$\DeclareMathOperator{\A}{\mathscr{A}}$
$\DeclareMathOperator{\B}{\mathscr{B}}$
$\DeclareMathOperator{\C}{\mathscr{C}}$
$\DeclareMathOperator{\kernel}{\mathrm{Ker}}$
$\DeclareMathOperator{\diag}{\mathrm{diag}}$
$\DeclareMathOperator{\span}{\mathrm{span}}$
$\DeclareMathOperator{\real}{\mathbb{R}^2}$
$\DeclareMathOperator{\rank}{\text{rank}}$
The question is:
Let $\A$ be a linear operator on the $n$-dimensional Euclidean space $V$. Prove there exists a partially orthogonal operator $\B$ and a semi-definite self-adjoint operator $\C$,
$\kernel(\B) = \kernel(\C)$, such that $\A = \B\C$, and operators $\B$ and $\C$ are unique; Prove the linear operator $\A$ is normal if and only if $\B$ commutes with $\C$.
An operator $\B$ is called partially orthogonal if there exists a $\B$-invariant subspace $U$ such that $\|\B(\alpha)\| = \|\alpha\|$ for any $\alpha \in U$, and $\B(\alpha) = 0$ for any
$\alpha \in U^\bot$. It is easy to show that an operator is partially orthogonal if and only there exists an orthonormal basis $\{\xi_1, \ldots, \xi_n\}$ such that $\B(\xi_1, \ldots, \xi_n) = (\xi_1, \ldots, \xi_n)\diag(O, 0)$, where $O$ is an order $r = \dim(U)$ orthogonal matrix.
Oddly enough, so far I am able to prove the uniqueness and commuting statement but the construction itself. Inspired by the polar decomposition, I tried using the singular value decomposition of $A$ (the matrix of $\A$ under a fixed orthonormal basis) as follows:
\begin{align*}
A = O_1\diag(M, 0)O_2 = O_1\diag(I_{(r)}, 0)O_2 \times O_2'\diag(M, 0)O_2
=:BC,
\end{align*}
where $M = \diag(\mu_1, \ldots, \mu_r)$ is a diagonal matrix with its diagonal elements of all eigenvalues. However, the $B$ defined in this way is not partially orthogonal. If enforcing $B$ to be partially orthogonal, then $C$ cannot be made symmetric.
So probably a new perspective is needed here. I appreciate any insights.
As pointed out by @Ben Grossmann, such decomposition is not universal. For example, consider the matrix $A = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$. Below we show this $A$ does not admit the required decomposition.
Suppose $A = BC$, where $B$ is partial orthogonal and $C \geq 0$. Since $\kernel(A) = \span({(1, 0)'}) \supset \kernel(C)$, $\kernel(C) = \span((1, 0)')$ or $\kernel(C) = \{0\}$. If $\kernel(C) = \{0\}$, then
$\kernel(B) = \{0\}$, which implies $\rank A = 2$, contradiction. Hence $\kernel(B) = \kernel(C) = \span((1, 0)')$.
Since $C \geq 0$ and $\rank C = 1$, we can accordingly assume $C$'s spectral decomposition is
$C = O\diag(\lambda, 0)O'$ where $\lambda \neq 0$ and $O$ is an order $2$ orthogonal matrix. By solving $O\diag(\lambda, 0)O'(1, 0)' = (0, 0)'$ and noting $O$ is orthogonal, it follows that $O = \begin{pmatrix} 0 & 1 \\
1 & 0 \end{pmatrix}$, thereby $C = \begin{pmatrix} 0 & 0 \\ 0 & \lambda \end{pmatrix}$.
Since $B$ is partially diagonal and $\rank B = 1$, there exists an order $2$ orthogonal matrix $P$ such that $B = P\diag(1, 0)P'$. In the same argument as above, $P = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, resulting
$B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$.
As a result, $A = BC = \begin{pmatrix} 0 & 0 \\ 0 & \lambda \end{pmatrix}$, this is a contradiction.
Best Answer
Either your understanding of the question is wrong, or the author has suggested something that isn't true. In either case, there exist operators $\mathscr A$ for which such a decomposition does not exist.
For example, take $$ A = \pmatrix{0&1\\0&0}. $$ We note that if $\ker \mathscr B = \ker \mathscr C$, then both operators must have kernel equal to that of $A$, i.e. the span of $(1,0)$. This means that the image of $\mathscr B$ must be the orthogonal complement to $(1,0)$, which would mean (assuming that the subspace on which $\|\mathscr B\alpha\| = \|\alpha\|$ is invariant) that the image of $\mathscr A$ is a subspace of the span of $(0,1)$.
However, this is not the case.
Here is an answer if we drop the requirement that $\ker \mathscr B = \ker \mathscr C$, but assume that your "partially orthogonal" is accurate.
Using the singular value decomposition, under a suitable choice of orthonormal basis $\mathscr A$ has the form $$ A = \pmatrix{\Sigma & 0\\0 & 0} U, $$ where $U$ is an orthogonal matrix and $\Sigma$ is diagonal with positive entries. Partition the matrix $U$ conformally to get $$ A = \pmatrix{\Sigma & 0\\0 &0}\pmatrix{U_{11} & U_{12}\\ U_{21} & U_{22}} = \pmatrix{\Sigma U_{11} & \Sigma U_{12}\\ 0 & 0}. $$ Note that $\mathscr B$, which contains the image of $\mathscr A$, must be such that the restriction of $\mathscr B$ to its image is an orthogonal operator. In other words, the matrix $B$ of $\mathscr B$ must be block diagonal. With that in mind, define $$ B = \pmatrix{U_{11} & 0\\0 & 0}, \\ C = \pmatrix{U_{11}^T \Sigma U_{11} & U_{11}^T \Sigma U_{12}\\ U_{12}^T \Sigma U_{11} & U_{12}^T \Sigma U_{12}} = \pmatrix{U_{11} & 0\\0 & U_{12}}^T \pmatrix{\Sigma & \Sigma\\ \Sigma & \Sigma} \pmatrix{U_{11} & 0\\0 & U_{12}}. $$