Is There a Quicker Way to Find All Possible Values of h for fh(a+bx+cx2+dx3)?

[haruspex]

ker(fh) = span {(0,-1,1,0) , (0,1,-1,0) , (1,1,1,0)}

One of those is redundant, another is wrong. Please post your working.

 

[says]

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)}

 

[SammyS]

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)}

(0,1,-1,0) and (0,-1,1,0) are redundant.

What does your matrix tell you about d ?

 

[says]

d=0

I don't understand how they are redundant if they are two different vectors?

 

[Samy_A]

I don't understand how they are redundant if they are two different vectors?

They are redundant because they are not linearly independent:
$(0,1,-1,0)=-1*(0,-1,1,0)$

 

[SammyS]

d=0

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.

Also, why does d need to be 0? isn't 0d = 0 no matter what the value of d ?

 

[says]

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]

Actually

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?

It's not, and Samy_A demonstrated that.

 

[says]

There is another vector (independent of this) in the kernel of f₀ .

What's the other vector in ker(fh) if that's not it though? There's only 3 equations there and your saying one is redundant, one is in the kernel, and the other isn't in there...

 

[SammyS]

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_h(a + bx + cx^2 + dx^3) = \pmatrix{a+b+c+hd&b+c\\-b-c-hd & hb} \$

so that

$\displaystyle \ f_0(a + bx + cx^2 + dx^3) = \pmatrix{a+b+c+0&b+c\\-b-c-0 & 0} \$

 

[says]

The only other vector I can think of is the zero vector. (0,0,0,0)

 

[SammyS]

The only other vector I can think of is the zero vector. (0,0,0,0)

Again I ask, What value must d have if 0⋅d = 0 ?

 

[says]

d can be any value 0*d=0

0*1=0
0*-1=0
0*0=0

 

[SammyS]

d can be any value 0*d=0

0*1=0
0*-1=0
0*0=0

Yes.

Therefore, the other vector is ____ ?

 

[says]

(1,0,0,-1)

 

[SammyS]

(1,0,0,-1)

That's not correct.

Why involve $a$ ?

 

[says]

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?

 

[SammyS]

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?

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]

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

 

[SammyS]

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

That's right.

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} \$

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?

 

[says]

(0,1,-1,1) and (0,1,-1,-1)

 

[SammyS]

(0,1,-1,1) and (0,1,-1,-1)

In the spirit of your previous submissions, why not

span{(0, 1, -1, 0), (0, 0, 0, 1)} ?

 

[says]

Yes!
I originally though this could be the span:
span{ (0,0,0,1) , (0,1,-1,0) , (0,1,-1,-1) , (0,1,-1,1) }

But then I put each vector into a matrix and row reduced them and got the (0,1,-1,0) and (0,0,0,1) vector.

 

[says]

There is a second part to this question asking if there are any values of h ∈ R such that fh is not an isomorphism? Not sure if I should ask this in a new post or not. I'm not too sure where to start. I've found this problem extremely difficult.

 

[Mark44]

There is a second part to this question asking if there are any values of h ∈ R such that fh is not an isomorphism? Not sure if I should ask this in a new post or not. I'm not too sure where to start. I've found this problem extremely difficult.

Please start a new thread. This one is now at 95 posts. Let's put this one out of its misery...