# Linear Transformation and Isomorphism

1. Dec 9, 2015

### says

1. The problem statement, all variables and given/known data
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?
2. Relevant equations
Ax=o

3. 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 ]

2. Dec 9, 2015

### 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 $f_h$ ?

3. Dec 9, 2015

### 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

4. Dec 9, 2015

### 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 ?

5. Dec 9, 2015

### says

They are linearly independent if h = 2

6. Dec 9, 2015

### geoffrey159

Ok, but they are linearly independent in many more cases

7. Dec 9, 2015

### 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 $\text{Im}(f_h)$.

8. Dec 29, 2015

### 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

9. Dec 29, 2015

### Ray Vickson

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!

10. Dec 29, 2015

### 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?

11. Dec 29, 2015

### Ray Vickson

Well, if you have h = 1, then Ker(fh) and Im(fh) have a certain form. If you have h = -1, then Ker(fh) and Im(fh) have some other form. And, if you have h ≠ -1,1, still other forms of Ker(fh) and/or Im(fh) occur. Since h is allowed to be anything, you need to figure these out in order to be sure you have covered all possibilities, and to be able to describe what happens as h varies.

And, while it is true that when h = 1 nobody forces you to have some particular value for x (for example), the issue is that if your point (x,y,z) is in Ker(f1), then you are, indeed, forced to take certain values; if you fail to take those values your point fails to fall into the set Ker(f1).

Last edited: Dec 29, 2015