For some reason I kept thinking what was the problem with doing it component wise...
In that case, how could you prove this then?
As a reference, they proved like this
Also gonna take a stab at the axioms they used...
1. Existence of 0
2. Scaling factor of 1
3. arithmetic (or just existence of 0 in field of scalars)
4. Scalar distributive law
5. Scaling factor of 1
Last edited by leehuan; 23 Aug 2016 at 10:12 AM.
They really aren't hard to remember.
The axioms are just annoying.
Actually knowing when and how to use them like there though is ...
S = {x is an element of R^4 | (x1)-5(x3) = 2(x4)}
Find one non-zero element in S.
Do you just sub in random values for x1,x3,x4 (also where's x2)?
Don't mind if this question isn't finished off for me but can someone please set it up? Finding it hard to set everything out neatly here.
i forgot the proper proof for these
U is subset of r3 which is known v.s <- not sure if u need to include this for first year, but im used to writing it anyway..
0 vector is obvious
Ax+Ay=A(x+y) y element of U etc i.e A(0)+A(0)=A(0+0) = 0
A*(Landa)(x)=(Landa) Ax landa is real, can be complex if u want A*Landa(0)-Landa A(0) =0
In other words, to show a set S is closed under addition, it means we have to show that if a and b are arbitrary elements of S, then so is a+b.
To show the set U is closed under scalar multiplication, it means we have to show that if c is an arbitrary scalar and v an arbitrary vector from U, then cv is also in U.
For this polynomial question, I get the same matrix but R1 and R3 swapped. Is this still okay, or is my method wrong?
^Feelsbad delaying the exercise
Anyway
Is this just because the zero polynomial can't be in S?
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