Change of basis and spectral theorem

Question

I've been having trouble with such a rudimentary problem. Let us define a matrix $A$: $$A = \begin{pmatrix} 3 & 0 & -i \\ 0 & 3 & 0 \\ i & 0 & 3 \end{pmatrix}$$ A is a 3 by 3 Hermitian matrix, so it has real eigenvectors and eigenvalues $|\phi_0 \rangle, |\phi_1 \rangle,|\phi_{2} \rangle $ and $\lambda_0, \lambda_1, \lambda_2$: $$\vec \phi_0 = \langle i, 0, 1 \rangle ; \lambda_0 = 2 \\ \vec \phi_1 = \langle 0, 1, 0 \rangle; \lambda_1 = 3 \\ \vec \phi_2 = \langle -i, 0, 1 \rangle; \lambda_2 = 4$$ I'm asked to transform the coefficients of the basis states in a state $|\psi (t) \rangle$, $\alpha$ and $\beta$ by multiplying by $e^{-i \lambda_j t}$, where $t$ is a number. $|\psi(0) \rangle = |0 \rangle$. This is simple, but my problem is I can't rewrite the $|0 \rangle$ state as a linear combination of the three eigenstates. When attempting to solve it normally, I get: $$|0 \rangle = -\frac{i}{\sqrt{2}} |\phi_0 \rangle + \frac{i}{\sqrt{2}} |\phi_2 \rangle $$ or $$\langle 1, 0 \rangle = \langle -\frac{i}{\sqrt{2}}, 0, \frac{i}{\sqrt{2}} \rangle$$ When transforming both coefficients by the exponentation function mentioned above when $t=\pi$: $$|0 \rangle = -\frac{i}{\sqrt{2}} |\phi_0 \rangle - \frac{i}{\sqrt{2}} |\phi_2 \rangle $$ or $$\vec 0(t) = \langle -\frac{i}{\sqrt{2}}, 0, -\frac{i}{\sqrt{2}} \rangle$$ I then rewrite in the 0, 1, 2 basis: $$-i |2 \rangle$$ or $\langle 0, 0, -i \rangle$ This is obviously wrong. The correct answer is $-|0 \rangle$. Where am I going wrong?