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

[says]

so ker(fh) = span {(1,0,0,h), (0,1,1,0), (0,h,0,0)}

 

[Samy_A]

so ker(fh) = span {(1,0,0,h), (0,1,1,0), (0,h,0,0)}

As we have $f_h(a,b,c,d)=(a+b+c+hd,b+c,-b-c-hd,hb)$ (writing everything as vectors), we get:
$f_h(1,0,0,h)=(1+h²,0,-h²,0)\neq (0,0,0,0)$
$f_h(0,1,1,0)=(2,2,-2,h)\neq (0,0,0,0)$
$f_h(0,h,0,0)=(h,h,-h,h²)\neq (0,0,0,0)$ unless $h=0$

So that seems wrong for $Ker(f_h)$.

Hint:

Okay, now reduce this step by step to find the kernel for all real values of h. As an example, what is the kernel for h=10?

 

[mfb]

1a+0+0+0=0
0+1b+1c+0=0
0+0+0+10d=0
0+10b+0+0=0

You are not fully done with simplifications here.

ker(f_h) = span {(1a,0,0,0), (0,1b,1c,0), (0,0,0,10d), (0,10b,0,0}

Is the kernel really given by that? That's a 4-dimensional subspace, which means it is the whole space. Is everything mapped to zero?

 

[says]

1a+0+0+0=0
0+0+1c+0=0
0+0+0+1d=0
0+1b+0+0=0

R4/10
R2-R4
R3/10

I'm confused by the second question.

 

[mfb]

Just simplify your equation system one step further, what can you say about a,b,c,d? Which set of vectors satisfies all those conditions at the same time?

 

[says]

1+0+0+0=0
0+0+1+0=0
0+0+0+1=0
0+1+0+0=0

a=b=c=d=0

 

[SammyS]

1+0+0+0=0
0+0+1+0=0
0+0+0+1=0
0+1+0+0=0

a=b=c=d=0

Yes, at least for most values of h.

Is there any value of h which gives a less restrictive answer?

 

[says]

I don't understand what you mean sorry

 

[SammyS]

I don't understand what you mean sorry

One result from your matrix equation is

hb = 0​

Doesn't this say that b = 0, unless h = 0? So, what if h is 0 ?

 

[says]

if h = 0 then b could be any real number

 

[SammyS]

if h = 0 then b could be any real number

Right.

What's the kernel for f₀ ?

 

[says]

What is f₀?

 

[SammyS]

What is f₀?

It's f_h when h = 0 .

 

[says]

ker f_h = span {(1,0,0,0), (0,1,1,0)}

 

[Samy_A]

ker f_h = span {(1,0,0,0), (0,1,1,0)}

Let's doublecheck:
$f_h(a,b,c,d)=(a+b+c+hd,b+c,-b-c-hd,hb)$
$f_0(a,b,c,d)=(a+b+c,b+c,-b-c,0)$

$f_0(1,0,0,0)=(1,0,0,0)\neq (0,0,0,0)$
$f_0(0,1,1,0)=(2,2,-2,0)\neq (0,0,0,0)$

 

[says]

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

when h=0

 

[SammyS]

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

when h=0

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

 

[says]

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)

 

[haruspex]

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)

How about you go back to the matrix equation you had in post #41, substitute h=0, and solve?

 

[says]

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

 

[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...