linear algebra - Intuitive perspective of eigenvalues and rank of a matrix

Assuming a matrix $A$, $n\times n$, with $n$ non-repeated and non-zero eigenvalues;

1. If we calculate the matrix $A-\lambda I$ for one of its $n$ eigenvalues, we see that its rank has been decreased by one. If the eigenvalue has repetitiveness of $k$, then the rank decreases again by $k$. What would be an intuitive explanation for it?




2. By $Ax=\lambda x$ one could argue that we try to find the values of $\lambda$ for which an $n\times n$ matrix with $\text{rank}(A)=n$ to have the same impact on $x$ as a:

   
   
   

   2a. a scalar $\lambda$?

   
   
   
   

   or

   
   
   

   2b. a $n\times n$ diagonal matrix of rank $n$?




3. In the relationship $(A-\lambda I)x=0$, given that we want a nontrivial solution for the vector $x$, could we declare the matrix $A-\lambda I$ as zero, without the determinant, following directly the above relationship?

Answer

For 1, it depends on what you mean by “intuitive”, but here’s a shot at it. A matrix either sends a vector to 0 or it doesn’t. The amount of vectors that it sends to 0 is related to the amount of vectors that it doesn’t by the rank it’s theorem. The more vectors it sends to 0, the fewer it sends to non-zero values. If A has an eigenvector then $A-\lambda I$ sends a vector to 0. Therefore, it can’t send as many to non-zero values (meaning it’s rank has been reduced).

EDIT: This is not always true as Widawensen points out in another answer.

For 2, the answer is a. But we normally think about it the other way around. We are trying to find a vector such that A operating on that vector simply scales it.

For 3, if $A-\lambda I$ is 0 then $A=\lambda I$. This is the special case where A has a single eigenvalue of multiplicity equal to its rank. But there are cases where A can have eigenvector a but A is not diagonal. Those cases can be found with the determinant because we know that the determinant of a singular matrix is 0 and we want $A-\lambda I$ to be singular.