# Diagonalizability of symmetric bilinear forms over fields of characteristic $2$

Theorem:

A symmetric bilinear form $$H$$ on a finite dimension vector space $$V$$ over a field $$\mathbb{F}$$, where it is not of characteristic two, is diagonalizable.

Proof:

By induction on $$\text{dim} \; V=n$$.

(induction base) $$n=0$$. Then it is trivial.

(induction hypothesis) Assume the above statement holds for all bilinear forms on vector spaces of dimension $$n-1$$.

(inductive step) Suppose the space $$\text{dim} \; V=n$$. If the bilinear form $$H=0$$, then it is trivial. Hence, assume that $$H\neq0$$, then there exists $$z \in V$$ such that $$H(z,z) \neq 0$$. Let $$W = \operatorname{span}\{z\}^\perp$$ and we have $$V = W \oplus \operatorname{span}\{z\}$$. As $$\text{dim} \; W=n-1$$, the theorem holds for this space and there is a basis $$\beta = \{v_1, ..., v_{n-1}\}$$ where $$H$$ is diagonal. Then, by extending the basis we have $$\gamma = \beta \cup \{z\} \subset V$$. Then we have: $$H(v_{i},z) = 0$$ for all $$i=0, ..., n-1$$. Hence, this implies that there exists a basis $$\gamma \subset V$$ such that the matrix corresponding to $$H$$ is diagonal.

Why is the field of characteristic two excluded?

• Can you say more about your proof? If you can find where it fails for characteristic $2$, you might be able to exploit that the find an explicit counterexample. Commented Nov 6, 2017 at 14:06
• I don't know the details of the proof, but it wouldn't surprise me if they somewhere divided by $2$, or assumed that $u$ and $-u$ are distinct vectors for $u\neq 0$. Commented Nov 6, 2017 at 14:06
• @Travis: I am writing the proof. Thanks.
– Amin
Commented Nov 6, 2017 at 14:08
• Related to math.stackexchange.com/questions/2341854/….
– lhf
Commented Nov 6, 2017 at 14:24
• I find your inductive step very confusing. See mathoverflow.net/q/23629/123740 for a counterexample.
– Bach
Commented Jul 29, 2019 at 4:48

... Hence, assume that $$H\neq0$$, then there exists $$z \in V$$ such that $$H(z,z) \neq 0$$. ...

This is not true in characteristic $$2$$. Let $$\Bbb F$$ be any field of that characteristic, and, for example, take $$V = \Bbb F^2$$ and the bilinear form $$H$$ with matrix representation $$[H] = \pmatrix{0&1\\1&0}$$ with respect to the standard basis; then, the quadratic form $$Q_H : {\bf x} \mapsto H({\bf x}, {\bf x})$$ is the zero form. Indeed, we can see directly that $$H$$ is not diagonalizable: Computing directly for any $$P \in \textrm{GL}(2, \Bbb F)$$ gives $$P^T [H] P = (\det P) [H],$$ which is not diagonal.

Put another way, it is not true in characteristic $$2$$ that the map $$H \mapsto Q_H$$ is injective. In other characteristics it is injective, as we can recover $$H$$ from $$Q$$ via the Polarization Identity $$H({\bf x}, {\bf y}) = \tfrac{1}{4}[Q_H({\bf x} + {\bf y}) - Q_H({\bf x} - {\bf y})] ,$$ but in characteristic $$2$$ one cannot divide by $$4$$ (which in that setting coincides with $$0$$).

Remark The above facts imply (since the spaces of symmetric bilinear forms on $$V$$ and quadratic forms on $$V$$ both have dimension $$\frac{1}{2} (\dim V)(\dim V + 1)$$) that (only) in characteristic $$2$$ there are quadratic forms that are not induced by symmetric bilinear forms (that is, are not in the image of the map $$H \mapsto Q_H$$); the simplest example is $$V = \Bbb F^2$$ and $$\pmatrix{x\\y} \mapsto x y .$$

• Thanks @Travis. I have two questions: first, based on the discussion of above in comments, here there is no such $P$ for which $P^{-1} [H] P$ is diagonal? The second question is how "the spaces of symmetric bilinear forms on $V$ and quadratic forms on $V$ both have dimension $1/2 (\text{dim} \; V)(\text{dim} \; V+1)$"?
– Amin
Commented Nov 7, 2017 at 2:20
• (1) As mentioned in the comments in Jose's answer, since we are asking about diagonalization of a bilinear form (rather than a linear transformation), the question is whether there is a $P \in \operatorname{GL}(2, \Bbb F)$ such that $P^T [H] P$ is diagonal. And the answer is no: Computing directly gives for our $H$ that $P^T [H] P = (\det P) \pmatrix{0&1\\1&0}$, which is not diagonal. Commented Nov 7, 2017 at 9:30
• (2) With respect to any basis we can specify any symmetric bilinear form $B$ by freely choosing the upper-triangular entries of the upper matrix $[B]$, but once we have done so symmetry (that is, that, $[B]^T = [B]$) implies determines the rest of the entries. So, the vector space of symmetric bilinear forms has the same dimension as the space of upper triangular matrices, namely, $\frac{1}{2}(\dim V)(\dim V + 1)$. For characteristic not $2$, the isomorphism $B \leftrightarrow Q$ then gives that the same is true for the dimension of the space of quadratic forms. Commented Nov 7, 2017 at 9:47
• For characteristic $2$ (and any characteristic), a choice of basis determines an isomorphism $V \cong \Bbb F^{\dim V}$, and we can write down an explicit basis of the space of quadratic forms, namely $\{(x_1, \ldots, x_{\dim V}) \mapsto x_i x_j : i \leq j\}$, and this basis has the claimed number of elements. Commented Nov 7, 2017 at 9:51
• I upvoted this back in the day. And would have done so for the final example alone! For a "simpler" argument of its non-diagonalizability I once used the following (natural to coding theorists). The form $xy$ has three zeros in $\Bbb{F}_2^2$, but the number of zeros of any diagonalizable is a power of two because it really is just a linear function. Commented Jul 31, 2021 at 19:30

Over $\mathbb{F}_2$, you can find symmetric bilinear forms which are not diagonalizable. Take, for instance, the bilinear form $B\colon{\mathbb{F}_2}^2\times{\mathbb{F}_2}^2\longrightarrow\mathbb{F}_2$ defined by$$B\bigl((x_1,x_2),(y_1,y_2)\bigr)=x_1y_1+x_1y_2+x_2y_1+x_2y_2.$$In ${\mathbb{F}_2}^2$ there are only three bases (without caring about the order of the vectors): $\bigl\{(1,0),(0,1)\bigr\}$, $\bigl\{(1,0),(1,1)\bigr\}$, and $\bigl\{(0,1),(1,1)\bigr\}$. You can check that the matrix of $B$ with respect to each one of these bases is not diagonal.

Note: This is a corrected version of my answer, made after Omnomnomnom telling, in the comments, that I was using the wrong concept of diagonalizable.

• $B\bigl((x_1,x_2),(y_1,y_2)\bigr)=x_1y_1+x_1y_2+x_2y_1+x_2y_2$?
– SddS
Commented Nov 6, 2017 at 15:16
• @SdidS Thanks. I've edited my answer. Commented Nov 6, 2017 at 15:17
• It's matrix with respect to the canonical basis is $\left(\begin{smallmatrix}1&1\\1&1\end{smallmatrix}\right)$. Since the characteristic of $F$ is $2$, the square of this matrix is $0$. But then it cannot be diagonalizable, because the only diagonal matrix whose square is $0$ is the null matrix. Commented Nov 6, 2017 at 15:22
• @JoséCarlosSantos Note that "diagonalizable" has a different meaning in the context of bilinear forms. In particular, we want to show that there is no invertible $P$ such that $P^T\left(\begin{smallmatrix}1&1\\1&1\end{smallmatrix}\right)P$ is diagonal. While you have correctly deduced that there is no invertible $P$ such that $P^{-1}\left(\begin{smallmatrix}1&1\\1&1\end{smallmatrix}\right)P$, this is not strictly relevant to our context. Commented Nov 6, 2017 at 16:38
• @JoséCarlosSantos Moreover, if your particular line of reasoning were valid, we would have trouble in any field of finite characteristic, as is briefly discussed here. Commented Nov 6, 2017 at 16:43