MHB Why is the characteristic different from 2?

[evinda]

Hello! (Wave)

In my notes, there is the following theorem:

Theorem:

Let $Q(X_1, X_2, \dots, X_n)=\sum_{i,j=1}^n a_{ij} X_i X_j$

$$a_{ij}=a_{ji}, \ \ \ \ ch K \neq 2 ( \text{ Characteristic of polynomial })$$

symmetric quadratic form of $n$ variables.

There is a linear transformation of the coordinates, so that, finally, $Q(X_1, X_2, \dots, X_n)$ is transformated to the form:

$$Q'(X_1, X_2, \dots, X_n)=\sum_{i=1}^r l_i X_i^2, i=1,2, \dots, r, l_i \neq 0$$How do we conclude that $ch K \neq 2$ ? (Thinking)

 

[I like Serena]

Hello! (Wave)

In my notes, there is the following theorem:

Theorem:

Let $Q(X_1, X_2, \dots, X_n)=\sum_{i,j=1}^n a_{ij} X_i X_j$

$$a_{ij}=a_{ji}, \ \ \ \ ch K \neq 2 ( \text{ Characteristic of polynomial })$$

symmetric quadratic form of $n$ variables.

There is a linear transformation of the coordinates, so that, finally, $Q(X_1, X_2, \dots, X_n)$ is transformated to the form:

$$Q'(X_1, X_2, \dots, X_n)=\sum_{i=1}^r l_i X_i^2, i=1,2, \dots, r, l_i \neq 0$$How do we conclude that $ch K \neq 2$ ? (Thinking)

Hey! (Happy)Can you clarify what you mean by $\text{ch K}$? (Wondering)I can tell you that the spectral theorem for real symmetric matrices says that:
$$A = B^T D B$$
where $D$ is a diagonal matrix of real eigenvalues and $B$ is an orthogonal matrix of eigenvectors. (Nerd)

It follows that:
$$Q = x^T A x = x^T (B^T D B) x = (Bx)^T D (Bx)$$

In other words, $B$ identifies the linear transformation and the $l_i$ are the eigenvalues. (Mmm)

 

[Opalg]

Hello! (Wave)

In my notes, there is the following theorem:

Theorem:

Let $Q(X_1, X_2, \dots, X_n)=\sum_{i,j=1}^n a_{ij} X_i X_j$

$$a_{ij}=a_{ji}, \ \ \ \ ch K \neq 2 ( \text{ Characteristic of polynomial })$$

symmetric quadratic form of $n$ variables.

There is a linear transformation of the coordinates, so that, finally, $Q(X_1, X_2, \dots, X_n)$ is transformated to the form:

$$Q'(X_1, X_2, \dots, X_n)=\sum_{i=1}^r l_i X_i^2, i=1,2, \dots, r, l_i \neq 0$$How do we conclude that $ch K \neq 2$ ? (Thinking)

The statement $\text{ch}\,K \ne2$ is not a conclusion, it is a condition. In fact, the result fails if the characteristic of the underlying field (not the polynomial!) is $2$.