A question about finding vectors

[transgalactic]

i added a link with the question and how i tried to solve it

i want to create a matrix using the vectors that i found

the problem is
that i got 4 vectors and two of them is (0,0,0)

http://img406.imageshack.us/my.php?image=img8228mh6.jpg

 

[HallsofIvy]

This is a problem about eigenvalues and eigenvectors, right?

Well, (0, 0, 0) is never an eigenvector since the whole point of an "eigenvalue", $\lambda$, of T is that there exist a non-zero vector v such that $Tv= \lambda v$.

You don't clearly state what the original matrix is but I think you mean
$$M=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1 &1 \\ 1 & 1 & 1\end{array}\right]$$

The characteristic equation for that is not at all what you give, however. I get $-\lambda^3+ 3\lambda^2= 0$ which has roots 0, 0, 3. That makes sense to me since the matrix is badly "singular". It's no surprise that it has 0 as a double eigenvalue. To find eigenvectors corresponding to the eigenvalue 0, we look at
$$M=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1 &1 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]= \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]$$

Obviously that gives us the single equation x+ y+ z= 0. The "solution space" is two dimensional- we can pick any values for x and y and solve for z. In particular if we take x= 1, y= 0, z= -1. An eigenvector corresponding to eigenvalue 0 is [1 0 -1]^T. If we choose x= 0, y= 1, we get z= -1. An independent eigenvector corresponding to eigenvalue 0 is [0 1 -1]^T.

To find an eigenvector corresponding to eigenvalue 3, we look at the equation
$$M=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1 &1 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]= \left[\begin{array}{c} 3x \\ 3y \\ 3z \end{array}\right]$$
That gives three equations, x+ y+ z= 3x, x+ y+ z= 3y, and x+ y+ z= 3z which are, again, dependent equations. This time they reduce to x= y= z. An eigenvector corresponding to eigenvalue 3 is [1 1 1]^T.

The matrix you are looking for, that diagonalizes M, is
$$M=\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 &1 \\ -1 & -1 & 1\end{array}\right]$$

 

[transgalactic]

thanks

how to prove that this matrix is orthogonal
and if not
how to transform it to an orthogonal matrix

??

 

[HallsofIvy]

See if it satisfies your definition of orthogonal, of course! There are several equivalent definitions (A^TA= I is one) and I don't know which you were given. However, either as a definition, or as a theorem, it is true that the individual columns of an orthogonal matrix, as vectors, form an orthonormal basis for the vector space: the dot product of two different columns must be 0 and the dot product of a column with itself must be 1. Here, I made no attempt to reduce those vectors to an orthonormal basis:
[1 0 -1], [0 1 -1], and [1 1 1] happen to be "orthogonal" (their dot products are 0) but are not normalized. (One can show that eigenvectors corresponding to different eigenvalues must be orthogonal. The two eigenvectors corresponding to 0 were not necessarily orthogonal, but my choice of x=1 y=0 and x=0 y=1 forced that.) Find the length of each vector and divide each vector by its length. Using those "normalized" eigenvectors as columns will give you an orthogonal matrix.

 

[transgalactic]

i tried to solve it again i didnd get your polinomial
in the start

http://img507.imageshack.us/my.php?image=img8252gd1.jpg

 

[transgalactic]

i have solved it again
but i don't get them to be orthogonal
what to do in that case??

i didnt get the dot product zero
(so its not orthogonal)

what to do??

and i have written bellow what what values i need to divide each vector in order to make
the matrix orthonormal
did i make it ok??

http://img267.imageshack.us/my.php?image=img8253ms5.jpg