# Matrix Ker(f) and Im(f)

1. Jan 18, 2016

### Kernul

1. The problem statement, all variables and given/known data
Being f : ℝ4 → ℝ4 the endomorphism defined by:
ƒ((x, y, z, t)) = (3x + 10z, 2y - 6z - 2t, 0, -y+3z+t)
Determine the base and dimension of Im(ƒ) and Ker(ƒ). Complete the base you chose in Im(ƒ) into a base of R4.

2. Relevant equations
Matrix A:
$$\begin {bmatrix} 3 & 0 & 10 & 0\\ 0 & 2 & -6 & -2\\ 0 & 0 & 0 & 0\\ 0 & 1 & 2 & 1 \end{bmatrix}$$

dim(Im(ƒ)) = rank(A)

dim(Ker(ƒ)) = number of columns - rank(A)

3. The attempt at a solution
So, first I have to know the rank of the matrix. The fourth row is actually the second row divided by -2. So:
A4 = (-1/2) * A2
At this point I know that the matrix has a rank of 3, but there is a null row. Does this mean that the rank is 2? When a row(or multiple row) is 0 the rank is the number of the remaining non-null rows?

This is no the only question so I'll continue by assuming it's a rank of 2, so:
rank(A) = 2 = dim(Im(ƒ))
At this point while I look at other exercises done by my professor I see that she write like this for the base of Im(ƒ):
BIm(ƒ) = [A1, A2, A3, ...] (This if it was my exercise: BIm(ƒ) = [A1, A2])
I want to know why this is the base because I really don't understand.

At this point we have to determine the dimension and the base of Ker(ƒ).
The dimension would be:
dim(Ker(ƒ)) = 4 - 2 = 2
To determine the base of Ker(ƒ), from what I understood, you have to take the rows and set 2(because the dimension of Ker(ƒ) is 2) of x, y, z and t, equal to a scalar(a and b), put them into a system and then solve it until you get the two columns of the base of Ker(ƒ). So it would be something like this:
{3x +10z = 0
{2y - 6z - 2t = 0
{z = b
{t = a
Solving it we have:
{x = (-10/3)b
{y= 3a + b
{z = b
{t = a
And this:
Ker(ƒ) = {$$\begin {bmatrix} 0\\ 3\\ 0\\ 1 \end{bmatrix}$$ * a +
$$\begin {bmatrix} -10/3\\ 1\\ 1\\ 0 \end{bmatrix}$$ * b : a, b ∈ ℝ}
And the base of Ker(ƒ) would be:
BKer(ƒ) = [$$\begin {bmatrix} 0\\ 3\\ 0\\ 1 \end{bmatrix}$$,
$$\begin {bmatrix} -10/3\\ 1\\ 1\\ 0 \end{bmatrix}$$
But why? I don't understand why this it how to calculate the bases.
I'm so sorry for all the confusion about the matrix but I don't know how to make them better. If you know, could you please tell me?

Last edited: Jan 18, 2016
2. Jan 18, 2016

### LCKurtz

I haven't checked your details yet but your method looks OK. Here's how you make a matrix. When you reply you will see the text:
$$\begin {bmatrix} a & b & c & d\\ e & f & g & h\\ i & j & k & l \end{bmatrix}$$

3. Jan 18, 2016

### PeroK

Your approach in general seems to be to memorise a technique for something without understanding it. The alternative is to learn to think on your feet a little more.

For example, if you have two simple basis vectors, it's not too hard to find two more linearly independent vectors. That means understanding more about what sort of vectors are independent.

Why not use the work you did to find x and z to finish off finding a basis for ker(f)?

4. Jan 18, 2016

### Kernul

Wow! Thank you a lot! I'm going to modify it a bit.

Anyway, I know the method is okay. The problem is that I did it "mechanically". Meaning that I did it without knowing what I was doing. Could you please explain to me why the exercise is done that way?

5. Jan 18, 2016

### Kernul

Wait, what do you mean to find x and z to finish off finding a basis for Ker(ƒ)? Didn't I already find the base?

Last edited: Jan 18, 2016
6. Jan 18, 2016

### PeroK

Did you check that f maps those vectors to 0?

7. Jan 18, 2016

### Kernul

You mean this system?

8. Jan 18, 2016

### PeroK

You claim, if I am not mistaken, that

$f(0,3,0,1) = 0$ and $f(-10/3, 1,1,0) = 0$

Did you check?

9. Jan 18, 2016

### Kernul

Oh, I got what you mean.
Yeah, because the definition of Ker(ƒ) says that the set of Ker(f) = {v ∈ V : ƒ(v) = 0}

10. Jan 18, 2016

### Ray Vickson

Do you agree that the kernel $\text{Ker}(f)$ is the set of vectors of the form $(x,y,z,t) = \langle -(10/3) b, 3a+b,b,a \rangle$? Can you not write any $\vec{v} \in \text{Ker}(f)$ in the form $\vec{v} = a \vec{u}_1 + b \vec{u}_2$, where $\vec{u}_1 = \langle 0,3,0,1 \rangle$ and $\vec{u}_2 = \langle -10/3, 1,1,0 \rangle$?

11. Jan 18, 2016

### nuuskur

Is matrix A the matrix of f corresponding to some fixed basis?
Easiest way to solve the problem in question is to apply the Rank-Nullity Theorem:
$\dim (Im f)+\dim (Ker f) = dim (V)$, where V is our $\mathbb{R}^4$

To determine the rank of A we can also use another theorem that states the rank of any matrix is equal to the highest degree non-singular minor of A. For example, you check the entire matrix A, if it is non-singular, we're done, its rank is four. If it's singular start checking the 3x3 cases, if ALL of them are singular, move on to the 2x2 cases. Its rank can't be 0, because evidently A is not a zero matrix.

Last edited: Jan 18, 2016
12. Jan 20, 2016

### Kernul

With a and b being t and z? But why should I write that? Isn't the base of Ker(ƒ) made of $\vec{u}_1$ and $\vec{u}_2$? Isn't that all I need?

Yeah, but I have to find both dim(Im(ƒ)) and dim(Ker(ƒ)) so I can't actually use that right now. I had to use the equations dim(Im(ƒ) = rank(A) and dim(Ker(ƒ)) = n - dim(Im(ƒ)) (or rank(A) since they are the same thing)

To check if it is singular or not you calculate the determinant, right? Or am I confusing it with something else?

13. Jan 20, 2016

### Ray Vickson

14. Jan 20, 2016

### nuuskur

You can do one of two things.
Determine which kind of vectors get turned into the zero vector, from that you will get the basis of the kernel and consequently the dimension of kernel subspace. Rank-Nullity then gives you the dimension of the image.
OR
Determine the basis of the image subspace which will also give you the dimension of the kernel subspace.

If matrix A is the matrix of f corresponding to a basis, then the column vectors of A are the images of those basis vectors - in other words, the column vectors Generate the image subspace. Determine the rank of the matrix, reduce the system to a linearly independent system of generators - a basis in the image subspace. One can accomplish this by performing elementary row operations on matrix A - I also said you can apply a theorem which gives you a way to get the rank of the matrix, however, you will not be able to determine the basis vectors with that trick.

Any set of linearly independent vectors can be completed into a basis of the entire vector space. One foolproof way to do that is to apply the Gram-Schmidt's process of orthogonalization.

dim(Ker(ƒ)) = number of columns - rank(A)
So, if a matrix has 4 columns, its rank is 2 and the kernel of f consists of only the zero vector? Then 0 = 4-2? Essentially what you have done is apply the Rank-Nullity theorem, but your corollary is sloppy which will make you remember it the wrong way.

15. Jan 20, 2016

### Kernul

Oh yeah! I'm sorry! I misunderstood you, Ray.

I determine those vectors by doing operations between the rows of the matrix? And what if there is already a zero vector, like in this case?

What do you mean by that? How can the kernel of f be only the zero vector? And even if it was, wouldn't it mean that there is at least 1 vector, so 1 = 4 - 2? The zero vector does count as a vector of the kernel of f, right?

My corollary is sloppy? Please, can you tell me a way to make me remember it correctly?

16. Jan 21, 2016

### nuuskur

If f was one to one, the kernel could only consist of the zero vector. Your corollary might be sufficient to hold for this particular problem, but it May or May NOT hold for every case. Remember instead, what rank nullity theorem states: $\dim (Im f) + \dim (Ker f) = \dim (V)$. Convince yourself first that the image and kernel actually are vectorspaces, maybe they're not? How can we talk about dimensions, then?

Another important part is to show that this particular f is a linear transformation, only then would you be able to apply Rank Nullity.

Last edited: Jan 21, 2016
17. Jan 22, 2016

### Kernul

One to one? What do you mean? Could you write an example, please?

The image and kernel are vectorsubspaces, right?

Yeah, I know that the Ran-Nullity theorem says:
If ƒ : Kn → Kn is a linear transformation so that:
ƒ(X) = AX
Then dim(Ker(ƒ)) + dim(Im(ƒ)) = dim(C)

18. Jan 22, 2016

### nuuskur

We say a function is "one-to-one" or injective when
$\forall x,y\in V [f(x) = f(y)\Rightarrow x=y]$
A result in linear algebra about linear transformations
$[f(x) = f(y)\Rightarrow x=y] \Leftrightarrow Ker f = \{0\}$
---

Yes, image and kernel are indeed subspaces. Prove it :)

19. Jan 22, 2016

### Kernul

Oh! An injective function! I didn't understand at first.
The result in linear algebra, is because ∀ x,y ∈ Ker(ƒ) ⇒ ƒ(x) = 0 = ƒ(y). Right?

In order to prove it, first we have to say:
ƒ : V → W is a linear transformation.
We have to see if the addition and the product by a scalar is possible in the kernel, since it inherits the properties of the space in which it is(in this case V, since Ker(ƒ) = {v ∈ V : ƒ(v) = 0}).
Being v1,v2 ∈ Ker(ƒ) ⇒ ƒ(v1) = 0 = ƒ(v2)
ƒ(v1 + v2) = ƒ(v1) + ƒ(v2) = 0 + 0 = 0v1 + v2 ∈ Ker(ƒ)
Being v ∈ Ker(ƒ) and α ∈ K ⇒ ƒ(v) = 0
ƒ(αv) = α ƒ(v) = α0 = 0 ⇒ αv ∈ Ker(ƒ)
For the image we have to do the same thing basically, but with the difference that Im(ƒ) = {w ∈ W : ∃ v ∈ V so that ƒ(v) = w} = {ƒ(v) : v ∈ V}.
Being w1,w2 ∈ Im(ƒ) ⇒ w1 = ƒ(v1) and w2 = ƒ(v2) for certain vectors v1,v2 ∈ V
w1 + w2 = ƒ(v1) + ƒ(v2) = ƒ(v1 + v2) ⇒ w1,w2 ∈ Im(ƒ)
Being w ∈ Im(ƒ), w = ƒ(v) for some v ∈ V and α ∈ K
αw = α ƒ(v) = ƒ(αv) ⇒ αw ∈ Im(ƒ)

Am I right?

20. Jan 22, 2016

### nuuskur

Exactly so :) , you should practice writing in TeX format, it helps to convey your problems in a more concise manner.