# Is the set of all 2 × 2 matrices whose trace is equal to 0 closed under scalar multiplication?

Sep 24, 2017

Let:

$\boldsymbol{A} = \left(\begin{matrix}{a}_{11} & {a}_{12} \\ {a}_{21} & {a}_{22}\end{matrix}\right)$

If $\setminus T r \left(\boldsymbol{A}\right) = 0$. then:

${a}_{11} + {a}_{22} = 0$ ..... [A]

Then for any real number $\mu$, we have:

$\mu \boldsymbol{A} = \mu \left(\begin{matrix}{a}_{11} & {a}_{12} \\ {a}_{21} & {a}_{22}\end{matrix}\right)$
$\setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}\mu {a}_{11} & \mu {a}_{12} \\ \mu {a}_{21} & \mu {a}_{22}\end{matrix}\right)$

And so:

$T r \left(\mu \boldsymbol{A}\right) = \mu {a}_{11} + \mu {a}_{22}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \mu \left({a}_{11} + {a}_{22}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \mu \times 0 \setminus \setminus \setminus$ by [A]
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 0$

Hence, a $2 \times 2$ matrix with trace $0$ is closed under scaler multiplication QED.