Spotting the linearly independent eigen vectors

Question

Hi all, I wanted to ask how complete part 1 of this past exam question. Considering the number of marks allocated to part 1, it seems that there is some sort of shortcut to get all six vectors

Answer 1

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Let $\bm e_j$ be the $j$th unit column vector, i.e. the column vector with all entries $0$ except the $j$th one being $1$. Then $$ \bm {Ce}_1 = \bm e_2, \bm {Ce}_2 = \bm e_1, \bm {Ce}_3 = \bm e_4, \bm {Ce}_4 = \bm e_3, \bm {Ce}_5 = \bm e_6, \bm {Ce}_6 = \bm e_5. $$ By observation $$ \bm C (\bm e_j \pm \bm e_{j+1}) =\bm e_{j+1} \pm \bm e_j\quad [j =1,3,5] $$ and hence the eigenvectors are $\bm e_j \pm \bm e_{j+1}$ for $j = 1,3,5$, and the corresponding eigenvalues are $1,-1, 1, -1, 1, -1$ respectively.

Answer 2

The matrix is block diagonal, so you can analyze each block separately. All three blocks are identical, so once you’ve found eigenvectors for one of them, generating similar-looking eigenvectors for the rest of the blocks should be trivial.

The problem is now reduced to finding a pair of linearly-independent eigenvectors for $$\begin{bmatrix}0&1\\1&0\end{bmatrix}.$$ The row sums are obviously the same, so you have $(1,1)^T$ as an eigenvector with eigenvalue $1$ (the sum). The matrix is symmetric, so $(-1,1)$ is also an eigenvector—it’s orthogonal to the first one.

So, a set of six linearly-independent eigenvectors of the original matrix is $(\pm1,1,0,0,0,0)^T$, $(0,0,\pm1,1,0,0)^T$ and $(0,0,0,0,\pm1,1)^T$.