Constructing several matrices that act as a same eigenvalue to a same eigenvector [closed]

Question

Suppose there is single eigenvector $x$. We want to form several matrices so that it works as a same eigenvalue to the eigenvector. How are these matrices related, how does one construct such ones?

Answer 1

Write $A$ in the form $A=S\Lambda S^{-1}$ where $S$ is the eigenvector matrix of $A$ and $\Lambda$ contains the corresponding eigenvalues on the diagonal. Then any matrix whose factorization has the eigenvector $x$ in a column of $S$ and its corresponding eigenvalue on the diagonal of $\Lambda$ will yield $Ax=\lambda x$.

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For example, say you want to construct several matrices with eigenvector $x=\begin{bmatrix}2 \\ 1 \end{bmatrix}$ and corresponding eigenvalue 2. Then the factorization of any such matrix will be of the form

$$A=S\Lambda S^{-1}=\begin{bmatrix} 2&x_1 \\ 1&x_2 \end{bmatrix}\begin{bmatrix} 2&0 \\ 0&\lambda_2 \end{bmatrix}\begin{bmatrix} 2&x_1 \\ 1&x_2 \end{bmatrix}^{-1}$$ We can choose arbitrary $x_1,x_2,\lambda_2$ (as long as $S$ remains invertible) and get an arbitrary matrix such that $Ax=2x$.

For the sake of demonstration let $$A_1=\begin{bmatrix} 2&1 \\ 1&1 \end{bmatrix}\begin{bmatrix} 2&0 \\ 0&1 \end{bmatrix}\begin{bmatrix} 1&-1 \\ -1&2 \end{bmatrix}=\begin{bmatrix} 3&-2 \\ 1&0 \end{bmatrix},x_1=1,x_2=1,\lambda_2=1$$ $$A_2=\begin{bmatrix} 2&5 \\ 1&3 \end{bmatrix}\begin{bmatrix} 2&0 \\ 0&3 \end{bmatrix}\begin{bmatrix} 3&-5 \\ -1&2 \end{bmatrix}=\begin{bmatrix} -3&10 \\ -3&8 \end{bmatrix},x_1=5,x_2=3,\lambda_2=3$$

Then $$A_1x=\begin{bmatrix} 3&-2 \\ 1&0 \end{bmatrix}\begin{bmatrix}2 \\ 1 \end{bmatrix}=\begin{bmatrix}4 \\ 2 \end{bmatrix}=2x$$ $$A_2x=\begin{bmatrix} -3&10 \\ -3&8 \end{bmatrix}\begin{bmatrix}2 \\ 1 \end{bmatrix}=\begin{bmatrix}4 \\ 2 \end{bmatrix}=2x$$

Answer 2

Let $A$ be one matrix so that

$$Ax=\lambda x $$

Then, for a matrix $B$ we have $Bx=\lambda x$ if and only if $(A-B)x =0$.

Let

$$V:= \{ C \in {\mathcal M}_n | Cx=0 \}$$

Then $V$ is a subspace of ${\mathcal M}_n$ and the matrices satisfying your relation are exactly

$$\lambda I+ V = \{ \lambda I+ C |C \in V \} \,.$$