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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: $$ A - 5I = \begin{bmatrix} 3-5 & 1 & 4 \\ 0 & 2-5 & 6 \\ 0 & 0 & 5-5 \\ \end{bmatrix} = \begin{bmatrix} 2 & 1 & 4 \\ 0 & -3 & 6 \\ 0 & 0 & 0 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & -3 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ \end{bmatrix} $$ $$ \begin{align*} x_1 = 3x_3 \qquad x_2 = 2x_3 \qquad x_3 = 1 \: (free \: variable) \\ \end{align*} $$ $$ v_3 = \begin{bmatrix} 3 \\ 2 \\ 1 \\ \end{bmatrix} \qquad (matches \: answer \: in \: text) $$

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    $\begingroup$ Both your and the text book's answer work. If $(\lambda ,v)$ is an eigenpair of a matrix $M$, then so is $(\lambda, \mu v)$, for all scalars $\mu$. $\endgroup$ – Git Gud Jan 23 '16 at 23:15
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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.

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  • $\begingroup$ So what you are saying is that both I and the book's answers are correct? So its just a matter of preference? $\endgroup$ – Jules Manson Jan 23 '16 at 23:24
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    $\begingroup$ Precisely, and in fact you could take for $x_2$ any nonzero real number. $\endgroup$ – John B Jan 23 '16 at 23:26
  • $\begingroup$ In your original post, you said that the problem was to find an eigenbasis. In other words, since there are three distinct eigenvalues, find one eigenvector for each eigenvalue. Which one you choose out of the infinite number of eigenvalues that correspond to each eigenvalue is "a matter of preference" but it is important that you understand that any multiple of an eigenvector is also an eigenvector. $\endgroup$ – user247327 Dec 13 '16 at 16:18

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