Does anyone have a nice way to create a $4\times 4$ matrix $A$ that has eigenvectors $$\vec v_1=\pmatrix{1\cr 1\cr 1\cr 1}\qquad\text{and}\qquad\vec v_2=\pmatrix{1\cr 0\cr 1\cr 0}$$ and eigenvalues $\lambda_1=1$ and $\lambda_2=2$? The eigenvalues both have algebraic multiplicity 2.
I'm looking for a method that I can use repeatedly; that is, suppose I want to find a $4\times 4$ matrix $A$ with eigenvectors $$\vec v_1=\pmatrix{1\cr 1\cr 1\cr 2}\qquad\text{and}\qquad\vec v_2=\pmatrix{1\cr 0\cr 1\cr 0}\qquad\text{and}\qquad\vec v_3=\pmatrix{-1\cr 1\cr 1\cr 1}$$ with eigenvalues $\lambda_1=-1$, $\lambda_2=1$, and $\lambda_3=2$, where the algebraic multiplicities are 1, 1, and 2.
Again, I don't just want answer for these specific examples, but a process that I can use repeatedly, say to find a $5\times 5$ matrix $A$ with 3 eigenvectors and 3 eigenvalues, one of algebraic multiplicity 1, the second with algebraic multiplicity 2, and the third with algebraic multiplicity 2.
Thanks.
Let $P$ be a non-singular matrix, s.t. $v_1, v_2 \in \text{im} P$, for example $$ P = \left(\begin{array}{rrrr} 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 1 \\ 1 & 0 & 0 & 1 \end{array}\right) $$ Let $B$ be a Jordan matrix of $A$: $$ B=\left(\begin{array}{rrrr} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{array}\right) $$
$$ A = PBP^{-1} = \left(\begin{array}{rrrr} -1 & 1 & 3 & -2 \\ -1 & 2 & 1 & -1 \\ -3 & 1 & 5 & -2 \\ -2 & 1 & 2 & 0 \end{array}\right) $$
It's important that first and third column of $P$ correspond to the beginnings of Jordan blocks in $B$.