On a Hilbert- (or otherwise inner-product-) space $\mathcal{H}$ with scalar product $S = \langle.|.\rangle_\mathcal{H}$, a self-adjoint operator is is readily defined as a linear mapping $A : \mathcal{H} \to \mathcal{H}$ with the property $$ S (A v, w) = S(v,A w) \quad \forall v,w\in \mathcal{H}. $$ Fair enough, but an inner product is a pretty specific structure on a space. Do we need it to define what self-adjoint means?

It seems that an operator which is self-adjoint with respect to $S$ is also self-adjoint with respect to another scalar product $T$, but I really don't see how one could go about proving this.

If so, would there be a definition of self-adjointness that makes no reference to any particular inner product? A useful candidate would be something like “a self-adjoint operator is one that has a system of eigenvectors $(\psi_i)_i$ which spans the entire space, such that $A(\psi_i) = \lambda_i\cdot \psi_i$ with real $\lambda$”. But is that actually equivalent to the usually given definition? The spectral theorem only goes one way, and makes itself reference to an inner product.

  • 3
    $\begingroup$ It is not true that self-adjointness is invariant under a change of the inner product. Consider $S(v,w) = v_1w_1+2v_2w_2$ on $\mathbb R^2$. $\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$ is not self-adjoint with respect to $S$. $\endgroup$
    – Lukas Betz
    Commented Jan 23, 2017 at 22:45
  • $\begingroup$ @LeBtz that's the answer then, isn't it? Though I wonder, would this also work if we consider only positive definite operators? $\endgroup$ Commented Jan 23, 2017 at 23:01

3 Answers 3


Self-adjointness is the operator-theoretic version of symmetry of bilinear forms. The notion of a symmetric bilinear form can be defined without reference to any extra structure but in order to talk about self-adjoint operators, we need some mechanism to convert operators to bilinear forms. An inner product $g$ allows us to do just that. Namely, let $V$ be a finite dimensional real vector space and let $g$ be an inner product. Given a linear map $T$, we can define the bilinear form associated to $T$ to be $\beta^g_T(v,w) := g(Tv,w)$. This gives us a map (in fact, a linear isomorphism) from $\operatorname{End}(V) \cong V^{*} \otimes V$ to $\operatorname{Bil}(V \times V, \mathbb{R}) \cong V \otimes V$.

The operator $T$ is $g$-self-adjoint if the associated bilinear form $\beta^g_T$ is symmetric. By changing $g$ we get different different maps between $\operatorname{End}(V)$ and $\operatorname{Bil}(V \times V, \mathbb{R})$ so $T$ might be $g$-self-adjoint but not $g'$-self-adjoint. This is best seen by invoking the spectral theorem.

In the real case, we know that $T$ is $g$-self-adjoint if and only if $T$ is $g$-orthogonally diagonalizable. Denote by $\lambda_1, \dots, \lambda_k$ the distinct eigenvalues of $T$. Then $V = \bigoplus_{i=1}^k \ker(T - \lambda_i I)$ is a direct sum orthogonal decomposition of $V$. If we change the inner product $g$ to an inner product $g'$ such that some $\ker(T - \lambda_i I)$ won't be $g'$-orthogonal to some $\ker(T - \lambda_j I)$ (with $i \neq j$) then by the spectral theorem, $T$ won't be self-adjoint. This is always possible as long as $k > 1$. In other words, the only operators $T$ which are self-adjoint with respect to all inner products are the scalar operators.

In fact, a real operator $T$ is $g$-self-adjoint with respect to some inner product $g$ if and only if $T$ is diagonalizable. If $T$ is self-adjoint, it is orthogonally diagonalizable and in particular diagonalizable. For the other direction, if $T$ is diagonalizable we can always take an arbitrary basis of eigenvectors of $T$ and define an inner product $g$ by declaring this basis to be orthonormal - then by the spectral theorem $T$ will be $g$-orthogonal.

  • 1
    $\begingroup$ “If we change the inner product $g$ to an inner product $g'$ such that $v_1,…,v_n$ are not orthonormal with respect to $g'$ then by the spectral theorem, $T$ won't be self-adjoint.” – really? The spectral theorem only says that for a given $g$, there exists an orthonormal eigensystem, not whether any eigensystem is orthonormal. $\endgroup$ Commented Jan 24, 2017 at 1:42
  • $\begingroup$ @leftaroundabout: You are of course correct, I've edited my answer to include a different argument. Thanks! $\endgroup$
    – levap
    Commented Jan 24, 2017 at 2:04
  • $\begingroup$ Ok, this makes sense now. In other words, self-adjointness is actually a much stronger and less natural condition than existance of a real-valued eigensystem? Then I wonder, why is it used all the time – is it just because self-adjointness is so useful for efficiently calculating an eigensystem? $\endgroup$ Commented Jan 24, 2017 at 10:34
  • $\begingroup$ In the presence of an inner product, it is often more convenient to work with an orthonormal basis instead of a non-orthonormal one (for example, you can calculate the expansion coefficients of a vector with respect to an orthonormal basis by projecting, no need to solve a linear system of equations, etc). It is interesting to know then when a linear operator $T$ admits an eigensystem of orthonormal vectors and it turns out that is the case iff the operator is self-adjoint. This is a condition that is easy to check given an orthonormal basis (just see if the matrix representing $T$ $\endgroup$
    – levap
    Commented Jan 24, 2017 at 19:59
  • $\begingroup$ with respect to the orthonormal basis is symmetric) unlike checking whether $T$ is diagonalizable which requires you to calculate the spectrum and check the multiplicities. $\endgroup$
    – levap
    Commented Jan 24, 2017 at 20:00

Start with a finite-dimensional space $X$ and a basis $\{b_1,b_2,\cdots,b_n\}$. You can define an inner-product between $x,y \in X$ by expanding $$ x = \alpha_1 b_1 + \alpha_2 b_2 + \cdots + \alpha_n b_n \\ y = \beta_1 b_1 + \beta_2 b_2 + \cdots \beta_n b_n $$ and setting $$ \langle x,y\rangle_{b} = \sum_{j=1}^{n}\alpha_j\beta_j. $$ In this inner product, $\{ b_1,b_2,\cdots, b_n \}$ is an orthonormal basis of $X$. A matrix $A$ is selfadjoint on the inner product space $(X,\langle\cdot,\cdot\rangle_b)$ if, for example, the elements of $b$ are eigenvectors of $A$ with real eigenvalues. Not every diagonalizable matrix $D$ with real eigenvalues is selfadjoint with respect to a given inner product. However, using this procedure, you can find an inner product with respect to which $D$ is selfadjoint.


That depends on the inner product, of course, because self-adjointness is all about preservation of inner product between two different bilinear transformations.

To illustrate, consider the Euclidean inner product space $\mathbb R^n$. Let $P$ be a positive definite matrix and define an inner product $\langle x,y\rangle_P=y^TPx$. Then a linear operator $L$ on $\mathbb R^n$ is self-adjoint with respect to $\langle\cdot,\cdot\rangle_P$ if and only if its matrix $A$ in the standard basis satisfies $PA=A^TP$.

Clearly not every matrix $L$ satisfies such a condition, and whether it is self-adjoint with respect to the Euclidean inner product (i.e. whether $A$ is real symmetric) or whether it is positive definite with respect to the Euclidean inner product (i.e. whether $A$ is a positive definite matrix) are completely irrelevant. It could happen that:

  • $L$ is not self-adjoint w.r.t. the Euclidean inner product but it is self-adjoint w.r.t. $\langle\cdot,\cdot\rangle_P$, e.g. when $A=\pmatrix{4&2\\ 3&6}$ and $P=\pmatrix{2&1\\ 1&2}$;
  • $L$ is self-adjoint and even positive definite w.r.t. the Euclidean inner product but it is not self-adjoint w.r.t. $\langle\cdot,\cdot\rangle_P$, e.g. when $A=\pmatrix{1\\ &2}$ and $P=\pmatrix{2&1\\ 1&2}$.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .