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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?

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    $\begingroup$ The matrix you have written is with respect to the basis $\{v_1,v_2,v_3\}$. That is all. The eigenvectors for the matrix you have written is $e_1,e_2$ and $e_3$. With $e_1,e_2$ corresponding to eigenvalue 1 and $e_3$ corresponding to 0. $\endgroup$
    – Soby
    Mar 15 '21 at 6:26
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    $\begingroup$ The matrix you have written down represents the transformation with respect to the basis $v_1,v_2,v_3$. With respect to that basis, $v_1$ is $(1,0,0)$ and $v_3$ is $(0,0,1)$, and those are orthogonal. (@Soby beat me by a few seconds!) $\endgroup$ Mar 15 '21 at 6:28
  • $\begingroup$ @Gerry Myerson , so we just consider the coordinate space in this which has standard basis and those are orthonormal? Can I say any symmetric matrix w.r.t non orthogonal basis has orthonormal eigenvectors in the coordinate space, but when I see in terms of the actual basis, it may not be so? I am confused when do we consider the coordinate space and when the actual vector space $\endgroup$ Mar 15 '21 at 6:46
  • $\begingroup$ Thanks @MorganRodgers so this is where I made a silly mistake, this arose my confusion.. $\endgroup$ Mar 15 '21 at 7:44
  • $\begingroup$ Yes this clears the picture thanks all.. $\endgroup$ Mar 15 '21 at 7:49
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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.

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  • $\begingroup$ Thank you for the clarification.. $\endgroup$ Mar 15 '21 at 8:06
  • $\begingroup$ Just to be clear, as we define these operators over $\mathbb{C}^n$, self adjoint and symmetric are equivalent right? $\endgroup$ Mar 15 '21 at 11:12
  • $\begingroup$ To be very clear, an operator $T$ on a finite-dimensional complex inner product space $V$ is self-adjoint if and only if $[T]_\beta^\beta$ is Hermitian (not symmetric!) where $\beta$ is an orthonormal basis for $V$. If $V = \Bbb{C}^n$, then the standard matrix for $V$ is the matrix for $T$ with respect to the standard basis, which is orthonormal, and hence $T$ is self-adjoint if and only if the standard matrix for $T$ is Hermitian. $\endgroup$ Mar 15 '21 at 11:17
  • $\begingroup$ Okay so orthonormal basis is the criteria here, any basis wont do $\endgroup$ Mar 15 '21 at 12:09
  • $\begingroup$ @roydiptajit Correct. $\endgroup$ Mar 15 '21 at 12:13

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