Linear algebra diagonalization

[bakin]

linear algebra diagonalization :(

Homework Statement

determine whether the given matrix is diagonalizable. where possible, find a matrix S such that
S^-1AS=Diag(λ₁+λ₂,...,λ_n)

Homework Equations

The Attempt at a Solution

I was able to find the eigenvalues, which are λ=1,-4,4. This is given in the back of the book as well, which matched up evenly.

Now, I'm having trouble finding the matrix S. I know I need to find the eigenvectors, and place them as columns in an n x n matrix.

When I plug 1 in, say, I come up with
0 0 0
0 2 7
1 1 -4

And for 4 I came up with
-3 0 0
0 -1 7
1 1 -7

Neither of them seem to be coming out right.

The answer in the back of the book is:

15 0 0
-7 7 1 = S
2 1 -1

Thanks, all.

 

[xaos]

let's see your rref. what is the nullspace for each of these matrices?

 

[HallsofIvy]

With $\lambda= 1$ your matrix requires that 0x+ 0y+ 0z= 0, 0x+ 2y+ 7z= 0, and x+ y- 4z= 0. The first equation is always true, of course, the second equation gives z= -(7/2)y and putting that into the third equation x- (7/2)z- 4z= x- (15/2)z= 0 so x= (15/2)z. Taking z= 2, an eigenvector corresponding to $\lambda= 1$ is <15, -7, 2>. That is exactly what you say your textbook gives as the first column.

With $\lambda= 4$ the corres0 7 ponding equations are -3x= 0, 0x- y+ 7z= 0, and x+ y- 7z= 0. The first equation obvioiusly gives x= 0. Putting x= 0, the second and third equations both reduce to y- 7z= 0 or y= 7z. Taking z= 1 y= 7 so an eigenvalue corresponding to $\lambda= -4$ is <0, 7, 1> , exactly what your textbook gives for the second column.

With $\lamba= 4$ the matrix $A- \lambda I$ must be
$$\begin{bmatrix}5 & 0 & 0 \\ 0 & 7 & 7 \\ 1 & 1 & 1\end{bmatrix}$$
which gives equations 5x+ 0y+ 0z= 0, 0x+ 7y+ 7z= 0, and x+ y+ z= 0. What eigevector do those give?