# Linear Transformation and Isomorphism

says

## Homework Statement

Given the transformation fh : R 3 → R 3 defined by fh(x, y, z) = (x−hz, x+y −hz, −hx+z), where h ∈ R is a parameter.
a) Find, for all possible values of h, Ker(fh), Im(fh), their bases and dimensions.
b) Is fh an isomorphism for some value of h?

Ax=o

## The Attempt at a Solution

[ x−hz ]
[ x+y −hz ]
[ −hx+z ]

Matrix associated to the linear transformation =

[ 1 0 -h ]
[ 1 1 -h ]
[ -h 0 1 ]

Reduced row echelon form =
[ 1 0 -h ]
[ 0 1 0 ]
[ -h 0 1 ]

x - hz = 0
y=0
-hx+z=0

Therefore h = 1.

ker(fn)

[ 1 0 -1 ]
[ 0 1 0 ] = 0
[ -1 0 1 ]

ker(fn) = span { (1,0,-1) , (-1,0,1) } = range(ker(fn))

To find the Im(fn) we will firstly set up the original matrix with h=1

[ 1 0 -1 ]
[ 1 1 -1 ]
[ -1 0 1 ]

Then transpose =

[ 1 1 -1 ]
[ 0 1 0 ]
[ -1 -1 1 ]

Row reduced echelon form of the transposed matrix =

[ 1 0 -1 ]
[ 0 1 0 ]
[ 0 0 0 ]

Im(fn) = span { (1,0,-1) }

Dimension of the ker(fn) and Im(fn) = 1

Because this is the only vector in the Im(fn) it also = the range(fn). I'm still a bit confused about what the range of a linear transformation is and how to find it.

I'm not really sure where to start with part b) of the question...
The only time the matrix will = 0 is when x,y,z = 0 (trivial solution) so I would say it is an isomorph, but I'm not totally sure.

[ 1 0 -1 ]
[ 1 1 -1 ]
[ -1 0 1 ]

geoffrey159
Call ##{\cal B} = \{ e_1, e_2, e_3 \}## the canonical base of ##\mathbb{R}^3##. Can you show that if ##h\notin \{-1,1\}##, then ##f_h({\cal B}) ## is linearly independent? Then why is ##f_h({\cal B})## is a base of ##\mathbb{R}^3##? Why does it prove that ##f_h## is an automorphism of ##\mathbb{R}^3## ?

If ##h\in\{-1,1\}##, then ##f_h({\cal B})## is linearly dependant but generates ##\text{Im}(f_h)##. Can't you easily find a free subfamily that generates ##\text{Im}(f_h)##? What does it tell you about the rank of ##f_h## ?

says
e1 e2 e3
[ 1 0 -h ]
[ 1 1 -h ]
[ -h 0 1 ]

iff h = 2

[ 1 0 -2 ]
[ 1 1 -2 ]
[ -2 0 1 ]

fh(B) is linearly independent if we row reduce the matrix above.

The Im(Fh) that I calculated above was linearly dependent.

Rank of fh is 1. for Im(fh) rank = 2

Can't you easily find a free subfamily that generates ##\text{Im}(f_h)##? What does it tell you about the rank of ##f_h## ?[/QUOTE said:
I'm not sure I understand what you're asking here, sorry.

geoffrey159
Unless I'm wrong, ##f_h(e_1) = (1,1,-h)##, ##f_h(e_2) = (0,1,0)##, and ##f_h(e_3) = (-h,-h,1)##.
First of all, at what condition ##f_h(e_1),f_h(e_2)## and ##f_h(e_3)## are linearly independant ? Unless you prove me wrong, this condition is ##h\notin \{ -1,1 \}##. Do we agree at that point ?

says
They are linearly independent if h = 2

geoffrey159
Ok, but they are linearly independent in many more cases

geoffrey159
Ok let's say we agree. The important result here, and it must be somewhere in your algebra notes in a more general form is :

##f_h## is bijective if and only ##\{ f(e_1), f(e_2), f(e_3) \}## is a base of ##\mathbb{R}^3##.

So that for ##h\notin \{-1,1\}##, your function is bijective, so it's kernel is ##\{0\}##, it's image is ##\mathbb{R}^3##. Bases and dimensions for the image and the kernel are very easy.

The less easy case is when ##h\in\{-1,1\}##. Here ##f_h## is not bijective. ##( f(e_1), f(e_2), f(e_3) )## is a linearly dependent family that spans the image of ##f_h##. You can find a linearly independent subfamily that also spans the image of ##f_h##, that is to say a base of ##\text{Im}(f_h)##.

says
revisiting an old post. Hope this is ok...

ker (fh) = span { ( 1,0,-1) , (-1,0,1) } dimension = 2
iff h=1
[ 1 0 -1 ]
[ 1 1 -1 ]
[ -1 0 1 ]

Transpose the matrix (above) and row reduce. We get:

[ 1 0 -1]
[ 0 1 0 ]
[ 0 0 0 ]

Im ((fh) = span { ( 1,0,-1) } dimension = 1

Homework Helper
Dearly Missed

## Homework Statement

Given the transformation fh : R 3 → R 3 defined by fh(x, y, z) = (x−hz, x+y −hz, −hx+z), where h ∈ R is a parameter.
a) Find, for all possible values of h, Ker(fh), Im(fh), their bases and dimensions.
b) Is fh an isomorphism for some value of h?

Ax=o

## The Attempt at a Solution

[ x−hz ]
[ x+y −hz ]
[ −hx+z ]

Matrix associated to the linear transformation =

[ 1 0 -h ]
[ 1 1 -h ]
[ -h 0 1 ]

Why bother with matrices and echelon forms, etc.? The kernel is just the set
$$\text{Ker}(fh) = \{(x,y,z): x-hz=0,x+y-hz=0, z-hx=0 \},$$
which can easily be solved/analyzed without using any matrix tools (which actually add nothing to the analysis). The image is
$$\text{Im}(fh) = \{(p,q,r): x-hz=p, x+y-hz=q, z-hx=r \:\text{for some} \: x,y,z \}.$$
Again, solving by high-school algebra is as fast a way as any.

Once you have solved both these questions the answers to the others drop out almost immediately.

Your conclusions about the value of h, etc., are ill-considered. For instance, in your first analysis you conclude that h = 1. Well, who says you need h = 1? What happens if you do not have h = 1? Nobody forces you to have h = 1 in the original transformation!

says
No one forces me to have x=1 either though do they? That's where I get confused. If one variable can take on any value then why can't another?