# Self-adjoint operator and inner products

Suppose $$V$$ is a nonzero finite dimensional vector space over $$F = R$$. If $$T: V \rightarrow V$$ is linear, and $$B = \{v_1, ..., v_n\}$$ is an ordered eigenbasis of $$V$$ with respect to $$T$$, show that there exists some inner product on $$V$$ where $$T$$ is self-adjoint.

My attempt:

By the orthogonality theorem, given $$B = \{v_1, ..., v_n\}$$, there is an orthonormal basis $$\{u_1, ..., u_n\}$$ of $$V$$.

By the spectral theorem, $$V$$ has an orthonormal basis with respect to $$T$$ if and only if $$T$$ is self-adjoint.

So, $$[T]_B = [T*]_B \implies T = T*$$.

From here, I'm not sure how to proceed. I can't think of how to prove that there is an inner product where $$T$$ is self-adjoint. Any tips or assistance is much appreciated!

• Why not use the scalar product on $V$ which makes $B$ orthonormal? (When using an other $u$-basis, what does it mean "there is" in the context? What should be the relation between the $v$-basis $B$ and the new $u$-basis? And the $u$-basis, claimed to be orthonormal, is orthonormal w.r.t. which scalar product on $V$? (There is no such product given...) Dec 18, 2020 at 13:48
• I'm sorry, I don't understand... Could you explain further? @dan_fulea Dec 18, 2020 at 13:51
• How do you use (which) orthogonality theorem?! Note that after when the attempt starts in the OP there is no scalar product on $B$. To be allowed to use the word orthonormal, there should be some scalar product around. Define one first. Just take the one that makes $B$ orthonormal! (As in the answer of @Christoph!) Dec 18, 2020 at 13:56

There is a unique inner product $$\langle -,-\rangle$$ on $$V$$ satisfying $$\langle v_i, v_j \rangle = \delta_{ij} = \begin{cases} 1 & \text{if i=j},\\ 0 & \text{otherwise.} \end{cases}$$ With respect to this inner product, your basis $$B$$ is an orthonormal basis.

Since $$B$$ is also an eigenbasis of $$T$$, the spectral theorem yields that $$T$$ is self-adjoint with respect to $$\langle - , -\rangle$$.

Note that the basis $$B=\{v_1,\dots,v_n\}$$ of $$V$$ yields an isomorphism of vector spaces $$\Phi\colon V\to \mathbb R^n$$ such that $$\Phi(v_i)=e_i$$ is the $$i$$-th standard basis vector of $$\mathbb R^n$$ for all $$i$$. The inner product given above is now obtained by transferring the standard dot product from $$\mathbb R^n$$ along this isomorphism: $$\langle v, v' \rangle := \Phi(v)\cdot\Phi(v').$$ We have $$\langle v_i, v_j\rangle = \Phi(v_i)\cdot\Phi(v_j) = e_i \cdot e_j = \delta_{ij}$$ so this does indeed give the desired product.

Since the dot product is an inner product on $$\mathbb R^n$$, this product is an inner product on $$V$$, rendering $$\Phi$$ an isometry.

• Oh I see! How would one write this as a proof though? Dec 18, 2020 at 13:54
• Write "Proof." in front of it and "$\square$" at the end. Dec 18, 2020 at 13:55
• Good one :) Thank you! Dec 18, 2020 at 13:57
• For the proof: Which is the matrix of $T$ w.r.t. the basis $B$? Which is the matrix for $T^*$? Are the eigenvalues of $T$ all real? Dec 18, 2020 at 13:59
• @TRONIIX I added a paragraph explaining how to obtain this inner product on $V$ from the standard dot product of $\mathbb R^n$, showing that it does in fact exists and is an inner product. Dec 18, 2020 at 14:04