da_{ii} = d\sum_{i=1}^n a_{ii} = 0$.

When closure as described above is satisfied, the other properties will fall in line. Let $c=0$, you have an additive identity, let $c=-1$, you have an inverse, everything else is inherited from the general properties of matrix addition and scalar multiplication.


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