I suppose a quantum state with density matrix like the following is not valid. $$ \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}. $$
Now, let's say I have a valid density operator representing the state $|\psi \rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1 \rangle)$. $$ |\psi \rangle \langle\psi | = \frac{1}{2}(|0\rangle \langle 0| + |0\rangle \langle 1| + |1\rangle \langle 0| + |1\rangle \langle 1|) = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}. $$
Now I send this state to depolarizing channel $\mathcal{E}$. Because $\mathcal{E}$ is linear: $$ \mathcal{E}(|\psi \rangle \langle\psi |) = \frac{1}{2}(\mathcal{E}(|0\rangle \langle 0|) + \mathcal{E}(|0\rangle \langle 1|) + \mathcal{E}(|1\rangle \langle 0|) + \mathcal{E}(|1\rangle \langle 1|)). $$
I'm wondering what the depolarization of $\mathcal{E}(|0\rangle \langle 1|)$ would mean. By definition of depolarizing channel, for noise parameter $p$,
$$ \mathcal{E}(\rho) = (1 - p)\rho + \frac{pI}{2}. $$
But then, what is the meaning of $\mathcal{E}(|0\rangle \langle 1|)$?