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!

  • $\begingroup$ 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...) $\endgroup$
    – dan_fulea
    Dec 18, 2020 at 13:48
  • $\begingroup$ I'm sorry, I don't understand... Could you explain further? @dan_fulea $\endgroup$ Dec 18, 2020 at 13:51
  • $\begingroup$ 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!) $\endgroup$
    – dan_fulea
    Dec 18, 2020 at 13:56

1 Answer 1


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.

  • $\begingroup$ Oh I see! How would one write this as a proof though? $\endgroup$ Dec 18, 2020 at 13:54
  • $\begingroup$ Write "Proof." in front of it and "$\square$" at the end. $\endgroup$
    – Christoph
    Dec 18, 2020 at 13:55
  • $\begingroup$ Good one :) Thank you! $\endgroup$ Dec 18, 2020 at 13:57
  • $\begingroup$ 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? $\endgroup$
    – dan_fulea
    Dec 18, 2020 at 13:59
  • $\begingroup$ @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. $\endgroup$
    – Christoph
    Dec 18, 2020 at 14:04

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