I Interpretation of "pseudo-diagonalisation"

[jk22]

I wanted to know what the usage of the following could be :

Let $A\in M_{n\times n}(K)$ a matrix over the field K.

Suppose we look for $x,\lambda\in M_{n\times 1}(K)$ such that

$$Ax=(\lambda_i x_i)$$

Hence instead of having a global eigenvalue we would have local ones.

I know the characteristic polynomial gives a relationship between the components of $\lambda$.What results are known about this problem, and can they have interpretations like quantum mechanics has with usual diagonalisation ?

 

[FactChecker]

It seems that this would hold for some $\lambda$ for any $x$ with all non-zero elements.

 

[jk22]

$$\exists x\neq 0$$

 

[FactChecker]

$\lambda_i = [Ax]_i/x_i$ will always work as long as $x_i \ne 0 \forall i$. Allowing different values for the $\lambda_i$ makes this property too easy to be significant.

 

[jk22]

So what if the condition that the $\lambda$ shall be independent of the $x$ were added ?

 

[Infrared]

If the $\lambda_i$ are given, define $\Lambda$ to be the diagonal matrix with the $\lambda_i$ on the diagonal. Then your equation is equivalent to $Ax=\Lambda x.$ Another way of saying this is that $x$ is in the nullspace of $A-\Lambda$.

Note that if all of the $\lambda_i=\lambda$ are equal then your question is the same as finding the $\lambda$-eigenspace of $A$. This agrees with the above because the $\lambda$-eigenspace of a matrix $A$ is the nullspace of $A-\lambda I$, and $\Lambda=\lambda I$ in this case.