Say we have matrix $\mathbf{A}$
$$
\mathbf{A}=\begin{pmatrix}
-3&2&0\\
4&-6&2\\
0&1&-1
\end{pmatrix}
$$
We now must show that $\mathbf{v}=\begin{pmatrix}-1.34&-0.8&1\end{pmatrix}^T$ is an approximate eigenvector for $\mathbf{A}$ to 2 decimal places, and find the corresponding eigenvalue. We know what $\mathbf{A}\mathbf{v}=\lambda\mathbf{v}$ for some vector. So multiplying $\mathbf{A}$ by $\mathbf{v}$ we get
$$
\begin{pmatrix}
-3&2&0\\
4&-6&2\\
0&1&-1
\end{pmatrix}
\begin{pmatrix}
-1.34\\
-0.8\\
1
\end{pmatrix}=
\begin{pmatrix}
2.42\\
1.44\\
-1.8
\end{pmatrix}
$$
So we need to find eigenvalue $\lambda$ such that
$$
\begin{pmatrix}
2.42\\
1.44\\
-1.8
\end{pmatrix}
=\lambda
\begin{pmatrix}
-1.34\\
-0.8\\
1
\end{pmatrix}
$$
Say $\lambda=-1.8$, this gives us
$$
-1.8
\begin{pmatrix}
-1.34\\
-0.8\\
1
\end{pmatrix}=
\begin{pmatrix}
2.412\\
1.44\\
-1.8\\
\end{pmatrix}
$$
Now $\lambda=-1.8$ gives a good approximation for the eigenvector, but we're getting 2.41 instead of 2.42 for the first value of the eigenvector (to 2 decimal places). Is this enough to say that $\mathbf{v}$ is an approximate eigenvector for $\mathbf{A}$? or am I missing something in my method?
Best Answer
Your method has a minor issue.
If you want numbers to match to a certain precision, you should use inequalities, e.g., solve $$|-1.34\lambda-2.42|<0.01,$$ $$|-0.8\lambda-1.44|<0.01,$$ $$|1\cdot\lambda-(-1.8)|<0.01.$$ If you want it to match exactly to 2 decimal places, you may need to shrink these margins (they are not equivalent to matching to 2 decimal places). Or you can add some more inequalities, since if they are going to match to 2 decimal places, we'll need $$-1.34\lambda \ge 2.42,$$ $$-0.8\lambda \ge 1.44,$$ $$1\cdot\lambda \le -1.8,$$ or else we could get something like $-1.34\lambda = 2.419$ or $1\cdot \lambda = -1.795$.
One example of such a value of $\lambda$ is $\lambda = -1.806$, which yields $$\lambda \begin{pmatrix} -1.34 \\ -0.8 \\ 1 \end{pmatrix} = \begin{pmatrix}2.42004\\ 1.4448\\ -1.806\end{pmatrix}.$$