The discussion revolves around finding all possible values of h in the context of a linear transformation represented by a 2x2 matrix derived from a polynomial of degree 3. The matrix is given as fh(a+bx+cx2+dx3) = [ a+b+c+hd b+c ] [ -b-c-hd hb ]. Participants are exploring the implications of h on the transformation and the relationships between the coefficients a, b, c, and d.
The discussion is ongoing, with various interpretations of the problem being explored. Some participants have provided insights into the nature of linear transformations and the implications of h, while others express confusion about how to determine the values of h and the relationship to the kernel and image of the transformation.
There are indications of confusion regarding the notation and the setup of the problem, particularly concerning the nature of fh and the dimensionality of the transformation. Participants are also grappling with the implications of h on the linear transformation and how it relates to the kernel and image.
Right.says said:if h = 0 then b could be any real number
It's fh when h = 0 .says said:What is f0?
Let's doublecheck:says said:ker fh = span {(1,0,0,0), (0,1,1,0)}
There is another vector (independent of this) in the kernel of f0 .says said:ker fh = span {(0,1,-1,0)}
when h=0
How about you go back to the matrix equation you had in post #41, substitute h=0, and solve?says said:but there is only one independent variable though. I thought that would mean only one vector? Unless the other vector is just (0,-1,1,0)
One of those is redundant, another is wrong. Please post your working.says said:ker(fh) = span {(0,-1,1,0) , (0,1,-1,0) , (1,1,1,0)}
(0,1,-1,0) and (0,-1,1,0) are redundant.says said:1+1+1+h=0
0-1-1-h=0
0+1+1+0=0
0+h+0+0=0
making h=0
1+1+1+0=0
0-1-1-0=0
0+1+1+0=0
0+0+0+0=0
if b=1 c=-1 then the first equation a =0
ker(fh)= span {(0,1,-1,0) , (0,-1,1,0)}
They are redundant because they are not linearly independent:says said:I don't understand how they are redundant if they are two different vectors?
One is simply -1 times the other. I.e. their sum is zero.says said:d=0
I don't understand how they are redundant if they are two different vectors?
It's not, and Samy_A demonstrated that.says said:Ok, I understand that, but how can the first equation (1,1,1,0) be in ker(fh) 1+1+1+0=0 as well?
SammyS said:There is another vector (independent of this) in the kernel of f0 .
Again I ask, What value must d have if 0⋅d = 0 ?says said:The only other vector I can think of is the zero vector. (0,0,0,0)
Yes.says said:d can be any value 0*d=0
0*1=0
0*-1=0
0*0=0
That's not correct.says said:(1,0,0,-1)
For h = 0, (1,1,1,h) → (1,1,1,0) so that if a=b=c=0, then d does not need to be zero.says said:Because the first equation was (1,1,1,h) = 1+1+1+hd. If d=-1 then can't b and c = 0, making a = 1?
That's right.says said:the transformation always has d as a product with h though. so even if d=1 or -1 in the polynom it will still be 0 in the matrix
There is no value of d which keeps this from being the 2×2 zero matrix.So, how to you write that as a vector?SammyS said:Actually, once you decide to look at h=0, then the original statement of the problem can be used to find Ker f0 by inspection.
##\displaystyle \ f_0(a + bx + cx^2 + dx^3) = \pmatrix{a+b+c+0&b+c\\-b-c-0 & 0} \ ##