## Pre-Jordan basis

Dear Prof Kim,

I had a couple of questions on Algebra3.

1) Pre-Jordan basis:

If you have your pre-jordan basis as

$B_1=\{(-1,1,0)^t\}, B_2=\{(-1,0,1)^t\}$

$V_3=Ker(0 \ \mbox{matrix})$

then how do you decide the pre jordan basis $B_3$? Is it always

$(1,0,0)^t$?

Another example, if

$B_1=\{(-1,0,3)^t,(-2,3,0)^t\}$

and vector space

$V_2=Ker (0\ \mbox{matrix})$

then

$B_2=\{(1,0,0) ^t\}$

or

$(0,1,0)^t?$

2) Canonical form:

In lecture notes it says, a symmetric bilinear form is positive definite iff its canonical form (over R) is $I_n$. But in solutions to sheet 7 it says canonical form is $I_{n+1}$. So is it trivial that $n$ can be any natural number depending on size original matrix corresponding to the sym bilinear form?

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Neither the pre-Jordan basis nor the Jordan basis are unique. This perhaps creates some difficulty in related questions owing to the answer not being completely determined.

Once $B_1, B_2, \ldots B_i$ has been chosen the only requirment of $B_{i+1}$ is that

$B_1\cup B_2\cup \cdots \cup B_i\cup B_{i+1}$

is a basis of $V_{i+1}$. In your first question $V_3$ is the whole space, so you need only choose $B_3$ so that you end up with a basis for the whole space. So your choice would work, but so would $(0,1,0)^t$, $(0,0,1)^t$, $(1,1,1)^t$, and so on. Any vector that’s independent of the first two will work. Similarly, in the second question, $B_2$ can consist of any vector independent of the two in $B_1$.

Regarding symmetric bilinear forms, it’s correct that $f$ is positive definite if and only if its real canonical form is

$I_n.$

But here, $n$ is the dimension of the vector space. So if the original form was written in terms of some variables, say $f(x_1,x_2, \ldots, x_n)$, then the number of variables $n$ is the same as the $n$ in $I_n$. Check the dimension of the vector space in that question on sheet 7.