Please check my work in finding an eigenbasis (eigenvectors) for the following problem. Some of my solutions do not match answers in my differential equations text (Advanced Engineering Mathematics by Erwin Kreyszig, 1988, John Wiley & Sons).
For reference the following identity is given because some textbooks reverse the formula having $\lambda$ subtract the diagonal elements instead of subtracting $\lambda$ from the diagonal elements:
$$ det(A – \lambda I) = 0 $$
$$
A =
\begin{bmatrix}
3 & 1 & 4 \\
0 & 2 & 6 \\
0 & 0 & 5 \\
\end{bmatrix}
$$
By inspection the eigenvalues are the entries along the diagonal for this upper triangular matrix.
$$
\begin{align*}
\lambda_1 = 3 \qquad \lambda_2 = 2 \qquad \lambda_3 = 5
\end{align*}
$$
When $\lambda_1 = 3$ we have:
$$
A – 3I =
\begin{bmatrix}
3-3 & 1 & 4 \\
0 & 2-3 & 6 \\
0 & 0 & 5-3 \\
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 4 \\
0 & -1 & 6 \\
0 & 0 & 2 \\
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
$$
$$
\begin{align*}
x_1 = 1 \: (free \: variable) \qquad x_2 = 0 \qquad x_3 = 0 \\
\end{align*}
$$
$$
v_1 =
\begin{bmatrix}
1 \\
0 \\
0 \\
\end{bmatrix} \qquad (matches \: answer \: in \: text)
$$
When $\lambda_2 = 2$ we have:
$$
A – 2I =
\begin{bmatrix}
3-2 & 1 & 4 \\
0 & 2-2 & 6 \\
0 & 0 & 5-2 \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 1 & 4 \\
0 & 0 & 6 \\
0 & 0 & 3 \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
$$
$$
\begin{align*}
x_1 = -x_2 \qquad x_2 = 1 \: (free \: variable) \qquad x_3 = 0 \\
\end{align*}
$$
$$
v_2 =
\begin{bmatrix}
-1 \\
1 \\
0 \\
\end{bmatrix} \qquad but \: answer \: in \: text \: is \qquad
\begin{bmatrix}
1 \\
-1 \\
0 \\
\end{bmatrix}
$$
What happened? Is it from a disagreement in what we should consider arbitrary or am I doing something fundamentally wrong?
When $\lambda_3 = 5$ we have:
$$
A – 5I =
\begin{bmatrix}
3-5 & 1 & 4 \\
0 & 2-5 & 6 \\
0 & 0 & 5-5 \\
\end{bmatrix}
=
\begin{bmatrix}
2 & 1 & 4 \\
0 & -3 & 6 \\
0 & 0 & 0 \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & -3 \\
0 & 1 & -2 \\
0 & 0 & 0 \\
\end{bmatrix}
$$
$$
\begin{align*}
x_1 = 3x_3 \qquad x_2 = 2x_3 \qquad x_3 = 1 \: (free \: variable) \\
\end{align*}
$$
$$
v_3 =
\begin{bmatrix}
3 \\
2 \\
1 \\
\end{bmatrix} \qquad (matches \: answer \: in \: text)
$$
Best Answer
Eigenvectors are never unique. In particular, for the eigenvalue $2$ you can take, for example, $x_2=-1$ which gives you the answer in the book.