1. ## [RESOLVED]diagonalizing a matrix

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

$

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

[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 $](-3I_3-A)=\left[ {\begin{array}{ccc}
1 \\
1 \\

1 \\
\end{array} } \right]
$

for $\lambda=3$

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

-2 & -2 & 4 \\
\end{array} } \right]$

$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 $\lambda=3$ is (again) $](-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.

$

$A =\left[ {\begin{array}{ccc} -1 & 2 & 2 \\ 2 & -1 & 2 \\2 & 2 & -1\\ \end{array} } \right]$" alt="

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

[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 $](-3I_3-A)=\left[ {\begin{array}{ccc}$ $
1 \\
1 \\

1 \\
\end{array} } \right]
" alt="
1 \\
1 \\

1 \\
\end{array} } \right]
" />

for $\lambda=3$

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

-2 & -2 & 4 \\
\end{array} } \right]" alt="
4 & -2 & -2 \\
-2 & 4 & -2 \\

-2 & -2 & 4 \\
\end{array} } \right]" />

$rref(3I_3-A)=](-3I_3-A)=\left[ {\begin{array}{ccc}$ $
1 & 0 & -1 \\
0& 1 & -1 \\

0 & 0 & 0 \\
\end{array} } \right]
" alt="
1 & 0 & -1 \\
0& 1 & -1 \\

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

r \\
\end{array} } \right]
" alt="
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 $\lambda =3$ and $\lambda =-3$

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

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

I get the two vectors

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

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

3. Originally Posted by TheEmptySet

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

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

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

I get the two vectors

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

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