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\}

and if your vector space

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?

Thank you for your help.

——————————————————————–

Reply:

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.

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