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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?

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    $\begingroup$ 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. $\endgroup$ – Travis Nov 6 '17 at 14:06
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    $\begingroup$ 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 $u\neq 0$. $\endgroup$ – Arthur Nov 6 '17 at 14:06
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    $\begingroup$ @Travis: I am writing the proof. Thanks. $\endgroup$ – Amin Nov 6 '17 at 14:08
  • $\begingroup$ Related to math.stackexchange.com/questions/2341854/…. $\endgroup$ – lhf Nov 6 '17 at 14:24
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... 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 true, 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 implies (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 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 $$\pmatrix{x\\y} \mapsto x y .$$

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  • $\begingroup$ 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 $1/2 (\text{dim} \; V)(\text{dim} \; V+1)$"? $\endgroup$ – Amin Nov 7 '17 at 2:20
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    $\begingroup$ (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. $\endgroup$ – Travis Nov 7 '17 at 9:30
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    $\begingroup$ (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. $\endgroup$ – Travis Nov 7 '17 at 9:47
  • $\begingroup$ Thanks @Travis for your complete answer. $\endgroup$ – Amin Nov 7 '17 at 9:48
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    $\begingroup$ 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. $\endgroup$ – Travis Nov 7 '17 at 9:51
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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.

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    $\begingroup$ $B\bigl((x_1,x_2),(y_1,y_2)\bigr)=x_1y_1+x_1y_2+x_2y_1+x_2y_2$? $\endgroup$ – SdidS Nov 6 '17 at 15:16
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    $\begingroup$ @SdidS Thanks. I've edited my answer. $\endgroup$ – José Carlos Santos Nov 6 '17 at 15:17
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    $\begingroup$ 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. $\endgroup$ – José Carlos Santos Nov 6 '17 at 15:22
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    $\begingroup$ @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. $\endgroup$ – Omnomnomnom Nov 6 '17 at 16:38
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    $\begingroup$ @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. $\endgroup$ – Omnomnomnom Nov 6 '17 at 16:43

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