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

[mfb]

The question states "Find all possible values of h" though

It does not.
Find, for all weekdays, the German word for those days.
Is finding all weekdays a part of the task? Well, you'll need to know them, but you are not expected to start research just to figure out the existence of 7 weekdays.

I've always found the kernel by row reducing a matrix, which I did originally. The problem was, I got to a point where I had to assume what the value of h was by also assuming what the value of b was.

You don't need any assumption about b. b is a free parameter in the transformation, and the equation has to be true for every b.

You get two different cases, one for h=0 and one for other (real) h, right. Find Ker and I am for both separately then.

 

[haruspex]

I don't know how to format this, but it's the original 2x2 matrix * (1,x,x²,x³) = 0, only 1,x,x²,x³ would be a column.

[ a+b+c+hd b+c ] * ((1,x,x²,x³) = 0
[ -b-c-hd hb ]

No, you don't need any powers of x here. P is a polynomial. The set of polynomials forms a ring, but in the present context all we are doing is applying linear transformations. This treats the set of polynomials as merely a vector space. We won't be multiplying polynomials at all. So we can express P as a vector [a b c d]^T.
In your OP, you present a 2x2 matrix illustrating f_h(P). Translating "Ay=0" to this context (avoiding any confusion over x) y is P and A is f_h, so what does Ay look like in terms of a, b, c, d, h? Equating that to 0, what equations do you get?

 

[haruspex]

I've always found the kernel by row reducing a matrix, which I did originally. The problem was, I got to a point where I had to assume what the value of h was by also assuming what the value of b was.

That's finding the kernel of the 2x2 matrix that results from applying f_h to some arbitrary polynomial. You are asked to find the kernel of f_h, not of f_h(P).

 

[says]

So a,b,c,d are all free parameters. They can have any value, but if I give them a value in the polynomial their value will be the same in the matrix, and then I will be able to find value/s of h by figuring out the kernel

 

[haruspex]

So a,b,c,d are all free parameters. They can have any value, but if I give them a value in the polynomial their value will be the same in the matrix, and then I will be able to find value/s of h by figuring out the kernel

No, you do not give a, b, c and d values. (I would call them dummy variables here, not free variables.) All they do is show you how f_h maps a polynomial to a 2x2 matrix.
The conceptual challenge with this whole problem is to recognise that the set of polynomials, on the one hand, and the set of 2x2 matrices on the other are each being treated merely as four dimensional vector spaces. I feel it might help if you represent them as such. The polynomial we can write as
$\begin{pmatrix} a \\ b \\ c \\ d\end{pmatrix}$ and applying f_h to this we get
$\begin{pmatrix} a+b+c+hd \\ b+c \\ -b-c -hd\\ hb\end{pmatrix}$.
Halls kindly wrote out the corresponding matrix form of f_h for you in post #21.
You can forget all the other info at that point and just find the kernel and image based on that matrix.

The question states "Find all possible values of h" though.

No it doesn't. It says for all possible values of h, find the kernel and the image. Since h can take any real value, we can drop the "possible".
Just find the kernel and the image, and note how these depend on the value of h.

 

[says]

So I can take this matrix:
$\begin{pmatrix} 1+1+1+h \\ 0+1+1+0 \\ 0-1-1-h\\ 0+h+0+0\end{pmatrix}$.

Row reduce it (R1+R3):
$\begin{pmatrix} 1+0+0+0 \\ 0+1+1+0 \\ 0-1-1-h\\ 0+h+0+0\end{pmatrix}$ = 0

From this row reduced matrix I can see h can equal -2 and 0. So now I put both values of h (0 and -2) back into the first matrix separately and find the kernel and image for both?

 

[says]

From the original matrix I can see h could also equal -3 (from the first row) of the matrix 1+1+1+h = 0

 

[mfb]

So I can take this matrix:
$\begin{pmatrix} 1+1+1+h \\ 0+1+1+0 \\ 0-1-1-h\\ 0+h+0+0\end{pmatrix}$.

Row reduce it (R1+R3):
$\begin{pmatrix} 1+0+0+0 \\ 0+1+1+0 \\ 0-1-1-h\\ 0+h+0+0\end{pmatrix}$ = 0

From this row reduced matrix I can see h can equal -2 and 0. So now I put both values of h (0 and -2) back into the first matrix separately and find the kernel and image for both?

You cannot add the columns like this.
-2 and 0 are possible values for h, but all other real numbers are possible as well.

 

[says]

Ax=0

I forgot to add

x = $\begin{pmatrix} a \\ b \\ c \\ d\end{pmatrix}$

So now I can represent them as four equations. And in those equations I can see by row reduction, and also looking at the original non-row reduced matrix, that h= -2,0,-3.

 

[mfb]

Can you show step by step how you get that result? It is not right, and I don't understand how you get it.

 

[says]

Ax = 0

$\begin{pmatrix} 1+1+1+h \\ 0+1+1+0 \\ 0-1-1-h\\ 0+h+0+0\end{pmatrix}$ $\begin{pmatrix} a \\ b \\ c \\ d\end{pmatrix}$ = 0

We can row reduce this matrix, and in doing so can see h=-2,0,-3.

 

[mfb]

Again, the addition in the matrix does not make sense.

 

[says]

formatting issue, I don't mean to have the + or minus in the matrix. But we can represent the matrix as 4 equations to find h:

1a+1b+1c+hd = 0
0+1b+1c+0 = 0
0-1b-1c-hd = 0
0+hb+0+0=0

 

[SammyS]

Ax = 0

$\begin{pmatrix} 1+1+1+h \\ 0+1+1+0 \\ 0-1-1-h\\ 0+h+0+0\end{pmatrix}$ $\begin{pmatrix} a \\ b \\ c \\ d\end{pmatrix}$ = 0

We can row reduce this matrix, and in doing so can see h=-2,0,-3.

So you meant to write:

$\displaystyle \begin{pmatrix} 1&1&1&h \\ 0&1&1&0 \\ 0&-1&-1&-h\\ 0&h&0&0\end{pmatrix}\begin{pmatrix} a \\ b \\ c \\ d\end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0\end{pmatrix} \$​

 

[says]

Yes, that's what I meant to write.

 

[mfb]

formatting issue, I don't mean to have the + or minus in the matrix. But we can represent the matrix as 4 equations to find h:

1a+1b+1c+hd = 0
0+1b+1c+0 = 0
0-1b-1c-hd = 0
0+hb+0+0=0

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?

 

[says]

1a+1b+1c+hd=0
0+1b+1c+0=0
0+0+0-hd=0
0+hb+0+0=0

1)R3+R2
2)R1-R2
3)R1+R3

1a+0+0+hd=0
0+1b+1c+0=0
0+0+0+0=0
0+hb+0+0=0

ker(f_h) = span {(1a,0,0,hd), (0,1b,1c,0), (0,hb,0,0)}

 

[says]

As an example, what is the kernel for h=10?

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

1)R1+R3
1)R1-R2

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

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

 

[haruspex]

So I can take this matrix:
$\begin{pmatrix} 1+1+1+h \\ 0+1+1+0 \\ 0-1-1-h\\ 0+h+0+0\end{pmatrix}$.

Row reduce it (R1+R3):
$\begin{pmatrix} 1+0+0+0 \\ 0+1+1+0 \\ 0-1-1-h\\ 0+h+0+0\end{pmatrix}$ = 0

From this row reduced matrix I can see h can equal -2 and 0. So now I put both values of h (0 and -2) back into the first matrix separately and find the kernel and image for both?

You reduction algorithm appears broken.
If you subtract a multiple of a column from other columns, you must subtract the same multiple from each. You appear to have subtracted the first column as is from the middle two columns, but subtracted h times it from the fourth.

 

[haruspex]

ker(f_h) = span {(1a,0,0,hd), (0,1b,1c,0), (0,hb,0,0)}

a, b, c and d are not specific values. They cannot appear in the answer.
The kernel is the set of vectors (polynomials) [a b c d] that is mapped to the zero 2x2 matrix by f_h.

 

[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