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

[says]

Homework Statement

Find all possible values of h.
fh(a+bx+cx²+dx³) =
[ a+b+c+hd b+c ]
[ -b-c-hd hb ]

Homework Equations

The Attempt at a Solution

1) Row reduce the matrix

[ a+b+c+hd b+c ]
[ -b-c-hd hb ]

to (R1+R2)

[ a hb+ b+c ]
[ 0 hb ]

to (R1-R2)

[ a b+c ]
[ 0 hb ]a+b+c = 0
hb = 0

From this row reduction I assume b=0, a=1 and c=-1
h could still be any number though.
So I got back to the original matrix.

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

If I plug in the values I assumed (above) then I get d=-1 and h=-1

Is there a better / quicker way to do this? I seem to come across a lot of similar problems.

 

[mfb]

I don't understand the notation. Is fh a function? If yes, does the h in the name matter and what do you know about the function? Or is that a simple multiplication?
The right side is a 2x2 matrix? If yes, is the equation true for a given x, or just for an arbitrary x?

 

[says]

It's a linear transformation

 

[mfb]

fh? Of a real number to 2x2 matrices of real numbers? An arbitrary one?
My crystal ball isn't working properly today, so I cannot know that if you don't write it.

 

[haruspex]

Just guessing here, is the matrix in the f_h equation supposed to be surrounded by vectors involving x, maybe [1 x], [1 x]^T?

 

[says]

It's a P³ to M_2x2 linear transformation. The first part of the question, which I'm having difficulty understanding, asks to find all possible values of h. I'm confused because h can have many many values, depending on what the values of x,y,z are and depending on what I set the polynomial of the matrix equations equal to.

i.e.
[ a+b+c+hd b+c ]
[ -b-c-hd hb ]

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

In this case I can row reduce the matrix to simplify both equations
a + hb + b+c = 0
0 + hb = 0

My confusion here is that I can see hb = 0, but h could be 0 or b could be 0 and h could be any number. So I don't really know how I'm supposed to answer this question. 'Find all possible values of h' ... Well h = 0, h ≠ 0 (essentially saying h= 1,2,3,4,5,6,7,...etc)

 

[says]

With my row reducing of the matrix (above) and setting both equations from the matrix = 0 I can show that h could = 0 or it could equal a positive number of a negative number. So is it enough for me to say all possible values of h are 0,1,-1 ?

 

[haruspex]

I must be missing something. It seems to me that f_h constitutes a linear transformation regardless of the value of h.
You know the definition of a linear transformation, right? How would you check that f_h is such?

 

[says]

Yes. The second part of the question asks for the kernel and image. I know how to do that, but I'm confused with how i find all possible values of h.

 

[haruspex]

Yes. The second part of the question asks for the kernel and image. I know how to do that, but I'm confused with how i find all possible values of h.

Well, if you apply the definition of a linear transformation, can you either deduce something about h or show that it is a linear transformation regardless of h?
For the kernel and image, they will depend on h, no?

 

[says]

That's what confuses me. If I take the original matrix and split it into two equations and make them = 0 then my value of h depends on my value of a,b,c,d and vice versa.

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

 

[haruspex]

That's what confuses me. If I take the original matrix and split it into two equations and make them = 0 then my value of h depends on my value of a,b,c,d and vice versa.

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

I don't understand why two equations. If P is in the kernel, what matrix would f_h(P) be?

 

[says]

Isn't writing the matrix of the linear transformation:

[ a+b+c+hd _ b+c | 0 ]
[ -b-c-hd _ hb | 0 ]

the same as writing them as two equations?

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

 

[haruspex]

Isn't writing the matrix of the linear transformation:

[ a+b+c+hd _ b+c | 0 ]
[ -b-c-hd _ hb | 0 ]

you've lost me. The linear transformation of interest is f_h. That effectively maps a four vector to a four vector, so its matrix would be written as a 4x4.
Try to answer my question. If P is in the kernel of f_h what does the 2x2 matrix f_h(P) look like?

 

[says]

I don't understand your question. sorry.

 

[haruspex]

I don't understand your question. sorry.

For the purposes of the question, the set of polynomials of degree 3 is just a vector space here. Likewise the set of 2x2 matrices over R. f_h is a linear transformation from the first to the second.
When we say that an element X is in the kernel of a linear transformation T, what equation expresses that?

 

[says]

f_h: P³ → M_2x2

Ax = 0

 

[haruspex]

f_h: P³ → M_2x2

Ax = 0

Right, so write that last equation out for the case where x is the polynomial a+bx+cx²+dx³.

 

[says]

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 ]

 

[says]

But we'd need a 4x4 matrix for A though for this to work...

 

[HallsofIvy]

So this is
[tex]f_h(a+ bx+ cx^2+ dx^3)= \begin{bmatrix} a+ b+ c+ hd & b+ c \\ -b- c- hd & hb \end{bmatrix}[/tex]
from the space of polynomials of degree 3 (or less) to the space of 2 by 2 matrices. And the problem is to find all possible values of h such that what?
As says says, since this is from a 4 dimensional vector space to a four dimensional vector space, it would be written
as a 4 by 4 matrix. Taking the "obvious" bases for the two spaces, [itex]\{1, x, x^2, x^3\}[/itex] and [itex]\{\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix}0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix}0 & 0 \\ 0 & 1 \end{bmatrix}\}[/itex], this maps [itex]1= 1+ 0x+ 0x^2+ 0x^3[/itex] (a= 1, b= c= d= 0) to [itex]\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}[/itex], it maps [itex]x= 0+ 1x+ 0x^2+ 0x^3[/itex] (b= 1, a= c= d= 0) to [itex]\begin{bmatrix}1 & 1 \\ -1 & h\end{bmatrix}= \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}+ \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}- \begin{bmatrix}0 & 0 \\ 1 & 0 \end{bmatrix}+ h\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}[/itex], it maps [itex]0+ 0x+ 1x^2+ 0x^3[/itex] (c= 1, a= b= d= 0) to [itex]\begin{bmatrix} 1 & 1 \\ -1 & 0\end{bmatrix}= \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}+ \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}- \begin{bmatrix}0 & 0 \\ 1 & 0 \end{bmatrix}[/itex], and maps [itex]0+ 0x+ 0x^2+ 0x^3[/itex] (d= 1, a= b= c= 0) to [itex]\begin{bmatrix} h & 0 \\ -h & 0\end{bmatrix}= h\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}- h\begin{bmatrix}0 & 0 \\ 1 & 0 \end{bmatrix}[/itex]

So the matrix representing this linear transformation, in these bases, is
[tex]\begin{bmatrix} 1 & 1 & 1 & h \\ 0 & 1 & 1 & 0 \\ 0 & -1 & -1 & -h \\ 0 & h & 0 & 0 \end{bmatrix}[/tex].

Now, again, exactly what is it you want to do with, or know about, this matrix?

 

[mfb]

Thanks Halls, that clarifies a lot.
A good question would be "which values of h make the kernel non-trivial".

 

[says]

I need to find all possible values of h and then find the kernel and image of each matrix with respect to the value/s of h.

 

[mfb]

I don't see any restriction on h.
Most values of h will lead to the same result for kernel and image.

 

[Ray Vickson]

Homework Statement

Find all possible values of h.
fh(a+bx+cx²+dx³) =
[ a+b+c+hd b+c ]
[ -b-c-hd hb ]

Homework Equations

The Attempt at a Solution

1) Row reduce the matrix

[ a+b+c+hd b+c ]
[ -b-c-hd hb ]

to (R1+R2)

[ a hb+ b+c ]
[ 0 hb ]

to (R1-R2)

[ a b+c ]
[ 0 hb ]a+b+c = 0
hb = 0

From this row reduction I assume b=0, a=1 and c=-1
h could still be any number though.
So I got back to the original matrix.

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

If I plug in the values I assumed (above) then I get d=-1 and h=-1

Is there a better / quicker way to do this? I seem to come across a lot of similar problems.

I don't think the question makes any sense as written. Assuming you mean
[tex] f_h(a + bx + cx^2 + dx^3) = \pmatrix{a+b+c+hd&b+c\\-b-c-hd & hb} [/tex]
or (better)
[tex] F_h ([a,b,c,d]) = \pmatrix{a+b+c+hd&b+c\\-b-c-hd & hb} [/tex]
then what prevents me from taking any value of ##h## that I want? Why can't I take ##h = 272.63##, or ##h = -107## or ##h = \sqrt{\pi/2}##? For any value of ##h## that I decide to use I will get a ##2 \times 2## matrix having that ##h## in it as a parameter.

If you had meant (but not stated) that you want all values of ##h## that give the mapping ##f_h## some desired property, that would be a different question, and it might well make sense. In that case, you need to spell out what that desired property might be. Or, perhaps, you want Ker(f_h) or Im(f_h), and want to figure out their dimensionalities and the like, as functions of h. Whatever it is you want to do, you need to tell us!

 

[says]

The question in full is:
Given the transformation f_h: P3 to M 2x2 defined by:
[tex] f_h(a + bx + cx^2 + dx^3) = \pmatrix{a+b+c+hd&b+c\\-b-c-hd & hb} [/tex]

where h ∈ R is a parameter.

Find, for all possible values of h, Ker(f_h), Im(f_h), their bases and dimensions.

I'm sorry if I've wasted your time by not writing this out in full. I know how to find Ker and Im, I just don't understand the first bit.

I still don't understand the process by which I can find all possible values of h.

 

[mfb]

The question in full is:

Finally. Sorry to be so direct, but we wasted 25 posts guessing around what the problem statement could be. That's the reason we ask for the full problem statement in post 1.

I know how to find Ker and Im

Then do it.
There are no values of h you have to find. h is a real number. You just have to find Ker and I am for all those real values.
Splitting this task in two cases depending on h will be useful in the process, but that is not a limit on h.

 

[says]

The question states "Find all possible values of h" though. I don't understand how I can find the kernel or I am without first knowing what the values of h can be.

 

[Ray Vickson]

The question states "Find all possible values of h" though. I don't understand how I can find the kernel or I am without first knowing what the values of h can be.

Just go ahead and do it; stop worrying and start computing.

During the course of you computations you will soon discover how the value of h affects things, but you will never understand this if you don't start doing the work.

 

[says]

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.

 

[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.