# Show vector is approximately an eigenvector of matrix, thus find eigenvalue

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?

• There is a calculation error in your first product. $1.4$ should read $1.44$ which is actually equal to $0.8 \times 1.8$. – Axel Kemper Jan 28 at 13:59
• amended, thanks – whitelined Jan 28 at 14:04
• Isn't that what "approximate" means? – user247327 Jan 28 at 14:08

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 $$\lambda \begin{pmatrix} -1.34 \\ -0.8 \\ 1 \end{pmatrix} = \begin{pmatrix}2.42004\\ 1.4448\\ -1.806\end{pmatrix}.$$