Find the Eigenvalues of the matrix and a corresponding eigenvalue

[suspenc3]

Find the Eigenvalues of the matrix and a corresponding eigenvalue. Check that the eigenvectors associated with the distinct eigenvalues are orthogonal. Find an orthogonal matrix that diagonalizes the matrix.

(1)[tex]\left(\begin{array}{cc}4&-2\\-2&1\end{array}\right)[/tex]

I found my eigenvalues to be 5 & 0, and the corresponding eigenvectors to be

[tex]\left(\begin{array}{cc}-2\\1\end{array}\right)[/tex]

and

[tex]\left(\begin{array}{cc}1/2\\1\end{array}\right)[/tex]

The book doesn't really explain this section well, can someone help me out with what to do next?

Also, how do you know if the eigenvectors produce an orthogonal set of vectors?what is orthonormal?

Thanks

 

[courtrigrad]

you are correct in the eigenvectorsWhat is the definition of orthogonal? An [itex] n\times n [/itex] invertible matrix [tex] Q [/tex] is called orthogonal, iff [tex] Q^{-1} = Q^{T} [/tex]

In other words, [tex] QQ^{T} = I [/tex]

An orthonormal set of vectors is basically a set of mutually orthogonal vectors such that: (1) [tex] u_{i}\cdot u_{i} = 1 [/tex] and (2) [tex] u_{i}\cdot u_{j} = 0 [/tex]

To see if the eigenvectors produce an orthogonal set of vectors, look at the definiton of orthogonal again.

What is a diagonalizable matrix?

 

[suspenc3]

A matrix is diagonalizable if there exists an n X n non-singular matrix P such that [tex]P^-^1AP[/tex] is a diagonal matrix.

If I take [tex]AA^t[/tex] I should get the Identity shouldn't I?I don't think it works..but it is a symmetric matrix..so does that mean that there is always an orthogonal matrix that diagonalizes A?

 

[HallsofIvy]

No, you shouldn't. Why would you think that? As courtigrad said, that's the definition of "orthogonal matrix" and you are not told that A is orthogonal. Yes, a matrix A is diagonalizable if there exist a matrix P such that PAP^-1 is diagonal. You are asked to find an orthogonal matrix that "diagonalizes" A. It is P you want to be orthogonal, not A.

The eigenvectors you got, (-2, 1) and (1/2, 1) are already orthogonal to one another: [itex](-2, 1)\cdot(1/2,1)= (-2)(1/2)+ 1(1)= 0][/itex]. Now can you find vectors of length 1 in the same direction? Those will give you the columns of an orthognal matrix.

 

[suspenc3]

Ohh, alright

Thanks