Is the set of $n \times n$ normal matrices a subspaces of $M_n(\mathbb{C})$ over $\mathbb{C} $ ?
My attempts : we know that every real Number is a complex number.
if $A,B \in M_n(\mathbb{C})$ and $\alpha \in \mathbb C$, then
$A+B \in M_n(\mathbb{C})$ amd $ \alpha A \in M_n(\mathbb{C})$
so the set of $n \times n$ normal matrices a subspaces of $M_n(\mathbb{C})$ over $\mathbb{C} $ .
Is its True/false
Any Hints/solution will be appreciated
Thanks u
The set $N$ of normal matrices is not a subspace of $M_n(\mathbb C)$
If $N$ was a subspace, you would have $$\begin{align}(A+B)(A+B)^* &=AA^* + AB^*+BA^*+BB^* \\ &=(A+B)^*(A+B)=A^*A+A^*B+B^*A+B^*B\end{align}$$ for any $A,B$ normal matrices.
Or equivalently $$AB^*+BA^* = A^*B+B^*A$$
For $A$ Hermitian and $B$ skew-Hermitian you get $$A(-B)+BA = AB+(-B)A$$
I.e. $AB=BA$. So if we find an Hermitian matrix $A$ and a skew-Hermitian matrix $B$ that doesn't commute, we'll have proven that $N$ is not a subspace of $M_n(\mathbb C)$.
Take $$A=\begin{pmatrix} 0 & 2+i\\ -(2-i) & 0\end{pmatrix} \text{ and } B=\begin{pmatrix} 1 & 0\\ 0 & 2\end{pmatrix}$$
and you're done finding a counterexample.