# Linear Algebra matrix linear transformation

• Rifscape
In summary: Yes, I did, I used T(1+ 0t+ 0t^2)= (-4(1)+ 2(0)+ 3(0))+ (2(1)+ 3(0)+ 3(0))t+ (-2(1)+ 4(0)+ 3(0))t^2= -4+ 2t- 2t^2. The first column of the matrix is \begin{pmatrix}-4 \\ 2 \\ -2\end{pmatrix}. That's what you wrote, but it's not correct. The first term is correct, but not the other two. ##T(1) = (-4+2) + (
Rifscape
Member warned about posting with no effort shown

## Homework Statement

Consider the linear transformation T from
V = P2
to
W = P2
given by
T(a0 + a1t + a2t2) = (−4a0 + 2a1 + 3a2) + (2a0 + 3a1 + 3a2)t + (−2a0 + 4a1 + 3a2)t^2
Let E = (e1, e2, e3) be the ordered basis in P2 given by
e1(t) = 1, e2(t) = t, e3(t) = t^2
Find the coordinate matrix [T]EE of T relative to the ordered basis E used in both V and W, that is, fill in the blanks below: (Any entry that is a fraction should be entered as a proper fraction, i.e. as either x/y or -x/y where x and y are positive integers with no factors in common.)
T(e1(t)) = _e1(t) + _ e2(t) + _e3(t)
T(e2(t)) = _e1(t) + _e2(t) + _e3(t)
T(e3(t)) = _e1(t) + _e2(t) + _e3(t)
and therefore :
[T]EE = the combined matrix of the coefficients of the above three equations.

Sorry for the poor formatting

No idea

## The Attempt at a Solution

I have no idea
I have no idea how to do this, if someone can give me a step by step solution so that I can understand each step and process I would appreciate it greatly.
Thanks for your help.

Rifscape said:

## Homework Statement

Consider the linear transformation T from
V = P2
to
W = P2
given by
T(a0 + a1t + a2t2) = (−4a0 + 2a1 + 3a2) + (2a0 + 3a1 + 3a2)t + (−2a0 + 4a1 + 3a2)t^2
You mean T(a0+ a1t+ a2t^2) on the left, right?

Let E = (e1, e2, e3) be the ordered basis in P2 given by
e1(t) = 1, e2(t) = t, e3(t) = t^2
Find the coordinate matrix [T]EE of T relative to the ordered basis E used in both V and W, that is, fill in the blanks below: (Any entry that is a fraction should be entered as a proper fraction, i.e. as either x/y or -x/y where x and y are positive integers with no factors in common.)
T(e1(t)) = _e1(t) + _ e2(t) + _e3(t)
T(e2(t)) = _e1(t) + _e2(t) + _e3(t)
T(e3(t)) = _e1(t) + _e2(t) + _e3(t)
and therefore :
[T]EE = the combined matrix of the coefficients of the above three equations.

Sorry for the poor formatting

No idea

## The Attempt at a Solution

I have no idea
I have no idea how to do this, if someone can give me a step by step solution so that I can understand each step and process I would appreciate it greatly.
Thanks for your help.
To find the matrix corresponding to a linear transformation in a given ordered basis, apply that linear transformation to each basis vector in turn and write the result as a linear combination of those basis vectors. The coefficients give a column of the basis. For example, to apply the linear transformation two the first basis vector, T(e1), note that e1= 1+ 0t+ 0t^2. That is, a0= 1, a1= 0, and a2= 0. Since T is defined by "T(a0 + a1t + a2t2) = (−4a0 + 2a1 + 3a2) + (2a0 + 3a1 + 3a2)t + (−2a0 + 4a1 + 3a2)t^2". T(1+ 0t+ 0t^2)= (-4(1)+ 2(0)+ 3(0))+ (2(1)+ 3(0)+ 3(0))t+ (-2(1)+ 4(0)+ 3(0))t^2= -4+ 2t- 2t^2. The first column of the matrix is $\begin{pmatrix}-4 \\ 2 \\ -2\end{pmatrix}$.

Now do the same to the second and third basis vectors.

T(f1(t)) = f1(t) + f2(t) + f3(t)
When I put in the first one which is f1(t) = 1, so it would be 1 + 0t + 0t^2. So when I plug it in, I should get (2(1) + 0 + 0) + (2(1) + 0 + 0)t + (-2(1) + 0 + 0)t^2
= (2,2,-2) However when I put this in it is wrong, does it have something to do with how f2(t) is now 1 + t, and f3(t) is now 1 + t + t^2?
T(f2(t)) = f1(t) + f2(t) + f3(t)
T(f3(t)) = f1(t) + f2(t) + f3(t)

Rifscape said:
T(f1(t)) = f1(t) + f2(t) + f3(t)
When I put in the first one which is f1(t) = 1, so it would be 1 + 0t + 0t^2. So when I plug it in, I should get (2(1) + 0 + 0) + (2(1) + 0 + 0)t + (-2(1) + 0 + 0)t^2
= (2,2,-2) However when I put this in it is wrong, does it have something to do with how f2(t) is now 1 + t, and f3(t) is now 1 + t + t^2?
T(f2(t)) = f1(t) + f2(t) + f3(t)
T(f3(t)) = f1(t) + f2(t) + f3(t)

So, you have ##T(e_1) = 2 e_1 + 2 e_2 - 2 e_3##. Why do you think this is wrong?

When I enter the numbers in it says it is wrong, except the - 2 at the end which is right. I feel like I am doing something wrong since f2(t) is now 1+t.

Rifscape said:
T(f1(t)) = f1(t) + f2(t) + f3(t)
When I put in the first one which is f1(t) = 1, so it would be 1 + 0t + 0t^2. So when I plug it in, I should get (2(1) + 0 + 0) + (2(1) + 0 + 0)t + (-2(1) + 0 + 0)t^2
= (2,2,-2) However when I put this in it is wrong, does it have something to do with how f2(t) is now 1 + t, and f3(t) is now 1 + t + t^2?
T(f2(t)) = f1(t) + f2(t) + f3(t)
T(f3(t)) = f1(t) + f2(t) + f3(t)
Where did f1, f2, and f3 come from? The same basis, i.e., ##e_1, e_2,## and ##e_3## is used in both V and W. In this problem V and W are the same spaces.
You aren't using the definition of the transformation, which I've copied below.
##T(a_0 + a_1t + a_2t^2) = (−4a_0 + 2a_1 + 3a_2) + (2a_0 + 3a_1 + 3a_2)t + (−2a_0 + 4a_1 + 3a_2)t^2##
Use this definition to see what T does to ##e_1## (HallsOfIvy already did this one), ##e_2##, and ##e_3##.

No sorry I didn't clarify. I finished the last question, this is a completely new question. I solved the previous question with hall of ivys help. However when I apply the same strategy to this question I cannot solve it. This question has f instead of e. Sorry for the confusion

Rifscape said:
No sorry I didn't clarify. I finished the last question, this is a completely new question.
I solved the previous question with hall of ivys help. However when I apply the same strategy to this question I cannot solve it. This question has f instead of e. Sorry for the confusion
Then you should start a new thread, with the complete problem statement of the new problem.

## 1. What is a matrix in linear algebra?

A matrix is a rectangular array of numbers, symbols, or expressions arranged in rows and columns. It is used to represent and manipulate linear transformations in linear algebra. Matrices are used to solve systems of linear equations, perform calculations in statistics, and represent geometric transformations.

## 2. What is a linear transformation in linear algebra?

A linear transformation is a function that maps one vector space to another, preserving the linear structure of the vector space. In simpler terms, it is a transformation that preserves straight lines and the origin. It can be represented by a matrix and can involve operations such as scaling, rotation, and reflection.

## 3. How do you perform matrix multiplication in linear algebra?

To perform matrix multiplication, the number of columns in the first matrix must match the number of rows in the second matrix. The product of two matrices is a new matrix where each element is calculated by multiplying the corresponding elements in the rows and columns of the two matrices and then adding them together.

## 4. How do you determine the inverse of a matrix in linear algebra?

The inverse of a matrix is a matrix that, when multiplied by the original matrix, results in the identity matrix. To find the inverse of a matrix, you can use the Gauss-Jordan elimination method or the adjugate matrix method. The inverse of a matrix is useful in solving systems of linear equations and performing other operations in linear algebra.

## 5. What is the difference between a row vector and a column vector in linear algebra?

A row vector is a matrix with a single row, while a column vector is a matrix with a single column. In linear algebra, row vectors are used to represent linear transformations that map from a one-dimensional vector space to a higher-dimensional vector space, while column vectors are used to represent linear transformations that map from a higher-dimensional vector space to a one-dimensional vector space.

### Similar threads

• Calculus and Beyond Homework Help
Replies
2
Views
928
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
9
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
3K
• Precalculus Mathematics Homework Help
Replies
3
Views
1K