A Linear Algebra Doubt Suppose we have a $3$-dim vector space $V$ and we choose any non orthogonal
set of basis ${v_1,v_2,v_3}$. Now we consider the linear transformation that
projects any vector in $V$ to span$(v_1,v_2)$. Clearly the matrix representation for this transformation is
$$\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0
\end{bmatrix}$$. The eigenvectors corresponding to eigen-value $1$
are ${v_1,v_2}$, and corresponding to $0$ is $v_3$. As the matrix is symmetric,
it should have orthogonal eigenvectors. But $\langle v_1,v_3\rangle$ should be non zero
as they are non orthogonal. Why is this?
 A: You're trying to invoke the following result:

Suppose $A$ is a symmetric real matrix (or a Hermitian complex matrix, if you prefer). Then the eigenspaces of $A$ are orthogonal.

This is true. The matrix you provided has orthogonal eigenspaces $\operatorname{span}\{(1, 0, 0), (0, 1, 0)\}$ and $\operatorname{span}\{(0, 0, 1)\}$. Note that these are subspaces of $\Bbb{R}^3$, which may or may not have anything to do with the inner product space $V$. Thus, it is not correct to say that these are eigenspaces of your projection operator $P$.
If you want the equivalent to the above result for general inner product spaces, this is it:

Suppose $V$ is a finite-dimensional inner product space, and $T : V \to V$ is self-adjoint. Then the eigenspaces of $T$ are orthogonal.

In our case, the projection map
$$P : a_1 v_1 + a_2 v_2 + a_3 v_3 \mapsto a_1 v_1 + a_2 v_2$$
is not self-adjoint if $v_3$ is not orthogonal to $v_1$ and $v_2$. Let's suppose it's not orthogonal to $v_1$. Then
$$\langle Pv_1, v_3 \rangle = \langle v_1, v_3 \rangle \neq 0 = \langle v_1, 0 \rangle = \langle v_1, Pv_3\rangle.$$
So, as you've observed, it's perfectly possible to take a linear operator that is not self-adjoint to a symmetric (or Hermitian) matrix, by way of a non-orthonormal basis. Just be aware of which theorems about inner product spaces require bases to be orthonormal to preserve inner product space properties.
