Caution while dealing with finite field in Linear Algebra by Hoffman & Kunze

181 Views Asked by At

In the first section $1.1$ of the book Linear Algebra by Hoffman & Kunze, author says that

In the examples and exercises of this book, the reader should assume that the field involved is a subfield of complex numbers, unless it is expressly stated that the field is more general.

I have two questions :

$1.$ So does this mean all the theorems in the book are true for any fields.

$2.$ If I want to be a little careful about finite fields, which steps I should do carefully while doing examples. Like in a subfield of $\mathbb{C}$, $x.y=0 \Rightarrow x=0 \ \text{or} \ y=0$, but this is not true for finite fields. Are there any other such steps, which I might do in flow which are true for a subfield of $\mathbb{C}$, but not for finite field.

1

There are 1 best solutions below

1
On BEST ANSWER

"Linear algebra" is really a broad term that covers diverse topics such as matrix theory, systems of linear equations, vector geometry comes up, we often see some quadratic forms... so it is somewhat difficult to say if everything will still hold over a finite field.

But you will find that almost all of the standard "linear algebra" theorems are either still true over a finite field, or have a close analog. There are a few exceptions.

  1. The most major exception is that we don't really have a notion of magnitude in a finite field (i.e. we can't say one finite field element is "larger than" another), and by extension, we don't consider finite field elements to be "positive" or "negative". So any statements about the distance between two vectors, or the length of a vector, or a quantity being positive or negative, are meaningless. This also means that we don't have a notion of an inner product (since it requires $\langle \bf{x}, \bf{x}\rangle \geq 0$ for all $\bf{x}$). Instead of inner products we typically deal with what are known as bilinear forms.
  2. A second exception is that finite fields are, well, finite; so a finite-dimensional vector space over a finite field has a finite number of vectors. Obviously this means that systems of equations cannot have infinitely many solutions. This also leads to some nice results that can be proven using counting arguments.
  3. A third exception has to do with fields of characteristic 2. In these fields, $1+1=0$ (really $a+a=0$ for any $a \in \mathbb{F}$). This can lead to some results that hold in most fields, but not those of characteristic 2; there are also some results that are still true over fields of characteristic 2, but require an alternate proof. The main example I can think of where characteristic 2 is an issue, is when looking at quadratic forms. You would typically represent a quadratic form $Q(x,y) = x^{2}+xy+y^{2}$ using a symmetric matrix $$Q(x,y) = \begin{bmatrix} x & y\end{bmatrix} \begin{bmatrix} 1 & \frac{1}{2}\\ \frac{1}{2} & 1 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}$$ but this does not work in characteristic 2, so modifications have to be made, typically by using an upper triangular matrix instead of a symmetric matrix, like so: $$Q(x,y) = \begin{bmatrix} x & y\end{bmatrix} \begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}.$$ It is worth noting that, unlike a symmetric matrix, an upper-triangular matrix cannot always be diagonalized.