# Homework Help: Find all possible values of h

1. Dec 29, 2015

### says

1. The problem statement, all variables and given/known data
Find all possible values of h.
fh(a+bx+cx2+dx3) =
[ a+b+c+hd b+c ]
[ -b-c-hd hb ]

2. Relevant equations

3. 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.

2. Dec 29, 2015

### Staff: Mentor

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?

3. Dec 29, 2015

### says

It's a linear transformation

4. Dec 29, 2015

### Staff: Mentor

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.

5. Dec 30, 2015

### haruspex

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

6. Dec 30, 2015

### says

It's a P3 to M2x2 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)

7. Dec 30, 2015

### 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 ?

8. Dec 30, 2015

### haruspex

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

9. Dec 30, 2015

### 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.

10. Dec 30, 2015

### haruspex

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?

11. Dec 30, 2015

### 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

12. Dec 30, 2015

### haruspex

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

13. Dec 30, 2015

### 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

14. Dec 30, 2015

### haruspex

you've lost me. The linear transformation of interest is fh. 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 fh what does the 2x2 matrix fh(P) look like?

15. Dec 30, 2015

### says

I don't understand your question. sorry.

16. Dec 30, 2015

### haruspex

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. fh 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?

17. Dec 30, 2015

### says

fh: P3 → M2x2

Ax = 0

18. Dec 30, 2015

### haruspex

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

19. Dec 30, 2015

### 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 ]

20. Dec 30, 2015

### says

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

21. Dec 30, 2015

### HallsofIvy

So this is
$$f_h(a+ bx+ cx^2+ dx^3)= \begin{bmatrix} a+ b+ c+ hd & b+ c \\ -b- c- hd & hb \end{bmatrix}$$
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, $\{1, x, x^2, x^3\}$ and $\{\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix}0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix}0 & 0 \\ 0 & 1 \end{bmatrix}\}$, this maps $1= 1+ 0x+ 0x^2+ 0x^3$ (a= 1, b= c= d= 0) to $\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}$, it maps $x= 0+ 1x+ 0x^2+ 0x^3$ (b= 1, a= c= d= 0) to $\begin{bmatrix}1 & 1 \\ -1 & h\end{bmatrix}= \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}+ \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}- \begin{bmatrix}0 & 0 \\ 1 & 0 \end{bmatrix}+ h\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}$, it maps $0+ 0x+ 1x^2+ 0x^3$ (c= 1, a= b= d= 0) to $\begin{bmatrix} 1 & 1 \\ -1 & 0\end{bmatrix}= \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}+ \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}- \begin{bmatrix}0 & 0 \\ 1 & 0 \end{bmatrix}$, and maps $0+ 0x+ 0x^2+ 0x^3$ (d= 1, a= b= c= 0) to $\begin{bmatrix} h & 0 \\ -h & 0\end{bmatrix}= h\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}- h\begin{bmatrix}0 & 0 \\ 1 & 0 \end{bmatrix}$

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

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

22. Dec 30, 2015

### Staff: Mentor

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

23. Dec 30, 2015

### 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.

24. Dec 30, 2015

### Staff: Mentor

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

25. Dec 30, 2015

### Ray Vickson

I don't think the question makes any sense as written. Assuming you mean
$$f_h(a + bx + cx^2 + dx^3) = \pmatrix{a+b+c+hd&b+c\\-b-c-hd & hb}$$
or (better)
$$F_h ([a,b,c,d]) = \pmatrix{a+b+c+hd&b+c\\-b-c-hd & hb}$$
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!

Last edited: Dec 30, 2015