Depolarization of density operator with zeros in diagonal

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 $|psi rangle = frac{1}{sqrt{2}}(|0rangle + |1 rangle)$.
$$
|psi rangle langlepsi | = frac{1}{2}(|0rangle langle 0| + |0rangle langle 1| + |1rangle langle 0| + |1rangle langle 1|) = frac{1}{2} begin{bmatrix} 1 & 1 1 & 1 end{bmatrix}.
$$

Now I send this state to depolarizing channel $mathcal{E}$. Because $mathcal{E}$ is linear:
$$
mathcal{E}(|psi rangle langlepsi |) = frac{1}{2}(mathcal{E}(|0rangle langle 0|) + mathcal{E}(|0rangle langle 1|) + mathcal{E}(|1rangle langle 0|) + mathcal{E}(|1rangle langle 1|)).
$$

I’m wondering what the depolarization of $mathcal{E}(|0rangle langle 1|)$ 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 $mathcal{E}(|0rangle langle 1|)$?