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The matrix $$ A := \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 4 \end{bmatrix}$$

has the eigenvalues $1$ and $4$. I am looking for the eigenspace corresponding to the eigenvalue $4$. So I need to find the solutions of $(A - \lambda I)v = 0$ with $\lambda = 4$, i.e.

$$(A - 4I)v = \begin{bmatrix} -3 & 1 & 0 \\ 0 & -3 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0$$ or equivalently $$\begin{bmatrix} 3 & -1 & 0 \\ 0 & 3 & -1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0$$

So $3y = z$ and $3x = y$, therefore $x = 9z$

Let $z = t$. Then

$$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = t \begin{bmatrix} 9 \\ 3 \\ 1 \end{bmatrix}$$

So the eigenspace is $\operatorname{span} \{ \begin{bmatrix} 9 & 3 & 1 \end{bmatrix}^T \}$.

But I have checked two online calculators and they give this eigenspace as $$\operatorname{span} \{\begin{bmatrix} 1 & 3 & 9 \end{bmatrix}^T \},$$ so where am I going wrong?

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1 Answer 1

up vote 5 down vote accepted

Mistake: what you actually have is $z=3y=3(3x)=9x$, not $x=9z$.

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