Consider matrices of the form $M$ = $$ \begin{bmatrix} 1 & s & 0 \\ t & 1 & s \\ 0 & t & 1 \\ \end{bmatrix} $$ where $s, t$ are both non-zero real numbers.
(a) Compute the determinant of M and deduce that M is invertible unless st = c where c is a constant that you should determine.
Answer
c = 1/2, st /= 1/2
(b) Why can we deduce from (a) that if $st = c$ then zero is an eigenvalue of $M$? Find a corresponding eigenvector in the case that $st = c$.
Answer
eigenvector = $$ \begin{bmatrix} -s \\ 1 \\ -t \\ \end{bmatrix} $$
(c) Show that $λ = 1$ is always an eigenvalue of M and find a corresponding eigenvector.
Answer
eigenvector = $$ \begin{bmatrix} s \\ 0 \\ -t \\ \end{bmatrix} $$
(d) Suppose now that $s = t = √ c$ where $c$ is as above. Use properties of real symmetric matrices to find a third eigenvector of $M$ and hence write down an orthogonal diagonalization of $M$
What I've got
Two unit vectors,$$ \begin{bmatrix} -1/2 \\ 1/(\sqrt 2) \\ -1/2 \\ \end{bmatrix} $$ \begin{bmatrix} 1/(\sqrt 2) \\ 0 \\ -1/(\sqrt 2) \\ \end{bmatrix}
orthogonal unit vector = $$ \begin{bmatrix} 1/2 \\ 1/(\sqrt 2) \\ 1/2 \\ \end{bmatrix} $$
This is where I get confused. The marking scheme says "Calculating, we find that M applied to this vector is twice the vector and so the eignevalue is 2." But when I tried to calculated the eigenvalue it was clearly incorrect but I cant figure out what went wrong. I did the matrix M $$ \begin{bmatrix} 1 & (\sqrt 2) & 0 \\ (\sqrt 2)& 1 & (\sqrt 2)\\ 0 & (\sqrt 2)& 1 \\ \end{bmatrix} $$ Multiplied by the orthogonal unit vector, which gave me $$ \begin{bmatrix} 3/2 \\ (\sqrt 2) + 1 \\ 3/2 \\ \end{bmatrix} $$
Would someone be able to do a step by step answer for d) because the marking scheme isn't very clear.