Matrix of linear transformation

In summary, the matrix of linear transformation in this conversation is given by the following equation:\mathsf{L}= \left[\begin{array}{rrr}1 & 0 & 2\\ 2 & 1 & 3\end{array}\right]
  • #1
Mathman23
254
0
Hi

I got a question regarding the matrix of linear transformation.

A linear transformation L which maps [itex]\mathbb{R}^{3} \rightarrow \mathbb{R}^2[/itex] implies that L(2,-1,-1) = (0,0) and L(-1,2,1) = (1,3) and L(2,2,1) = (4,9).

My question is: The matrix of linear transformation is that then?

[itex]
\left[\begin{array}{ccc}
0 & 1 & 4\\
0 & 3 & 9\\
\end{array}\right]
[/itex]


Sincerely
Fred
 
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  • #2
Well, you may to remember that the matrix of a homomorphism or linear application, or transformation between to vectorial spaces is defined by the results of applying trasformation to basis vectors of any basis of the origin space (in this case a basis of [tex]\mathbb{R}^{3}[/tex]).

This is not acompplished by the vectors that you give (they are NOT lineary independent). The most simple form to obtain a matrix of the transformation is using the property of linearity, it let you to make some linear combinations between the functions that you are given to get the vectors of the canonical basis and their transforms under the linear transformation.
 
  • #3
Hi and thanks for Your answer.

I know that the definition of linear transformation[itex]\mathrm{L}:\mathbb{R}^m \rightarrow \mathbb{R}^n[/itex] is a follows.

[itex]L(u+v) = L(u) + L(v) \ \ \mathrm{and} \ \ \mathrm{L(\alpha u)} = \alpha \mathrm{L(u)}[/itex]

Do I then apply this definition to the given pre-conditions in my first post?

In order to obtain the matrix of linear transformation?

Sincerely

Fred

p.s.

Is this matrix of linear transformation then ??
[itex]
\mathrm{L} \left(\begin{array}{ccc}
2\\
-1\\
-1\\
\end{array}\right) +
\mathrm{L} \left(\begin{array}{ccc}
-1\\
2\\
1\\
\end{array}\right)
=
\left(\begin{array}{cc}
1\\
3
\end{array}\right)
[/itex]

[itex]
\mathrm{L} \left(\begin{array}{ccc}
-1\\
2\\
1\\
\end{array}\right) +
\mathrm{L} \left(\begin{array}{ccc}
2\\
2\\
1\\
\end{array}\right)
=
\left(\begin{array}{cc}
5\\
12
\end{array}\right)
[/itex]

The matrix A of linear transformation then being:

[itex]
A= \left [\begin{array}{cc}
1 & 5 \\
3 & 12
\end{array}\right]
[/itex]
 
Last edited:
  • #4
Mathman23 said:
Hi

I got a question regarding the matrix of linear transformation.

A linear transformation L which maps [itex]\mathbb{R}^{3} \rightarrow \mathbb{R}^2[/itex] implies that L(2,-1,-1) = (0,0) and L(-1,2,1) = (1,3) and L(2,2,1) = (4,9).

My question is: The matrix of linear transformation is that then?

[itex]
\left[\begin{array}{ccc}
0 & 1 & 4\\
0 & 3 & 9\\
\end{array}\right]
[/itex]


Sincerely
Fred
SOLUTION HINTS:
To obtain the Linear Transformation "L" for which:
a) L(2,-1,-1) = (0,0)
b) L(-1,2,1) = (1,3)
c) L(2,2,1) = (4,9)
the following simultaneous system of 6 equations in 6 unknowns (grouped to correspond to conditions "a", "b", and "c" above) must be solved:

[tex] 1: \ \ \ \
\begin{array}{rrrrr}
\hline
(2)L_{1,1} \ + & (-1)L_{1,2} \ + & (-1)L_{1,3} & = & 0 \\
(2)L_{2,1} \ + & (-1)L_{2,2} \ + & (-1)L_{2,3} & = & 0 \\
\hline
(-1)L_{1,1} \ + & (2)L_{1,2} \ + & (1)L_{1,3} & = & 1 \\
(-1)L_{2,1} \ + & (2)L_{2,2} \ + & (1)L_{2,3} & = & 3 \\
\hline
(2)L_{1,1} \ + & (2)L_{1,2} \ + & (1)L_{1,3} & = & 4 \\
(2)L_{2,1} \ + & (2)L_{2,2} \ + & (1)L_{2,3} & = & 9 \\
\hline
\end{array}
[/tex]

The Linear Transformation "L" would then be represented by the following matrix:

[tex] 2: \ \ \ \ \ \ \mathsf{L} \ \ = \ \ \left [
\begin{array}{ccc}
L_{1,1} & L_{1,2} & L_{1,3} \\
L_{2,1} & L_{2,2} & L_{2,3} \\
\end{array} \right ]
[/tex]

Equation System #1 above consists of 6 equations in 6 unknowns. The first step towards its solution is formation of the 6x6 coefficient matrix "M":

[tex] 3: \ \ \ \ \ M \ \ = \ \ \left [
\begin{array}{rrrrrr}
2 & -1 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 & -1 & -1 \\
-1 & 2 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 2 & 1 \\
2 & 2 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 & 2 & 1 \\
\end{array} \right ]
[/tex]

such that:

[tex] 4: \ \ \ \ \ \ \ \ \left [
\begin{array}{rrrrrr}
2 & -1 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 & -1 & -1 \\
-1 & 2 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 2 & 1 \\
2 & 2 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 & 2 & 1 \\
\end{array} \right ]
\left [
\begin{array}{c}
L_{1,1} \\
L_{1,2} \\
L_{1,3} \\
L_{2,1} \\
L_{2,2} \\
L_{2,3} \\
\end{array} \right ] \ \ = \ \ \left [
\begin{array}{c}
0 \\
0 \\
1 \\
3 \\
4 \\
9 \\
\end{array} \right ]
[/tex]

Solve for "L" by determining M(-1).
(Hint #1: The latter inverse exists if det(M) ≠ 0).
(Hint #2: det(M) = -9)
(Hint #3: L = [ 1 0 2 ; 2 1 3 ] )



~~
 
Last edited:
  • #5
xanthym said:
SOLUTION HINTS:
To obtain the Linear Transformation "L" for which:
a) L(2,-1,-1) = (0,0)
b) L(-1,2,1) = (1,3)
c) L(2,2,1) = (4,9)
the following simultaneous system must be solved:

[tex] 1: \ \ \ \
\begin{array}{rrrrr}
(2)L_{1,1} \ + & (-1)L_{1,2} \ + & (-1)L_{1,3} & = & 0 \\
(2)L_{2,1} \ + & (-1)L_{2,2} \ + & (-1)L_{2,3} & = & 0 \\
(-1)L_{1,1} \ + & (2)L_{1,2} \ + & (1)L_{1,3} & = & 1 \\
(-1)L_{2,1} \ + & (2)L_{2,2} \ + & (1)L_{2,3} & = & 3 \\
(2)L_{1,1} \ + & (2)L_{1,2} \ + & (1)L_{1,3} & = & 4 \\
(2)L_{2,1} \ + & (2)L_{2,2} \ + & (1)L_{2,3} & = & 9 \\
\end{array}
[/tex]

The Linear Transformation "L" would then be represented by the following matrix:

[tex] 2: \ \ \ \ \mathsf{L} \ \ = \ \ \left [
\begin{array}{ccc}
L_{1,1} & L_{1,2} & L_{1,3} \\
L_{2,1} & L_{2,2} & L_{2,3} \\
\end{array} \right ]
[/tex]

Equation System #1 above consists of 6 equations in 6 unknowns. The first step towards its solution is formation of the 6x6 coefficient matrix:

[tex] 3: \ \ \ \ M \ \ = \ \ \left [
\begin{array}{rrrrrr}
2 & -1 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 & -1 & -1 \\
-1 & 2 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 2 & 1 \\
2 & 2 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 & 2 & 1 \\
\end{array} \right ]
[/tex]

such that:

[tex] 4: \ \ \ \ \left [
\begin{array}{rrrrrr}
2 & -1 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 & -1 & -1 \\
-1 & 2 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 2 & 1 \\
2 & 2 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 & 2 & 1 \\
\end{array} \right ]
\left [
\begin{array}{c}
L_{1,1} \\
L_{1,2} \\
L_{1,3} \\
L_{2,1} \\
L_{2,2} \\
L_{2,3} \\
\end{array} \right ] \ \ = \ \ \left [
\begin{array}{c}
0 \\
0 \\
1 \\
3 \\
4 \\
9 \\
\end{array} \right ]
[/tex]

Solve for "L" by determining M(-1).
(Hint #1: The latter inverse exists if det(M) ≠ 0).
(Hint #2: det(M) = -9)
(Hint #3: L = [ 1 0 2 ; 2 1 3 ] )



~~


Hello and many thanks for your answer :smile:

Please correct me if I understand You incorrectly.

By determining the inverse matrix of M does that then give me the matrix of Linear transformation in [itex]\mathbb{R}^2[/itex] ??

Sincerely Fred
p.s. Again many thanks for Your answer.
 
  • #6
Mathman23 said:
Hello and many thanks for your answer :smile:

Please correct me if I understand You incorrectly.

By determining the inverse matrix of M does that then give me the matrix of Linear transformation in [itex]\mathbb{R}^2[/itex] ??

Sincerely Fred
p.s. Again many thanks for Your answer.
The inverse M(-1) of matrix "M" is NOT itself the matrix representing Linear Transformation "L". Rather, the inverse M(-1) enables computation of elements Lj,k of the matrix "L" representing Linear Transformation "L". This is accomplished by solving the matrix equation:

[tex] 5: \ \ \ \ M \cdot
\left [
\begin{array}{c}
L_{1,1} \\
L_{1,2} \\
L_{1,3} \\
L_{2,1} \\
L_{2,2} \\
L_{2,3} \\
\end{array} \right ] \ \ = \ \ \left [
\begin{array}{c}
0 \\
0 \\
1 \\
3 \\
4 \\
9 \\
\end{array} \right ]
[/tex]

so that elements Lj,k are determined from:

[tex] 6: \ \ \ \ \
\left [
\begin{array}{c}
L_{1,1} \\
L_{1,2} \\
L_{1,3} \\
L_{2,1} \\
L_{2,2} \\
L_{2,3} \\
\end{array} \right ] \ \ = \ \ M^{-1} \cdot \left [
\begin{array}{c}
0 \\
0 \\
1 \\
3 \\
4 \\
9 \\
\end{array} \right ]
[/tex]

The Linear Transformation matrix "L" is then given by:

[tex] 7: \ \ \ \ \ \ \mathsf{L} \ \ = \ \ \left [
\begin{array}{ccc}
L_{1,1} & L_{1,2} & L_{1,3} \\
L_{2,1} & L_{2,2} & L_{2,3} \\
\end{array} \right ]
[/tex]

Or in this case:

[tex] \color{red} 8: \ \ \ \ \ \mathsf{L} \ \ = \ \ \left [
\begin{array}{ccc}
1 & 0 & 2 \\
2 & 1 & 3 \\
\end{array} \right ]
[/tex]


~~
 
Last edited:

1. What is a matrix of linear transformation?

A matrix of linear transformation is a rectangular array of numbers that represents a linear transformation between two vector spaces. It is used to perform operations on vectors, such as rotation, scaling, and shearing.

2. How is a matrix of linear transformation used in computer graphics?

In computer graphics, a matrix of linear transformation is used to transform objects in a three-dimensional space. It is used to rotate, scale, and translate objects to create realistic 3D animations and graphics.

3. What is the difference between a matrix of linear transformation and a standard matrix?

A matrix of linear transformation is a specific type of matrix that represents a linear transformation between vector spaces. A standard matrix, on the other hand, can represent any type of mathematical operation, not just linear transformations.

4. How do you perform multiplication with a matrix of linear transformation?

To multiply a matrix of linear transformation by a vector, you first need to ensure that the number of columns in the matrix is equal to the number of rows in the vector. Then, you multiply each element of the vector by the corresponding elements in the matrix and add them together to get the resulting vector.

5. What are some real-world applications of matrix of linear transformation?

A matrix of linear transformation has many real-world applications, such as in computer graphics, robotics, and data analysis. It is also used in physics and engineering to model and analyze systems and processes. Additionally, it is used in machine learning and artificial intelligence algorithms to perform operations on large datasets.

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