Eigenvalues/vector question

1. Apr 24, 2005

toffee

why are the eigenvectors in a square matrix which is diagonalised (only has numbers in the M_jj elements), just each column of the matrix? And why are the eigenvalues the actual number in each column.

I can understand if its diagonal, the matrix consists of linearly indepednent therefore orthogal vectors. Are these necessarily the eigenvectors of the matrix?

Many thanks

2. Apr 24, 2005

matt grime

Which of the many matrices you're alluding to but not mentioning are you talking about?

Let M be the original matrix, and let D be its diagonalized form, and let P and Q=P^{-1} be the matrix satisfying

M = PDQ, or D=QMP

When you refer to the rows/columns, you ought to be talking about those of Q/P, when it should be clear what's going on if you're ok with the notion of "change of basis".

3. Apr 24, 2005

Gokul43201

Staff Emeritus
It sounds like toffee is talking about the eigenvectors and eigenvalues of a diagonal matrix, D.

(that the eigenvalues are the diagonal elements themselves comes directly from the characteristic equation : $det|D - \lambda I| = 0 => \Pi _{i=1}^N (D_{ii} - \lambda) = 0$)

4. Apr 26, 2005

HallsofIvy

We write any linear transformation as a matrix by choosing a specific basis and seeing what the transformation does to the basis vectors. That is, if v1, v1, . . ., v1 is a basis, we can write any vector v= a1v1+ a1v1+ anv1 and write v as the "n-tuple" (a1, a2, . . ., an). Each basis vector is represented by (1, 0, ..., 0), (0, 1, . . ., 0), etc.
Applying the linear transformation to v1 is the same as multiplying the matrix times (1, 0, . . ., 0) which gives whatever the coefficients of that vector are.

Suppose T is a linear transformation, over vector space V, with eigenvalues &lambda1, &lambda2, . . ., &lambdan, having corresponding eigenvectors v1, v1, . . ., v1 which form a basis for V. Applying T to v1 ((represented as (1, 0, . . ., 0)) gives &lambda;1: that would be written (&lamba;1, 0, . . ., 0) so the first column is simply that: (&lamba;1, 0, . . ., 0) . That's why the matrix is a diagonal matrix with the eigenvalues on the diagonal.
The "matrix with eigenvectors as columns" you are talking about is, I think, the "change of basis" matrix. If you have a linear transformation written as a matrix in a given basis and want to change to another basis, then you need to multiply by a "change of basis" matrix which is just the same as applying a linear transformation. It's columns are, again, just the result of applying the transformation to the basis vectors. In particular, the "change of basis" matrix from the basis of eigenvectors to your original take (1, 0,..., 0) to the first eigenvector (written in the orginal basis) and so has first column exactly what that eigenvector is.