Consider $\mathbb{R}^n$ with standard inner product. Prove or give counterexample:
For any normal matrix $A$, the left-multiplication operator $L_A$ is also normal.
Isn't it sufficient enough to say $A^*Ax = AA^*x$, therefore $L_A$ is normal? (obviously not, I am missing some crucial point, can someone point it out?)
Thank you for any kind of help!
I don't think this argument works. Letting $L_{A} : \mathbb{R}^{n\times n} \rightarrow \mathbb{R}^{n \times n}$ stand for the left-multiplication operator your goal is to show that $L_{A}L_{A}^* = L_{A}^* L_{A}$. You can do this by finding a matrix representation of $L_A$.
$L_A$ will be represented by $n^2 \times n^2$ matrix. To apply $L_A$ to a matrix $M$ you'll need to represent it as a column vector in $\mathbb{R}^{n^2}$. The easiest way to do this is to just take the columns of $M = [m_1,\ldots,m_n]$ and stack them on top of each other yielding
$$ \left[\begin{array}{c}m_1 \\ \vdots \\ m_n \end{array}\right] \in \mathbb{R}^{n^2} $$
Now you want a matrix to represent left multiplication by $A$. Since $AM = [Am_1,\ldots,Am_n]$ you need to describe a matrix $L_A$ such that
$$L_A\left[\begin{array}{c}m_1 \\ \vdots \\ m_n \end{array}\right] = \left[\begin{array}{c}Am_1 \\ \vdots \\ Am_n \end{array}\right] $$
Can you see how $L_A$ can be formed? Can you see why this matrix is normal when $A$ is normal?