I would like to collect references (or direct quotations) about as many "simultaneous diagonalization" results in linear algebra as possible.

Let $V$ be an $n$-dimentional ($n$ finite) vector space over the field $\mathbb{K}$. Note that if the characteristic of $\mathbb{K}$ is $\neq$ from $2$ then speaking of quadratic forms is essentially the same as speaking of bilinear forms.

Theorem 1. Let $\mathcal{S}$ be a set of diagonalizable operators on $V$. Then $\mathcal{S}$ is simultaneously diagonalizable if and only if it commutes (i.e. $AB=BA\;\;\; \forall A,B\in\mathcal{S}$).

Theorem 2. Let $\mathbb{K}=\mathbb{C}$. Let $\langle\;|\;\rangle$ be a non-degenerate symmetric bilinear form and $b$ any symmetric bilinear form on $V$. Write $b(\cdot,\cdot)=\langle A\cdot|\cdot\rangle$ for a $\langle\;|\;\rangle$-symmetric operator $A$. Suppose $A$ is diagonalizable with distinct eigenvalues $\lambda_1,\dots , \lambda_n$. Then there is a $\langle\;|\;\rangle$-orthonormal basis $\mathcal{B}$ in which the matrix of $b$ is diagonal with diagonal entries $\lambda_1,\dots , \lambda_n$.

Proof 2. Let $\mathcal{B}=\{ v_1,\dots , v_n \}$ a basis of eigenvectors for $A$, with distinct eigenvalues $\lambda_1, \dots , \lambda_n$. Then $\lambda_l\cdot\langle v_l|v_k \rangle=\langle \lambda_l\cdot v_l|v_k \rangle=\langle A(v_l) |v_k \rangle=\langle v_l |A^{T}(v_k) \rangle=\langle v_l| A(v_k)\rangle=\lambda_k\cdot\langle v_l|v_k \rangle$, hence $(\lambda_l-\lambda_k)\cdot\langle v_l|v_k \rangle=0$ for all $l,k$. Then eigenvectors of distinct eigenvalues are orthogonal. Furthermore, $\langle v_i|v_i \rangle\neq 0$ for all $i$ or $v_i$ would be orthogonal to $\mathcal{B}$ for some $i$. Now let $\xi_i$ be a square root of $\lambda_i$ in $\mathbb{C}$, and set $w_i:=\frac{1}{\xi_i}\cdot v_i$. Then $\{ w_1,\dots , w_n \}$ is the basis we are seeking.

Theorem 3. Let $\mathbb{K}=\mathbb{R}$. Let $q$ be an indefinite quadratic form and $q'$ any quadratic form on $V$. Suppose that, for every $v\in V$, $q(v)=0$ implies $q'(v)=0$. Then there is $\alpha\in\mathbb{R}$ such that $q'=\alpha \cdot q$. Furthermore, if $\alpha\neq 0$, also $q'$ is indefinite and, for all $v \in V$, $q(v)=0$ if and only if $q'(v)=0$. Note that in particular, by Sylvester's theorem, $q$ and $q'$ are simultaneously diagonalizable.

Proof 3. See Elton, Indefinite quadratic forms and the invariance of the interval in Special Relativity.

Theorem 4. Let $\mathbb{K}=\mathbb{R}$. Let $q$ and $q'$ be two semidefinite quadratic forms on $V$. Suppose that for all $v\in V$ $q(v)=0$ implies $q'(v)=0$. Then $q$ and $q'$ are simultaneosly diagonalizable.

Proof 4. See Elton, Indefinite quadratic forms and the invariance of the interval in Special Relativity.

Theorem 5. Let $\mathbb{K}=\mathbb{R}$. Let $\langle\;|\;\rangle$ be a non-degenerate symmetric bilinear form and $b$ any symmetric bilinear form on $V$. Then there is a $\langle\;|\;\rangle$-orthonormal basis $\mathcal{B}$ of $V$ in which the matrix of $b$ is diagonal.

Proof 5. Write $b(\cdot,\cdot)=\langle A\cdot|\cdot\rangle$ for a $\langle\;|\;\rangle$-symmetric operator $A$. Then by the Spectral Theorem for real symmetric operators there is a $\langle\;|\;\rangle$-orthogonal basis $\mathcal{B}=\{ e_1, \dots, e_n \}$ in which the matrix of $A$ is diagonal with diagonal entries $\lambda_1,\dots , \lambda_n$. Then $b(e_i,e_j)=\langle Ae_i|e_j \rangle=\lambda_i\cdot\langle e_i|e_j \rangle=\lambda_i\cdot\delta_{ij}$ ($\delta_{ij}$ is Kronecker's symbol), hence $\mathcal{B}$ diagonalizes $b$.

Theorem 6. Let $\mathbb{K}=\mathbb{C}$. Let $\langle\;|\;\rangle$ be a Hermitian (i.e. a sesquilinear positive definite) form and $b$ any sesquilinear form on $V$. Then there is a $\langle\;|\;\rangle$-orthonormal basis $\mathcal{B}$ of $V$ in which the matrix of $b$ is diagonal.

Proof 6. As in Theorem 5. but using the Spectral Theorem for Hermitian operators instead.

Edit: I have collected two more theorems.

Theorem 7. Let $\mathbb{K}$ be of characteristic $\neq 2$. Let $\langle\; |\; \rangle_1$, $\langle\; |\; \rangle_2$ be two non degenerate symmetric bilinear forms on $V$. Let $\Phi_i:V\to V^*, v\mapsto \langle v | \;\rangle_i$, $i=1,2$. Then $\langle\; | \;\rangle_1$ and $\langle\; | \;\rangle_2$ are simultaneously diagonalizable if and only if $\Phi_2^{-1}\circ\Phi_1\in \mathrm{Aut}(V)$ is diagonalizable as an operator.

Proof 7. See M.J.Wonenburger, Simultaneous Diagonalization of Symmetric Bilinear Forms.

Remark: Note that $\Phi_2^{-1}\circ\Phi_1$ is symmetric w.r.t. $\langle\; | \;\rangle_2$, and $\Phi_1^{-1}\circ\Phi_2$ is symmetric w.r.t. $\langle\; | \;\rangle_1$. So, in case $\mathbb{K}=\mathbb{R}$, we recover part of Theorem 5.

Theorem 8. Let $\mathbb{K}$ be a real closed field. Let $f_1, f_2$ be two quadratic forms on $V$ such that there is no $v\neq 0$ in $V$ for which $f_1(v)=f_2(v)=0$. Suppose $n=\dim V>2$, then $f_1$ and $f_2$ are simultaneously diagonalizable.

Proof 8. See M.J.Wonenburger, Simultaneous Diagonalization of Symmetric Bilinear Forms.

Q1. I would like to know other theorems like the above ones (not deriving from them in a tautological way).

Q2. Is there a more general theorem on the topic that treats all fields at once and reduces to (a subset of) the above theorems by specialization?

Q3. What happens in characteristic $2$?

• In theorem 5 you want the bilinear form $b$ to be symmetric. – Marc van Leeuwen Oct 3 '12 at 16:24
• Of course. Edited accordingly. – Simplicius Oct 3 '12 at 18:03