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.