Testing Change of Basis in Linear Operator

In summary, the conversation discusses the concept of change of basis in a linear operator. It explains the process of expressing a linear operator in a different basis and provides an example using a 2x2 matrix. It also shows the calculation for finding the new linear operator in the new basis and verifies its correctness.
  • #1
psholtz
136
0
I just want to test/verify my knowledge of change of basis in a linear operator.. (it's not a homework question).

Suppose I have linear operator mapping R^2 into R^2, and expressed in the canonical basis (1,0), (0,1). Suppose (for the sake of discussion) that the linear operator is given by:

[tex]A=\left(\begin{array}{ccc}5 & 6 \\ 7 & 8\end{array}\right) [/tex]

Suppose now that I want to express this linear operator in a different basis, say (2,1), (1,2). To do this, I first apply the linear operator to the two (original) basis vectors:

[tex]A \cdot \left(\begin{array}{c}1 \\ 0 \end{array}\right) = \left(\begin{array}{c}5 \\ 7\end{array}\right)[/tex]

[tex]A \cdot \left(\begin{array}{c}0 \\ 1 \end{array}\right) = \left(\begin{array}{c}6 \\ 8\end{array}\right)[/tex]

To find the linear operator A' in the new basis, we express these two vectors in terms of the new basis vectors. Solving the equation for x1, x2 gives:

[tex]\left(\begin{array}{cc}2 & 1 \\ 1 & 2 \end{array}\right) \left(\begin{array}{c}x_1 \\ x_2\end{array}\right) = \left(\begin{array}{c}5 \\ 7\end{array}\right)[/tex]

[tex]x_1 = 1, x_2=3[/tex]

and similarly:

[tex]\left(\begin{array}{cc}2 & 1 \\ 1 & 2 \end{array}\right) \left(\begin{array}{c}x_1 \\ x_2\end{array}\right) = \left(\begin{array}{c}6 \\ 8\end{array}\right)[/tex]

[tex]x_1 = 4/3, x_2=10/3[/tex]

So the expression for the linear operator in the new basis will be given by:

[tex]A' = \left(\begin{array}{cc}1 & 4/3 \\ 3 & 10/3\end{array}\right)[/tex]

Is this correct?
 
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  • #2
Let
[tex]
B = \left(\begin{array}{cc}2 & 1 \\ 1 & 2 \end{array}\right)
[/tex]

Now

[tex]
A' = B^{-1}AB = \frac{1}{3}\left(\begin{array}{cc}10 & 11 \\ 28 & 29 \end{array}\right)
[/tex]
 

Related to Testing Change of Basis in Linear Operator

What is a linear operator?

A linear operator is a mathematical function that maps one vector space to another, while preserving the basic algebraic structure of the vector space.

What is a basis?

A basis is a set of linearly independent vectors that can be used to represent all other vectors in a vector space through linear combinations.

Why is it important to test change of basis in linear operators?

Testing change of basis in linear operators is important because it allows us to understand how the operator behaves when represented in different coordinate systems. This is crucial for applications in physics, engineering, and other fields where different coordinate systems are used.

How do you test change of basis in linear operators?

The most common method for testing change of basis in linear operators is to use matrix representations. This involves representing the linear operator as a matrix in one basis, then transforming the basis and finding the matrix representation in the new basis. The two matrices can then be compared to test for consistency.

What are some potential sources of error when testing change of basis in linear operators?

Some potential sources of error when testing change of basis in linear operators include rounding errors in calculations, incorrect implementation of the transformation, and mistakes in the original matrix representation of the operator. It is important to carefully double-check all calculations and code to minimize these errors.

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