Find two linearly independent eigenvectors for eigenvalue 1

[Potatochip911]

Homework Statement

A linear transformation with Matrix A = $\begin{pmatrix} 5&4&2\\ 4&5&2\\ 2&2&2 \end{pmatrix}$ has eigenvalues 1 and 10. Find two linearly independent eigenvectors corresponding to the eigenvalue 1.

Homework Equations

3. The Attempt at a Solution [/B]
I know from the trace of matrix A that eigenvalue 1 has a multiplicity of 2 and that eigenvalue 10 has a multiplicity of 1. Solving for eigenvectors corresponding for $\lambda = 1$, pretend it's an augmented matrix...
$\begin{pmatrix} 4&4&2\\ 4&4&2\\ 2&2&1 \end{pmatrix} R_2 ->R_2 - R_1; R_3 ->R_3 - \dfrac{1}{2} \cdot R_2 = \begin{pmatrix} 4&4&2\\ 0&0&0\\ 0&0&0 \end{pmatrix} R_1-> \dfrac{1}{2} \cdot R_1 = \begin{pmatrix} 2&2&1\\ 0&0&0\\ 0&0&0 \end{pmatrix}$
Therefore $x_2$ and $x_3$ are arbitrary. Solving for $x_1$ gives $$x_1=-x_2-\dfrac{1}{2} x_3$$ $$x_2=x_2$$ $$x_3=x_3$$
$$v= \begin{pmatrix}
-x_2-\dfrac{1}{2} x_3\\
x_2\\
x_3
\end{pmatrix} $$
$v= x_2 \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} + x_3 \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}$ Now those are the two eigenvectors I got however the answer in my book says the two eigenvectors are $v_1=<1,0,-2>$ and $v_2=<0,1,-2>$

 

[Mark44]

Homework Statement

A linear transformation with Matrix A = $\begin{pmatrix} 5&4&2\\ 4&5&2\\ 2&2&2 \end{pmatrix}$ has eigenvalues 1 and 10. Find two linearly independent eigenvectors corresponding to the eigenvalue 1.

Homework Equations

3. The Attempt at a Solution [/B]
I know from the trace of matrix A that eigenvalue 1 has a multiplicity of 2 and that eigenvalue 10 has a multiplicity of 1. Solving for eigenvectors corresponding for $\lambda = 1$, pretend it's an augmented matrix...
$\begin{pmatrix} 4&4&2\\ 4&4&2\\ 2&2&1 \end{pmatrix} R_2 ->R_2 - R_1; R_3 ->R_3 - \dfrac{1}{2} \cdot R_2 = \begin{pmatrix} 4&4&2\\ 0&0&0\\ 0&0&0 \end{pmatrix} R_1-> \dfrac{1}{2} \cdot R_1 = \begin{pmatrix} 2&2&1\\ 0&0&0\\ 0&0&0 \end{pmatrix}$
Therefore $x_2$ and $x_3$ are arbitrary. Solving for $x_1$ gives $$x_1=-x_2-\dfrac{1}{2} x_3$$ $$x_2=x_2$$ $$x_3=x_3$$
$$v= \begin{pmatrix}
-x_2-\dfrac{1}{2} x_3\\
x_2\\
x_3
\end{pmatrix} $$
$v= x_2 \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} + x_3 \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}$ Now those are the two eigenvectors I got however the answer in my book says the two eigenvectors are $v_1=<1,0,-2>$ and $v_2=<0,1,-2>$

Your eigenvectors are correct, which I verified by multiplying them on the left by your matrix, with both multiplications resulting in 1 times the eigenvector.
The book's first eigenvector is the -1 multiple of your second eigenvector. Their second eigenvector is not a scalar multiple of your first eigenvector, but it still is an eigenvector associated with the eigenvalue 1. Your two vectors and the book's two vectors span the same two-dim. subspace of R³, so all is good.

 

[Potatochip911]

Your eigenvectors are correct, which I verified by multiplying them on the left by your matrix, with both multiplications resulting in 1 times the eigenvector.
The book's first eigenvector is the -1 multiple of your second eigenvector. Their second eigenvector is not a scalar multiple of your first eigenvector, but it still is an eigenvector associated with the eigenvalue 1. Your two vectors and the book's two vectors span the same two-dim. subspace of R³, so all is good.

Okay thanks!

 

[vela]

In the book's solution, they solved for $x_3 = -2 x_1 -2 x_2$, so $x_1$ and $x_2$ ended up being arbitrary instead of $x_2$ and $x_3$ as in your case.

 

[Potatochip911]

In the book's solution, they solved for $x_3 = -2 x_1 -2 x_2$, so $x_1$ and $x_2$ ended up being arbitrary instead of $x_2$ and $x_3$ as in your case.

Oh, so that's where they're getting that eigenvector from.