1. ## [RESOLVED]diagonalizing a matrix

Ortogonally diagonalize the given matrix A, giving the diagonal matrix D and the diagonalizing orthogonal matrix P.

$\displaystyle $A =\left[ {\begin{array}{ccc} -1 & 2 & 2 \\ 2 & -1 & 2 \\2 & 2 & -1\\ \end{array} } \right]$$Find the eigenvalues using charactreistic polynomial$\displaystyle det(\lambdaI_3-A)=(\lambda-3)(\lambda+3)^2$For $\displaystyle \lambda=-3$

[tex](-3I_3-A)=\left[ {\begin{array}{ccc}
-2 & -2 & -2 \\
-2 & -2 & -2 \\

-2 & -2 & -2 \\
\end{array} } \right]
\]

so this means the eigenvector is $\displaystyle ](-3I_3-A)=\left[ {\begin{array}{ccc} 1 \\ 1 \\ 1 \\ \end{array} } \right]$

for $\displaystyle \lambda=3$

$\displaystyle (3I_3-A)=](-3I_3-A)=\left[ {\begin{array}{ccc} 4 & -2 & -2 \\ -2 & 4 & -2 \\ -2 & -2 & 4 \\ \end{array} } \right]$

$\displaystyle rref(3I_3-A)=](-3I_3-A)=\left[ {\begin{array}{ccc} 1 & 0 & -1 \\ 0& 1 & -1 \\ 0 & 0 & 0 \\ \end{array} } \right]$ so the eigvenvector for $\displaystyle \lambda=3$ is (again) $\displaystyle ](-3I_3-A)=\left[ {\begin{array}{c} r \\ r \\ r \\ \end{array} } \right]$

I can't diagonalize a matrix with only one eigenvector, where am I going wrong?

2. Originally Posted by Jskid
Orthogonally diagonalize the given matrix A, giving the diagonal matrix D and the diagonalizing orthogonal matrix P.

$\displaystyle $A =\left[ {\begin{array}{ccc} -1 & 2 & 2 \\ 2 & -1 & 2 \\2 & 2 & -1\\ \end{array} } \right]$$Find the eigenvalues using charactreistic polynomial$\displaystyle det(\lambdaI_3-A)=(\lambda-3)(\lambda+3)^2$For $\displaystyle \lambda=-3$

[tex](-3I_3-A)=\left[ {\begin{array}{ccc}
-2 & -2 & -2 \\
-2 & -2 & -2 \\

-2 & -2 & -2 \\
\end{array} } \right]
\]

so this means the eigenvector is $\displaystyle ](-3I_3-A)=\left[ {\begin{array}{ccc}$$\displaystyle 1 \\ 1 \\ 1 \\ \end{array} } \right] for \displaystyle \lambda=3 \displaystyle (3I_3-A)=](-3I_3-A)=\left[ {\begin{array}{ccc}$$\displaystyle 4 & -2 & -2 \\ -2 & 4 & -2 \\ -2 & -2 & 4 \\ \end{array} } \right]$

$\displaystyle rref(3I_3-A)=](-3I_3-A)=\left[ {\begin{array}{ccc}$$\displaystyle 1 & 0 & -1 \\ 0& 1 & -1 \\ 0 & 0 & 0 \\ \end{array} } \right] so the eigvenvector for \displaystyle \lambda=3 is (again) \displaystyle ](-3I_3-A)=\left[ {\begin{array}{c}$$\displaystyle r \\ r \\ r \\ \end{array} } \right]$

I can't diagonalize a matrix with only one eigenvector, where am I going wrong?

2nd the matrix admits to eigenvalues $\displaystyle \lambda =3$ and $\displaystyle \lambda =-3$

The repeated eigenvalue $\displaystyle \lambda =-3$ admits two linearly independent eigenvectors

Please double check your calculation for $\displaystyle \lambda =-3$

I get the two vectors

$\displaystyle \begin{bmatrix}-1 \\ 0 \\ 1 \end{bmatrix}$ and

$\displaystyle \begin{bmatrix}-1 \\ 1\\ 0 \end{bmatrix}$

3. Originally Posted by TheEmptySet

2nd the matrix admits to eigenvalues $\displaystyle \lambda =3$ and $\displaystyle \lambda =-3$

The repeated eigenvalue $\displaystyle \lambda =-3$ admits two linearly independent eigenvectors

Please double check your calculation for $\displaystyle \lambda =-3$

I get the two vectors

$\displaystyle \begin{bmatrix}-1 \\ 0 \\ 1 \end{bmatrix}$ and

$\displaystyle \begin{bmatrix}-1 \\ 1\\ 0 \end{bmatrix}$
I have no clue how the Latex got so messed up, thanks for reading it in such a poor state.