Let $V,H$ be Hilbert spaces with embedding $V\hookrightarrow H$ compact and $V$ dense in $H$. Let $a:V\times V\to\mathbb{R}$ be a symmetric, continuous bilinear form and define $A:V\to V'$ by $\left<Au,v\right> = a(u,v),u,v\in V$ with $\left<.,.\right>$ being the duality pairing between $V$ and its dual space $V'$. Define $\left<.,.\right>_V$ by
$$\left<u,v\right>_V = a(u,v) + t\left<u,v\right>_H$$
for a fixed $t\in\mathbb{R}$. Take it as granted that $\left<.,.\right>_V$ is an inner product on $V$ and suppose that $\{u_k\}_k$ is an orthonormal basis of $H$ such that $u_k$s are also eigenvectors of $A$ with corresponding eigenvalues $\lambda_k$. Lastly, define $P_mu := \sum_{k=1}^m\left<u,u_k\right>_Hu_k$.
I am trying to understand why then $\left<u - P_mu,u_j\right>_V = 0$. That is, why
$$\left<u - P_mu,u_j\right>_V = \left<u,u_j\right>_V - \sum_{k=1}^m\left<u,u_k\right>_H\left(a(u_k,u_j) + t\left<u_k,u_j\right>_H\right)\leftrightarrow$$
$$\left<u - P_mu,u_j\right>_V = a(u,u_j) + t\left<u,u_j\right>_H - \sum_{k=1}^m\left<u,u_k\right>_H\left(\lambda_k\delta_k^j + t\delta_k^j\right)\leftrightarrow$$
$$\left<u - P_mu,u_j\right>_V = a(u,u_j) + t\left<u,u_j\right>_H - \left<u,u_j\right>_H\left(\lambda_j + t\right)\leftrightarrow$$
$$\left<u - P_mu,u_j\right>_V = a(u,u_j) - \lambda_j \left<u,u_j\right>_H = 0.$$
Edit: My exact question is the following: Why is $a(u,u_j) = \lambda_j \left<u,u_j\right>_H = 0$? By definition $a(u,u_j) = \left<Au,u_j\right> \equiv \left(Au\right)(u_j)$ as per the duality pairing. We know that $\lambda_j$ is an eigenvalue of $A^*$ but why would $u_j$ be an eigenvector of $A$ in the general case that $A$ is an unbounded operators?