I am likely going to tutor a course in linear algebra in the coming semester, so I am brushing up on my concepts right now, which are a little rusty as I last touched linear algebra more than three years ago.
I was practicing a question on diagonalization and was stuck halfway.
I would like to find the eigenvectors of
$$A = \begin{bmatrix}
-2 & 6 \\
-2 & 5
\end{bmatrix}.$$
I start by finding the eigenvalues of $A$, which are $\lambda = 1$ and $\lambda = 2$.
Next, I find the eigenvector associated with each of the eigenvalues, which I recall to be the solutions to the nullspace of $(\lambda I – A)$.
Now, I let $\lambda = 1$ and obtain
$$\begin{bmatrix}
3 & 6 \\
-2 & -4
\end{bmatrix},$$
which solve to give
$$\begin{pmatrix}
-2 \\
1
\end{pmatrix}.$$
However, the eigenvector associated with $\lambda = 1$ is, instead,
$$\begin{pmatrix}
2 \\
1
\end{pmatrix}.$$
I realised that the solution solved for the nullspace of $(A – \lambda I)$, whereas I solved for the nullspace of $(\lambda I – A)$. As far as I recall, in terms of finding eigenvalues and eigenvectors, it does not matter whether I use $(\lambda I – A)$ or $(A – \lambda I)$.
Where have I gone wrong?
Any intuitive explanations or suggestions will be greatly appreciated 🙂
Edit
As it turns out, my concepts have not failed me but I was just a little careless in my calculations. Thank you to those who took their time to point out where I have gone wrong!
Best Answer
There is no difference, as $A_{\lambda}x=0$ is equivalent to $(-A_{\lambda})x=0$ for any matrix $A_{\lambda}$.
In terms of computation, $A-\lambda I$ is easier as it keeps the off-diagonal elements unchanged, and in theory $\det(\lambda I - A)$ is better as it's always a monic polynomial.
You made the mistake in computing $\lambda I - A$ for $\lambda =1$: You forgot to put a minus sign before the off-diagonal elements.