Matrix of linear transformation

AI Thread Summary
The discussion centers on finding the matrix representation of a linear transformation L that maps from R^3 to R^2, given specific transformation results for three vectors. It is clarified that the provided vectors are not linearly independent, which complicates directly forming the transformation matrix. To derive the matrix, a system of equations based on the properties of linear transformations must be solved, leading to the formation of a coefficient matrix M. The inverse of this matrix is used to compute the elements of the transformation matrix L, ultimately resulting in L being represented as a specific 2x3 matrix. The conversation emphasizes the importance of linear independence and the correct application of linear transformation properties in matrix formation.
Mathman23
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Hi

I got a question regarding the matrix of linear transformation.

A linear transformation L which maps \mathbb{R}^{3} \rightarrow \mathbb{R}^2 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?

<br /> \left[\begin{array}{ccc}<br /> 0 &amp; 1 &amp; 4\\<br /> 0 &amp; 3 &amp; 9\\<br /> \end{array}\right]<br />


Sincerely
Fred
 
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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 \mathbb{R}^{3}).

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.
 
Hi and thanks for Your answer.

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

L(u+v) = L(u) + L(v) \ \ \mathrm{and} \ \ \mathrm{L(\alpha u)} = \alpha \mathrm{L(u)}

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 ??
<br /> \mathrm{L} \left(\begin{array}{ccc}<br /> 2\\<br /> -1\\<br /> -1\\<br /> \end{array}\right) +<br /> \mathrm{L} \left(\begin{array}{ccc}<br /> -1\\<br /> 2\\<br /> 1\\<br /> \end{array}\right)<br /> = <br /> \left(\begin{array}{cc}<br /> 1\\<br /> 3<br /> \end{array}\right)<br />

<br /> \mathrm{L} \left(\begin{array}{ccc}<br /> -1\\<br /> 2\\<br /> 1\\<br /> \end{array}\right) +<br /> \mathrm{L} \left(\begin{array}{ccc}<br /> 2\\<br /> 2\\<br /> 1\\<br /> \end{array}\right)<br /> = <br /> \left(\begin{array}{cc}<br /> 5\\<br /> 12<br /> \end{array}\right)<br />

The matrix A of linear transformation then being:

<br /> A= \left [\begin{array}{cc}<br /> 1 &amp; 5 \\<br /> 3 &amp; 12<br /> \end{array}\right]<br />
 
Last edited:
Mathman23 said:
Hi

I got a question regarding the matrix of linear transformation.

A linear transformation L which maps \mathbb{R}^{3} \rightarrow \mathbb{R}^2 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?

<br /> \left[\begin{array}{ccc}<br /> 0 &amp; 1 &amp; 4\\<br /> 0 &amp; 3 &amp; 9\\<br /> \end{array}\right]<br />


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:

1: \ \ \ \ <br /> \begin{array}{rrrrr}<br /> \hline<br /> (2)L_{1,1} \ + &amp; (-1)L_{1,2} \ + &amp; (-1)L_{1,3} &amp; = &amp; 0 \\<br /> (2)L_{2,1} \ + &amp; (-1)L_{2,2} \ + &amp; (-1)L_{2,3} &amp; = &amp; 0 \\<br /> \hline<br /> (-1)L_{1,1} \ + &amp; (2)L_{1,2} \ + &amp; (1)L_{1,3} &amp; = &amp; 1 \\<br /> (-1)L_{2,1} \ + &amp; (2)L_{2,2} \ + &amp; (1)L_{2,3} &amp; = &amp; 3 \\<br /> \hline<br /> (2)L_{1,1} \ + &amp; (2)L_{1,2} \ + &amp; (1)L_{1,3} &amp; = &amp; 4 \\<br /> (2)L_{2,1} \ + &amp; (2)L_{2,2} \ + &amp; (1)L_{2,3} &amp; = &amp; 9 \\<br /> \hline<br /> \end{array}<br />

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

2: \ \ \ \ \ \ \mathsf{L} \ \ = \ \ \left [<br /> \begin{array}{ccc}<br /> L_{1,1} &amp; L_{1,2} &amp; L_{1,3} \\<br /> L_{2,1} &amp; L_{2,2} &amp; L_{2,3} \\<br /> \end{array} \right ] <br />

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":

3: \ \ \ \ \ M \ \ = \ \ \left [<br /> \begin{array}{rrrrrr}<br /> 2 &amp; -1 &amp; -1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 2 &amp; -1 &amp; -1 \\<br /> -1 &amp; 2 &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; -1 &amp; 2 &amp; 1 \\<br /> 2 &amp; 2 &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 2 &amp; 2 &amp; 1 \\<br /> \end{array} \right ] <br />

such that:

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

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:
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:

1: \ \ \ \ <br /> \begin{array}{rrrrr}<br /> (2)L_{1,1} \ + &amp; (-1)L_{1,2} \ + &amp; (-1)L_{1,3} &amp; = &amp; 0 \\<br /> (2)L_{2,1} \ + &amp; (-1)L_{2,2} \ + &amp; (-1)L_{2,3} &amp; = &amp; 0 \\<br /> (-1)L_{1,1} \ + &amp; (2)L_{1,2} \ + &amp; (1)L_{1,3} &amp; = &amp; 1 \\<br /> (-1)L_{2,1} \ + &amp; (2)L_{2,2} \ + &amp; (1)L_{2,3} &amp; = &amp; 3 \\<br /> (2)L_{1,1} \ + &amp; (2)L_{1,2} \ + &amp; (1)L_{1,3} &amp; = &amp; 4 \\<br /> (2)L_{2,1} \ + &amp; (2)L_{2,2} \ + &amp; (1)L_{2,3} &amp; = &amp; 9 \\<br /> \end{array}<br />

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

2: \ \ \ \ \mathsf{L} \ \ = \ \ \left [<br /> \begin{array}{ccc}<br /> L_{1,1} &amp; L_{1,2} &amp; L_{1,3} \\<br /> L_{2,1} &amp; L_{2,2} &amp; L_{2,3} \\<br /> \end{array} \right ] <br />

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

3: \ \ \ \ M \ \ = \ \ \left [<br /> \begin{array}{rrrrrr}<br /> 2 &amp; -1 &amp; -1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 2 &amp; -1 &amp; -1 \\<br /> -1 &amp; 2 &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; -1 &amp; 2 &amp; 1 \\<br /> 2 &amp; 2 &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 2 &amp; 2 &amp; 1 \\<br /> \end{array} \right ] <br />

such that:

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

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 \mathbb{R}^2 ??

Sincerely Fred
p.s. Again many thanks for Your answer.
 
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 \mathbb{R}^2 ??

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:

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

so that elements Lj,k are determined from:

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

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

7: \ \ \ \ \ \ \mathsf{L} \ \ = \ \ \left [<br /> \begin{array}{ccc}<br /> L_{1,1} &amp; L_{1,2} &amp; L_{1,3} \\<br /> L_{2,1} &amp; L_{2,2} &amp; L_{2,3} \\<br /> \end{array} \right ] <br />

Or in this case:

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


~~
 
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