If an operator $T$ is self-adjoint, why is the matrix of $T$ symmetric if and only if the basis is orthonormal?

I'm seeing that this is what is being said in Artin's Algebra, Treil's Linear Algebra Done Wrong, and Hoffman and Kunze's Linear Algebra, but no justification is given.

How do I show that:

Let $T$ be a self-adjoint operator on a complex inner product space $V$ and let $\beta$ be a basis for $V$ . The matrix of $T$ is Hermitian if and only if $\beta$ is orthonormal.

  • 1
    $\begingroup$ This is false unless you add some hypothesis to make $T$ nontrivial. As it stands, you can take $T$ to be the zero operator (which is self-adjoint) and take $\beta$ to be any basis that isn't orthonormal. The matrix of $T$ with respect to $\beta$ is the zero matrix, which is Hermitian. The same issue arises if $T$ is the identity operator; its matrix is the identity matrix with respect to any basis, even if the basis is not orthonormal. To avoid all such problems, you need to assume that $T$ has no multiple eigenvalues. $\endgroup$ Apr 7, 2018 at 2:41
  • $\begingroup$ @AndreasBlass Actually, that condition is not enough: if $f:\mathbb{C}^2\to \mathbb{C}^2$ is the operator with matrix $\begin{pmatrix} 1&3\\3&0\end{pmatrix}$ on the basis $\{(6,6);(6,4)\}$, by the change of basis formula one easily gets that the matrix of $f$ with respect to the canonical basis is still $\begin{pmatrix} 1&3\\3&0\end{pmatrix}$, and thus $f$ is symmetric. $\endgroup$
    – user515010
    Apr 23, 2020 at 9:49
  • $\begingroup$ @Caffeine The operator in your coment doesn't seem to be self-adjoint, because the basis $\{(6,6),(6,4)\}$ isn't orthonormal --- unless you mean self-adjoint with respect to an unusual inner product that makes that basis orthonormal, but then the canonical basis won't be orthonormal. $\endgroup$ Apr 23, 2020 at 13:29
  • $\begingroup$ @AndreasBlass the operator is self adjoint with respect to the usual inner product: to see why, it is sufficient to note that the matrix of $f$ with respect to the canonical basis is $\begin{pmatrix}6&6\\6&4\end{pmatrix}\begin{pmatrix}1&3\\3&0\end{pmatrix}\begin{pmatrix}6&6\\6&4\end{pmatrix}^{-1}=\begin{pmatrix}1&3\\3&0\end{pmatrix}$. Thus, since the matrix of $f$ is symmetric with an orthonormal basis (the canonical one), $f$ is simmetric $\endgroup$
    – user515010
    Apr 23, 2020 at 13:38
  • $\begingroup$ @Caffeine You're right; thanks. $\endgroup$ Apr 23, 2020 at 13:43

2 Answers 2


I'm assuming you are working over complex numbers (I'll denote the complex conjugation by $*$) and let $(-,-)$ be the inner product. Let $M$ be the matrix of $T$ in basis $\beta=\{\beta_1, \cdots, \beta_n\}$, i.e. $T\beta_j = \sum_{i=1}^n M_{ij}\beta_i$.

Suppose $\beta$ is orthonormal, then by taking the inner product of the above equation with $\beta_i$, you find $$ (\beta_i, T\beta_j)=M_{ij}=(T\beta_j, \beta_i)^* $$ But $T$ is self-adjoint, so $M_{ij}=(\beta_i, T\beta_j)=(T\beta_i, \beta_j)={M_{ji}}^*$, meaning $M$ is hermitian.

Conversely, note that the whole argument is completely reversible, so you find that if $\beta$ is a basis such that $M$ is hermitian then, given any pair $i,j$ we have $$ M_{ij} = M_{ji}^*\Longrightarrow (\beta_i, T\beta_j)=(T\beta_i,\beta_j) $$ Therefore, by bilinearity of inner product and since $\beta$ is a basis, given any two vectors $v,w$, one has $(v,Tw)=(Tv,w)$, i.e. $T$ is self-adjoint.


The claim is false, as the condition on the basis is sufficient but not necessary.

An easy counterexample, as noted by Andreas Blass in the comments to the question, is the identity operator, which has the same symmetric matrix ($I_n$) for every basis, even those which are not orthonormal. However, one may think that such a case can be avoided by imposing stricter conditions on $T$: this is usually not the case, as we will prove with a general counterexample.

Let $T:\mathbb{C}^n\to \mathbb{C}^n$ be a self-adjoint operator of maximal rank which is not an orthonormal transformation on of rank $n$, and let $M$ be its matrix in the canonical basis. If we define the basis $\beta=\{v_1,\dots,v_n\}$ as the basis composed by the column vectors of $M$, by the formula for the change of basis we obtain

$$T_{\beta}=(v_1|\dots |v_n)^{-1}M(v_1|\dots |v_n)=M^{-1}M M=M$$

Thus the matrix of $T$ in the basis $\beta$ (which is not orthonormal, for otherwise $T$ would be orthonormal) is symmetric, and we obtained a general counterexample


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