Linear Functions dimensions and kernels

In summary: Maybe somebody can help me out here?In summary, if T:Mnxn ----> Mnxn then dim(T(Mnxn))+dim(ker(T))=dim(Mnxn)
  • #1
Badger33
36
0
If T:Mnxn ----> Mnxn then dim(T(Mnxn))+dim(ker(T))=dim(Mnxn)
I chose matrices because I thought they would be hardest and I am looking to understand concepts here. Suppose n=2. The dim(Mnxn)=4. Now I need to be able to find the other two values. I also know dim(T(Mnxn))=RANK, well at least I think this is correct. How do I find the rank and the ker. Then once I have the ker how do I find the dim of the ker?
I know this to be linear so let's use this for sake of example:
T:M2x2 ---> M2x2
[a b] |___\ [a a+b]
[c d] | / [c c+d]

Hopefully you can understand all my notation.


 
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  • #2
Choose a basis for M2x2 and then find the 4x4 matrix that represents T. Once you have the matrix, you do the usual things to find the rank and null space.
 
  • #3
So I can chose any values I want then for a,b,c,d
[1 2]
[2 3]
is a basis then? and then by computing the RREF of that I get
[1 0]
[0 1]
So that would mean the rank of the is 2 and then dim(T(Mnxn))=2 and then because 2+dim(ker(T))=4 The dim(ker(T))=2.
Is this all correct?
How can I find the ker(T)? Then once I have that how can I find the dim of that?
 
  • #4
Badger33 said:
So I can chose any values I want then for a,b,c,d
[1 2]
[2 3]
is a basis then? and then by computing the RREF of that I get
[1 0]
[0 1]
So that would mean the rank of the is 2 and then dim(T(Mnxn))=2 and then because 2+dim(ker(T))=4 The dim(ker(T))=2.
Is this all correct?
No, that's completely wrong.

As you noted M2x2 is a 4-dimensional vector space, so you should have 4 basis vectors. Each of these vectors is an element of M2x2, in other words, a 2x2 matrix. So find 4 matrices that span M2x2 and are linearly independent. You need to get this part straight before going on with the rest.
 
  • #5
Ok so I would need four matrices.
[1 0] [0 1] [0 0] [0 0]
[0 1],[1 0],[1 0],[0 1]
Would these be the correct four matrices needed? I guess I do not know how to find them. I have four spanning sets then in the original 2x2 matrix? I am very lost ... as you can see.
 
  • #6
Nope, 4x4 is correct because you're mapping from a 4-dimensional space to a 4-dimensional space.
 
  • #7
Nope, 4x4 is correct because you're mapping from a 4-dimensional space to a 4-dimensional space.

You lost me at mapping.
 
  • #8
vela said:
Nope, 4x4 is correct because you're mapping from a 4-dimensional space to a 4-dimensional space.

you lost me at mapping.
 
  • #9
Badger33 said:
Ok so I would need four matrices.
[1 0] [0 1] [0 0] [0 0]
[0 1],[1 0],[1 0],[0 1]
Would these be the correct four matrices needed? I guess I do not know how to find them. I have four spanning sets then in the original 2x2 matrix? I am very lost ... as you can see.
I think you have a typo in the second matrix, but yes, that's a perfectly fine basis because

[tex]\begin{pmatrix}a & b \\ c & d\end{pmatrix} = a\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix} + b\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} + c\begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix} + d\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}[/tex]

and it's clear they're linearly independent.

Now do you remember how to find the columns of the matrix representing T once you have a basis?
 
  • #10
Badger33 said:
you lost me at mapping.
That was a reply to sethric's post, though apparently he deleted his post while I was replying. You can ignore it.
 
  • #11
Ehh, I make mistakes when I'm tired. Realized my error essentially immediately. I preferred deletion to edit so as not to confuse the thread starter. Which you cancelled. :p
 
  • #12
vela said:
Now do you remember how to find the columns of the matrix representing T once you have a basis?

What exactly do you mean?
(Im a beginner)
And Yes it is very clear that they are lin indep
 
  • #13
I am not looking at
[a a+b]
[c c+d]
am I?
 
  • #14
Sethric said:
Ehh, I make mistakes when I'm tired. Realized my error essentially immediately. I preferred deletion to edit so as not to confuse the thread starter. Which you cancelled. :p
No worries. If I had noticed you had deleted your post quickly enough, I'd have deleted mine. :wink:
 
  • #15
Badger33 said:
I am not looking at
[a a+b]
[c c+d]
am I?
Not really.

How much linear algebra have you had so far? It might be better to start with a simpler example to make things less confusing if you're trying to nail down basic concepts.
 
  • #16
Yea sadly I am trying to nail the basic concept. I have had a full semesters worth and I have the exam tomorrow so I am just reviewing and this is one of the things I am having a very hard time grasping. I get eigen stuff and also Rank. I am just not getting stuff with dim and ker. I just didn't have them explained well by my professor in class or one on one and the book we use doesn't give enough examples if any on the subject and that is not helping me out.
I have to get some sleep now so that I can function for the exam...but because it is mostly concepts(I think) that I need help with please post as much about the concepts so that you think a novice(AKA me) will really grasp it. I will be back in about 6hrs and reading the posts and hopefully not needing to add anymore to clear anything up. Thanks for the help so far and also in advance.
 
  • #17
Also as you can probably tell this is my first introduction to proofs of this kind so I am trying hard to grasp how these things work.
 

FAQ: Linear Functions dimensions and kernels

What is a linear function?

A linear function is a mathematical function that can be written in the form f(x) = mx + b, where m and b are constants. It is called "linear" because when graphed, it forms a straight line.

How do you determine the dimensions of a linear function?

The dimensions of a linear function can be determined by looking at the number of variables in the function. For example, a function with one variable (x) has one dimension, while a function with two variables (x and y) has two dimensions.

What is the kernel of a linear function?

The kernel of a linear function is the set of all inputs (or values of x) that produce an output of 0. In other words, it is the set of all solutions to the equation f(x) = 0. The kernel is also known as the null space.

How is the kernel related to the dimensions of a linear function?

The dimension of the kernel is related to the dimension of the linear function through the rank-nullity theorem. This theorem states that the dimension of the kernel plus the dimension of the range (or the number of output values) equals the dimension of the input space (or the number of variables).

How can linear functions be used in real-world applications?

Linear functions have many real-world applications, including in economics, physics, and engineering. They can be used to model relationships between variables, such as the relationship between income and expenses or the relationship between distance and time. Linear functions are also used in data analysis and machine learning to make predictions and solve problems.

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