How do I calculate the bases for Im(f) and Ker(f)?

[Kernul]

Homework Statement

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.

Homework 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)

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):
B_Im(ƒ) = [A¹, A², A³, ...] (This if it was my exercise: B_Im(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:
B_Ker(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?

 

[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:
$$\begin {bmatrix}
a & b & c & d\\
e & f & g & h\\
i & j & k & l
\end{bmatrix}$$

 

[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)?

 

[Kernul]

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:
$$\begin {bmatrix}
a & b & c & d\\
e & f & g & h\\
i & j & k & l
\end{bmatrix}$$

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?

 

[Kernul]

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)?

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?

 

[PeroK]

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

Did you check that f maps those vectors to 0?

 

[Kernul]

Did you check that f maps those vectors to 0?

You mean this system?

{3x +10z = 0
{2y - 6z - 2t = 0
{z = b
{t = a

 

[PeroK]

You mean this system?

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?

 

[Kernul]

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?

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}

 

[Ray Vickson]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

Ker:
{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

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 $\vec{u}_2 = \langle -10/3, 1,1,0 \rangle$?

 

[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.

 

[Kernul]

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 $\vec{u}_2 = \langle -10/3, 1,1,0 \rangle$?

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?

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.

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?

 

[Ray Vickson]

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?

***********************
Isn't that exactly what I said? Any vector in Ker(f) is a linear combination of u1 and u2; that is the definition of what it means for {u1, u2} to form a basis (along with the linear independence of u1 and u2, of course).

***********************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?

 

[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.

Some comments:
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.

 

[Kernul]

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

Determine which kind of vectors get turned into the zero vector, from that you will get the basis of the kernel

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?

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?

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?

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.

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

 

[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.

 

[Kernul]

If f was one to one, the kernel could only consist of the zero vector.

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

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(Imf)+dim(Kerf)=dim(V)dim⁡(Imf)+dim⁡(Kerf)=dim⁡(V)\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?

The image and kernel are vectorsubspaces, right?

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

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)

 

[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 :)

 

[Kernul]

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

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?

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

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 = 0 ⇒ v₁ + 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?

 

[nuuskur]

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

 

[Kernul]

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

Yeah, I'm sorry for all the confusion.
Anyway, there is one more thing that I am completely clueless about. The exercise, at the end, says:

Complete the base you chose in Im(ƒ) into a base of R⁴.

What does it mean? What do I have to do with Im(ƒ)?

 

[nuuskur]

First you will need to determine the dimension of the image. You need the image basis vectors which you will then complete into a basis of $\mathbb{R}^4$.
For example, assume you have two basis vectors in the image, you would need a total of four linearly independent vectors to span all of $\mathbb{R}^4$.

Recall what Rank-Nullity says: If you add the dimension of the image and the dimension of nullity, you will get the dimension of the entire space. That must mean if you take the union of the bases of image and nullity you will get the desired basis for $\mathbb{R}^4$. Convince yourself that the only common element in image and nullity is the zero vector, then the said method will work.

The problem does say, however: complete the basis you CHOSE in Im f. You don't have to work with the basis you determined at first. If you manipulate the basis with elementary row operations (in matrix form), it will be Equivalent to the one you started with.
Remember when you were solving system of linear equations and you put coefficients in a matrix form and started reducing into a triangle shape? It's almost the same scene, only now you are working backwards

Alternatively you can also determine the basis vectors of image, put them in columns in matrix form and generate ANY other two vectors such that the 4 x 4 matrix is non-singular - that will mean the column vectors are linearly independent and whammo, you have a basis completed.

 

[Kernul]

So I have to put both the vectors of the image and of the kernel, right? So, since I said that the basis of image is made of the first two columns and the basis of kernel is the one I calculated, I would have a base like this:
$$\begin{pmatrix}
3\\
0\\
0\\
0
\end{pmatrix}
\begin{pmatrix}
0\\
2\\
0\\
1
\end{pmatrix}\begin{pmatrix}
0\\
3\\
0\\
1
\end{pmatrix}\begin{pmatrix}
-10/3\\
1\\
1\\
0
\end{pmatrix}$$
Am I right? And if yes, I should write just this?

 

[nuuskur]

Calculate the determinant of this 4 x 4 matrix, if it's not zero then you are finished :)

 

[Kernul]

Calculate the determinant of this 4 x 4 matrix, if it's not zero then you are finished :)

The determinant is 3, if I'm not wrong.
Thank you for all! You really helped me understand!

 

[nuuskur]

Well done, young Padawan :)

Just kidding, good job, regardless. Practice that TeX!

 

[Kernul]

Hahahaha! Thanks, master. I'll practice the force of TeX.