# 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?

• 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.
– Soby
Mar 15 '21 at 6:26
• 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!) Mar 15 '21 at 6:28
• @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 Mar 15 '21 at 6:46
• Thanks @MorganRodgers so this is where I made a silly mistake, this arose my confusion.. Mar 15 '21 at 7:44
• Yes this clears the picture thanks all.. Mar 15 '21 at 7:49

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.

• Thank you for the clarification.. Mar 15 '21 at 8:06
• Just to be clear, as we define these operators over $\mathbb{C}^n$, self adjoint and symmetric are equivalent right? Mar 15 '21 at 11:12
• 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. Mar 15 '21 at 11:17
• Okay so orthonormal basis is the criteria here, any basis wont do Mar 15 '21 at 12:09
• @roydiptajit Correct. Mar 15 '21 at 12:13