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

[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

 

[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)}