What is the difference between using $(\lambda I - A)$ and $(A - \lambda I)$ when finding eigenvalues and eigenvectors?

Question

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!

Answer 1

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.

Answer 2

It shouldn't change anything as what you're looking for are the solutions of $Av_i=\lambda_iv_i$, with $A$ a square matrix and $v_i$/$\lambda_i$ the eigenvectors/eigenvalues. So the sign of $A-\lambda I$ doesn't matter.

I guess that when you computed $\lambda I-A$, you accidentally made a mistake.