Finding the matrix of a linear transformation from ℝ³ to ℝ²

[Mathman23]

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?

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


Sincerely
Fred

 

[elessar_telkontar]

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.

 

[Mathman23]

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

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

The matrix A of linear transformation then being:

$A= \left [\begin{array}{cc}<br /> 1 & 5 \\<br /> 3 & 12<br /> \end{array}\right]$

 

[xanthym]

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?

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


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:

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

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

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

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":

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

such that:

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

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 ] )



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[Mathman23]

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:

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

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

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

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

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

such that:

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

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 ] )



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Hello and many thanks for your answer

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.

 

[xanthym]

Hello and many thanks for your answer

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 L_j,k of the matrix "L" representing Linear Transformation "L". This is accomplished by solving the matrix equation:

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

so that elements L_j,k are determined from:

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

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

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

Or in this case:

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


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