Is $A$ a linear transformation on $\mathbb{C}^n$, where $A(u) = \| u\| u$? Is  $A$  a  linear transformation  or  not  on $\mathbb{C^n}$, where
$A(u) = \|u\| u$ and $\|u\| $ is the  length of  the  vector $u$.
I know  the definition  of linear  transformations: $T$ is linear if it satisfies
$$
T(\mathbf x+\mathbf y) = T(\mathbf x) + T(\mathbf y)\\
T(c\mathbf x) = cT(\mathbf x)
$$
But  here  how can I apply it? I am confused. . .
 A: Since there is a factor of $\| u \|$ in the definition of $A(u)$, it means that $A$ is not scaling linearly. The magnitude of the scaling increases as the magnitude of the vector $u$ increases. But this contradicts the second property of a linear transformation, which essentially says that scaling is uniform.
So, we expect that $A$ is not linear. A good choice of $u$ to verify this is to take a nonzero vector $u$ and multiply it by a factor $c$ which is not equal to $1$, and then check. So, let's fix a nonzero $u \in \mathbb{C}^n$, say $u = (1,\dots,1)$, and $c = 2$. Then, we get
$$
A(2u) = \| 2u \| \cdot 2u = 4 \| u \| u.
$$
But if $A$ were linear, then we would have had $$A(2u) = 2 A(u) = 2 \|u \| u.$$ Since this is not the case, we have shown that $A$ is not linear.

Also note that the other answer by @GinoCHJ is incomplete. Merely knowing that $$A(u+v) \leq A(u) + A(v) + \| v\| u + \|u\|v$$ does not imply that $A$ is not a linear transformation. Perhaps by some wicked magic it turns out that $\|v\| u + \|u\|v = 0$ for all $u,v \in \mathbb{C}^n$? Their answer would be satisfactorily completed if specific vectors $u$ and $v$ are produced such that the inequality is strict.
A: Let $u, v \in \mathbb{C}^n$, then
\begin{align*}
  &A(u + v) = \lVert u + v \rVert(u + v) \\
\Rightarrow &A(u + v) = \lVert u + v \rVert u + \lVert u + v \rVert v \leq \lVert u \rVert u + \lVert v \rVert u + \lVert u \rVert v + \lVert v \rVert v \\
\Rightarrow &A(u + v) \leq A(u) + \lVert v \rVert u + \lVert u \rVert v + A(v) 
\end{align*}
Hence, $A$ is not a linear transformation.
