# Is it possible to define an inner product such that an arbitrary operator is self adjoint?

Given a vector space $V$ (possibly infinite dimensional) with inner product $(.,.)$. We say an operator $A$ is self adjoint if $(Af,g)=(f,Ag)$.

The definition as stated require us to start with an inner product $(.,.)$ in $V$ and check if the operator $A$ satisfies the equality.

My question is:

If we start with an operator $B$ on a vector space $W$ what are the necessary and sufficient conditions such that we can define an inner product such that $B$ is self adjoint with that inner product?

• I think in the f.d case, every operator has an adjoint. I think Riesz representation can be used to prove this, so it is left for the infinite-dimensional case. May 14, 2014 at 1:05
• No, even in a finite-dimensional vector space, an self-adjoint operator T is diagonalizable. In fact, in this case, an inner product can be defined such that an operator T is self-adjoint if and only if T is diagonalizable. May 14, 2014 at 1:16
• @user43378: Ouch, yes, I was thinking of the existence of an adjoint and not about self-adjointness. In the f.d case, when representing with a matrix M, we just need M=M^T , where M^T is the transpose of M. May 14, 2014 at 1:51
• Note that in the infinite-dimensional case, the same result is not true. You can still prove that if $T$ is diagonalizable, then $T$ is self-adjoint with respect to certain inner product by using the same method. However, even if $T$ is self-adjoint, it may not have any eigenvectors at all. An example is the multiplication operator $T$ on $L^2[0, 1]$, that is $(Tf)(t) = tf(t)$. May 14, 2014 at 2:15

I will address the case when $V$ is finite dimensional (in infinite dimensions, things may be more subtle). Let me also assume that the base field is either $\mathbb{R}$ or $\mathbb{C}$.

I claim that the following is a necessary and sufficient condition for the existence of an inner product for which $B$ is self-adjoint: There exists a basis of $V$ consisting of eigenvectors of $B$ with real eigenvalues (i.e., $B$ is diagonalizable with real eigenvalues).

This condition is necessary by the spectral theorem.

It is also sufficient: given a basis of eigenvectors, we can construct an inner product by declaring that basis to be orthonormal, and extending the inner product (sesqui-)linearly. If the eigenvalues are real, then $B$ will be self-adjoint with respect to this inner product.

• Doesn't the f.d case reduce to having , for a matrix representation M for the operator, that M=M^T , where M^T is the transpose of M? May 14, 2014 at 1:53
• Of course, the matrix representation of $B$ depends on a choice of basis. If you mean that there exists some basis in which the matrix for $B$ is symmetric (or Hermitian, if the base field is $\mathbb{C}$), then yes, you're correct. But the matrix representation for $B$ will not generally be symmetric with respect to every basis. May 14, 2014 at 2:14
• Ah yes, for some reason I was assuming a fixed basis .Serenity now!!!! May 14, 2014 at 2:30
• This is a very nice answer. Unfortunatly, I am particularly interested in the infinite dimensional case, specially when $B=\square$ and $W=W^{k,2}$ is the Sobolev spaces $(k,2)$. If you have any insights that would be much appreciated.
– yess
May 14, 2014 at 15:59
• What do you mean by $\Box$? Also, there's a comment on your question above that illustrates why this argument doesn't work in infinite dimensions. I don't know what the answer to your question is in infinite dimensions (or if there is a good answer). @Nate Eldredge has demonstrated knowledge of various spectral theorems in infinite dimensions to me in the past, so he might be able to help. May 14, 2014 at 16:27