Answer:
Subspace; Closed
Step-by-step explanation:
The trace of a square n x n matrix [tex]A = (a_{ij})[/tex] is the sum [tex]a_{11} + a_{22} +...+ a_{nn}[/tex] of the entries on its main diagonal.
Let V be the vector space of all 2 x 2 matrices with real entries.
Let H be the set of all 2 x 2 matrices with real entries that have trace 0.
Theorem: H is a subspace of the vector space V, if
1) for every [tex]A,\ B\in H: \ \ A+B\in H;[/tex]
2) for each [tex]A\in H[/tex] and [tex]\lambda \in R:\ \ \lambda A\in H.[/tex]
Check these two conditions:
1) Let [tex]A=(a_{ij}),\ B=(b_{ij})\in H[/tex]
This means
[tex]a_{11}+a_{22}=0\\ \\b_{11}+b_{22}=0[/tex]
Consider the matrix [tex]A+B=(a_{ij}+b_{ij})=\left(\begin{array}{cc}a_{11}+b_{11}&a_{12}+b_{12}\\a_{21}+b_{21}&a_{22}+b_{22}\end{array}\right)[/tex]
This matrix sum has the trace
[tex](a_{11}+b_{11})+(a_{22}+b_{22})=(a_{11}+a_{22})+(b_{11}+b_{22})=0+0=0[/tex]
So, [tex]A+B\in H[/tex]
2) Consider [tex]\lambda A=\lambda\left(\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right)=\left(\begin{array}{cc}\lambda a_{11}&\lambda a_{12}\\\lambda a_{21}&\lambda a_{22}\end{array}\right)[/tex]
Its trace is [tex]\lambda a_{11}+\lambda a_{22}=\lambda (a_{11}+a_{22})=\lambda \cdot 0=0[/tex]
So, [tex]\lambda A\in H[/tex]
Therefore, H is a subspace of the vector space V and is closed under addition.
