# Guessing the eigenvectors knowing the eigenvalues of a 3x3 matrix

So I'm given a matrix $$A$$

$$A=\begin{pmatrix}0.4 & 0.3& 0.3\\ 0.3 & 0.5 & 0.2 \\ 0.3& 0.2 & 0.5 \end{pmatrix}$$

Find the eigenvector corresponding to the eigenvalue $$\lambda_1=1$$ then proceed to find the remaining eigenvalues using a built in command in MatLab

So far I've found the eigenvector for $$\lambda_1=1$$ which is $$v_1=\begin{pmatrix}1\\1\\1\end{pmatrix}$$ and after i've used the eig command I found the remaining eigenvalues to be $$\lambda_2=0.3,\lambda_3=0.1$$ now it wants me to literary "guess" the corresponding eigenvectors.And this is where i'm stuck. I have read lots of guides and stuff about how to find eigenvectors by inspection but that was for 2x2 matrices and for 3x3 matrices they just mentioned that they need a 3rd fact which is a bit more complicated. Besides I can't see any obvious eigenvectors.....

• Where does it say "guess"? eig will give you both eigenvalues and eigenvectors. Commented Apr 8, 2019 at 19:50
• @RobertIsrael In my problem sheet. (Use "eig" in matlab to find the remaining eigenvalues). From what the command returns, guess the corresponding eigenvectors then verify them (the verifying part is not an issue) but the part that it says "guess" got me pretty confused Commented Apr 8, 2019 at 19:55
• hmm maybe I've misunderstood the problem? As you said you'll get both eigenvectors and values if you use [v,b] = eig maybe it's referring to guess which one of the eigenvalues is for which eigenvectors as in you get both of them so you just somehow "see" which of them is for which? Commented Apr 8, 2019 at 20:00

Note that $$A - 0.3 I = \pmatrix{0.1 & 0.3 & 0.3\cr 0.3 & 0.2 & 0.2\cr 0.3 & 0.2 & 0.2\cr}$$ has its second and third columns equal. That says that $$\pmatrix{0\cr 1 \cr -1\cr}$$ will be in its null space, i.e. an eigenvector of $$A$$ for eigenvalue $$0.3$$.
Now $$A$$ being a symmetric matrix, its eigenvectors are orthogonal, so the eigenvector for the remaining eigenvalue $$0.1$$ has sum $$0$$ and its second and third entries equal. Thus it should be (a multiple of) $$\pmatrix{2 \cr -1\cr -1\cr}$$.