Linear Algebra: show it is diagonalizable and find eigenbasis

In summary, we have shown that J is diagonalizable with eigenvalues \pi, 2i, and -2i and corresponding eigenvectors (1, 0, 0), (0, 1, i), and (0, 1, -i) respectively. Therefore, the eigenbasis for J is {(1, 0, 0), (0, 1, i), (0, 1, -i)}.
  • #1
braindead101
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Let T1 be the C-vector space with basis B = (1, cosx, sinx). Define J: T1->T1 by (Jf)(x) = integ(0->pi) f(x-t)dt. Show that J is diagonalizable and find an eigenbasis.

J(1) = integ(0->pi) 1 dt
J(1) = t | (0->pi)
J(1) = pi

J(cos(x)) = integ(0->pi) cos(x-t) dt
J(cos(x)) = - sin (x-t) | (0->pi)
J(cos(x)) = - sin (x-pi) + sin(x)
J(cos(x)) = - (sin(x) cos(pi) - cos(x) sin (pi)) + sin(x)
J(cos(x)) = - (-sin(x)) + sin(x)
J(cos(x)) = 2sin(x)

J(sin(x)) = integ(0->pi) sin(x-t) dt
J(sin(x)) = cos (x-t) | (0->pi)
J(sin(x)) = cos (x-pi) - cos (x)
J(sin(x)) = cos(x) cos (pi) + sin(x) sin(pi) - cos (x)
J(sin(x)) = -2cos(x)

[J]T1 = pi 0 0
0 0 -2
0 2 0

v = lambda
So, det [J-vI] = (pi-v)(v^2 +4)
So, pi - v = 0 or v^2 +4 = 0
v = pi, v = 2i v = -2i

how do i go about looking for eigenbasis?
 
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  • #2
Apply the definition of "eigenvector": any eigenvector, v, of A, corresponding to eigenvalue [itex]\lambda[/itex] satisfies [itex]Av= \lambda v[/itex].
For eigenvector [itex]\lambda= \pi[/itex]
[tex]\left(\begin{array}{ccc} \pi & 0 & 0 \\ 0 & 0 & 2 \\ 0 & -2 & 0\end{array}\right)\left(\begin{array}{c} x \\ y \\ z \end{array}\right)= \left(\begin{array}{c} \pi x \\ \pi y \\ \pi z\end{array}\right)[/tex]
Which gives the equations [itex]\pi x= \pi x[/itex], [itex]2z= \pi y[/itex], and [itex]-2y= \pi z[/itex]. The first is true for any x, the last two are only true if y= z= 0. Any eigenvector corresponding to [itex]\lambda= 1[/itex] is of the form (x, 0, 0) and the eigenspace is spanned by (1, 0, 0).

For eigenvalue 2i,
[tex]\left(\begin{array}{ccc} \pi & 0 & 0 \\ 0 & 0 & 2 \\ 0 & -2 & 0\end{array}\right)\left(\begin{array}{c} x \\ y \\ z \end{array}\right)= \left(\begin{array}{c} 2i x \\ 2i y \\ 2i z\end{array}\right)[/tex]
which gives us equations [itex]\pi x= 2i x[/itex], 2z= 2i y, and -2y= 2i z. The first equation is true only for x= 0 and the last two are true as long as z= i y. The eigenspace is any vector of the form (0, y, iy) and is spanned by (0, 1, i).

I'll leave [itex]\lambda= -2i[/itex] to you.
 

Related to Linear Algebra: show it is diagonalizable and find eigenbasis

1. What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of linear equations and their properties. It involves the use of matrices, vectors, and other mathematical concepts to solve problems related to systems of linear equations.

2. How do you show that a matrix is diagonalizable?

To show that a matrix is diagonalizable, we need to find a basis of eigenvectors for the matrix. This can be done by finding the eigenvalues of the matrix and then finding the corresponding eigenvectors. If the matrix has n distinct eigenvalues, then it is diagonalizable.

3. What is an eigenbasis?

An eigenbasis is a set of linearly independent eigenvectors that span the vector space. This means that any vector in the space can be expressed as a linear combination of the eigenvectors in the eigenbasis.

4. How do you find the eigenbasis of a matrix?

To find the eigenbasis of a matrix, we first need to find the eigenvalues of the matrix. Then, for each eigenvalue, we find the corresponding eigenvector. These eigenvectors will form the eigenbasis for the matrix.

5. What is the significance of finding the eigenbasis of a matrix?

Finding the eigenbasis of a matrix is significant because it allows us to simplify the matrix and make it easier to work with. It also helps us understand the behavior of the matrix and how it affects the vectors in the space. Additionally, it is useful in solving systems of linear equations and in applications such as computer graphics and data analysis.

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