# Linear Algebra - Change of Basis

• raay
In summary, the equation analogous to "T(t) = t + 1 = 1(1+t)+ 0(1-t) + 0(t^2)" for problem 3b is ##[S]_{C \to C} = [S]_{C \oplus C}##.
raay

Hi please i need help in number 3 of the tutorial questions. It is not an assignment its just a tutorial (read title in the image). I am currently studying for my final and i need help in (3b). the only way I am thinking of solving this questions is to use the equation given in part (d). But after reading the whole question i think i have to use a different method to solve part (b). Any ideas or hints ?

thanks

Can you explain in words what ##[ S]_{C \to C}## is?

Hi, if I understand, you are looking for the matrix of ##S## in the basis ##{\cal C}## at source and destination, are you ?

I have a review question for you that will probably answer your question too :
In general, what does it mean that ## A = (a_{i,j})_{1\le i \le n , 1 \le j \le p}## is the matrix of a linear map ##f## relative to a source basis ##{\cal B}## and destination basis ##{\cal C }## ?

vela said:
Can you explain in words what ##[ S]_{C \to C}## is?
I can do an example like 2c:
T(1) = 1 + 0 = 1 = 1/2(1+t) + 1/2(1- t) + 0(t^2) = (1/2 1/2 0)
T(t) = t + 1 = 1(1+t) 0(1-t) + 0(t^2) = (1 0 0 )
T(2) = t^2 + 2t = 1(1+t) + (-1)(1-t) + 1(t^2) = (1 -1 1)

so [T](B-->C) is equal to the matrix :
[1/2 1 1]
[1/2 0 -1]
[ 0 0 1]

geoffrey159 said:
Hi, if I understand, you are looking for the matrix of ##S## in the basis ##{\cal C}## at source and destination, are you ?

I have a review question for you that will probably answer your question too :
In general, what does it mean that ## A = (a_{i,j})_{1\le i \le n , 1 \le j \le p}## is the matrix of a linear map ##f## relative to a source basis ##{\cal B}## and destination basis ##{\cal C }## ?
an example of what i mean is :
I can do an example like 2c:
T(1) = 1 + 0 = 1 = 1/2(1+t) + 1/2(1- t) + 0(t^2) = (1/2 1/2 0)
T(t) = t + 1 = 1(1+t) 0(1-t) + 0(t^2) = (1 0 0 )
T(2) = t^2 + 2t = 1(1+t) + (-1)(1-t) + 1(t^2) = (1 -1 1)

so [T](B-->C) is equal to the matrix :
[1/2 1 1]
[1/2 0 -1]
[ 0 0 1]

raay said:
I can do an example like 2c:
T(1) = 1 + 0 = 1 = 1/2(1+t) + 1/2(1- t) + 0(t^2) = (1/2 1/2 0)
T(t) = t + 1 = 1(1+t) 0(1-t) + 0(t^2) = (1 0 0 )
T(2) = t^2 + 2t = 1(1+t) + (-1)(1-t) + 1(t^2) = (1 -1 1)

so [T](B-->C) is equal to the matrix :
[1/2 1 1]
[1/2 0 -1]
[ 0 0 1]
I didn't check all the details, but it looks like you're using the right method. So you just need to do the same in problem 3. (You seem to think that you need a different method when the two bases are the same, but the method you just used works for arbitrary bases).

Fredrik said:
I didn't check all the details, but it looks like you're using the right method. So you just need to do the same in problem 3. (You seem to think that you need a different method when the two bases are the same, but the method you just used works for arbitrary bases).

I don't know how to imply it. working with matrices confuses me :S

I don't understand what is the problem. If you have managed to do q1, q2, q3-a , what is different in q3-b ?
What about my question: what is the matrix of a linear map relative to two basis ?

I know that q3a is just the standard matrix. I am used to solving q3b by using the equation in part (d). but q3b is asking to solve by a method similar to the method in q2 like the example i have written. the thing is I am not used to solving it using matrices i know how to solve polynomials and integrals using the method in q2. for your question i really have no idea at the moment lol because I am studying calculus and my final is tmrw when i finish i will go through algebra again. I just need an explanation so when i reach this topic i can just move on.

You seem to be imagining a difficulty where there is none. The method that you used to solve problem 2c, also solves problems 3a to 3c.

The FAQ post may be useful:

Last edited by a moderator:
geoffrey159
raay said:
I can do an example like 2c:
T(1) = 1 + 0 = 1 = 1/2(1+t) + 1/2(1- t) + 0(t^2) = (1/2 1/2 0)
T(t) = t + 1 = 1(1+t) 0(1-t) + 0(t^2) = (1 0 0 )
T(2) = t^2 + 2t = 1(1+t) + (-1)(1-t) + 1(t^2) = (1 -1 1)

so [T](B-->C) is equal to the matrix :
[1/2 1 1]
[1/2 0 -1]
[ 0 0 1]
This isn't really an explanation. It seems, perhaps, that you understand the mechanics of solving some types of problems, but you get stuck because you don't completely understand conceptually what you're doing.

You wrote "T(t) = t + 1 = 1(1+t) + 0(1-t) + 0(t^2) = (1 0 0)." Express in words the answer to the following two questions:

1. Why did you calculate T(t)?
2. Why did you express the result t+1 as 1(1+t) + 0(1-t) + 0(t^2)?

What would be the equation analogous to "T(t) = t + 1 = 1(1+t)+ 0(1-t) + 0(t^2) = (1 0 0)" for problem 3b?

geoffrey159 and Fredrik
Fredrik said:
The FAQ post may be useful

I would say 'The FAQ post is your last chance'

## 1. What is the concept of "Change of Basis" in Linear Algebra?

The concept of "Change of Basis" in Linear Algebra refers to the process of representing a vector or a linear transformation in a different coordinate system. It involves finding a new basis for the vector space and expressing the vector or linear transformation in terms of the new basis vectors. This allows us to analyze and understand the vector or linear transformation in a different way, making it easier to solve certain problems.

## 2. How is "Change of Basis" related to matrix transformations?

"Change of Basis" is closely related to matrix transformations in Linear Algebra. When we change the basis of a vector space, the transformation matrix that represents the linear transformation also changes. This is because the new basis vectors have different coordinates compared to the original basis vectors, and this affects how the linear transformation is represented in matrix form.

## 3. Can "Change of Basis" be applied to any vector space?

Yes, "Change of Basis" can be applied to any vector space. This concept is not limited to a specific type of vector space or a specific dimension. It can be applied to both finite and infinite dimensional vector spaces, as long as the vector space has a well-defined basis.

## 4. What is the importance of "Change of Basis" in Linear Algebra?

"Change of Basis" is an important concept in Linear Algebra because it allows us to simplify and solve complex problems. By changing the basis, we can transform a problem into a more manageable form and gain a better understanding of the underlying structure. It also helps in finding alternative methods to solve a problem and can be applied in various fields such as physics, engineering, and data science.

## 5. How is "Change of Basis" used in real-world applications?

"Change of Basis" is used in many real-world applications, especially in fields that involve data analysis and manipulation. It is commonly used in computer graphics, where it helps in transforming and manipulating images. It is also used in machine learning algorithms, where changing the basis of data can improve the accuracy of predictions. In physics and engineering, "Change of Basis" is used to simplify and solve problems involving vectors and matrices.

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