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

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?

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## closed as not a real question by DonAntonio, Henry T. Horton, Micah, Austin Mohr, Brandon CarterDec 28 '12 at 6:28

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Related question: math.stackexchange.com/q/256158/48763 –  Learner Dec 28 '12 at 2:59
I'm not sure I understand the question... if they have both the same eigenvalues and same eigenvectors then both matrices are the same, right? What does "so that it works as a same eigenvalue to the eigenvector" mean ? –  WhitAngl Dec 28 '12 at 3:18
The question makes no sense:"that it works" refers to the eigenvector, but it is required to work as "a same eigenvalue". Besides this, it makes no sense to talk of an eigenvector without referring it to some determined matrix/operator. –  DonAntonio Dec 28 '12 at 3:31

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 $Ax=\lambda x$.

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

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What is "eigenvector matrix"? –  DonAntonio Dec 28 '12 at 3:32
@DonAntonio sorry. I meant to say the eigenvector matrix *of $A$*. Each column of $S$ contains the eigenvectors of $A$. –  E.O. Dec 28 '12 at 3:37

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

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

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