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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?

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    $\begingroup$ The eigenvectors of $\lambda$ are exactly the kernel of $A-\lambda I$. What does this say about the rank of this matrix? $\endgroup$ – amd Jun 2 at 19:23
  • $\begingroup$ The answer to 2 is a. 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. $\endgroup$ – NicNic8 Jun 2 at 19:30
  • $\begingroup$ "The eigenvectors of 𝜆 are exactly the kernel of 𝐴−𝜆𝐼. What does this say about the rank of this matrix?" I understand that the kernel is the vector that if we let the 𝐴−𝜆𝐼 act upon, it will give zero. However, I unfortunately cannot see for the moment the causal relationship between the kernel and how the repeated number of eigenvalues influence the rank. I can see it on paper when i calculate it by hand, but i am missing something I guess. I would be grateful if you developed your thought $\endgroup$ – Alex Jun 2 at 21:23
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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.

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  • $\begingroup$ Helpful! About the first part, you concluded: "it can’t send as many to non-zero values (meaning it’s rank has been reduced)". This implies that the rank of a square matrix, that is: - the number of linearly independent columns - the number of vectors that form the basis - the number of nonzero eigenvalues is the same as the number of vectors, that a matrix can send to 0. Is this correct? Could you please name the theorem concerning the rank you talked about? $\endgroup$ – Alex Jun 2 at 21:04
  • $\begingroup$ @Alex It's the rank-nullity theorem. $\endgroup$ – user326210 Jun 2 at 21:18
  • $\begingroup$ @Alex. It's called the "Rank-Nullity Theorem". The theorem states what I said above in words: rank(A) + Nullity(A) = number of columns of A. $\endgroup$ – NicNic8 Jun 2 at 23:19
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Consider matrix ( so called Jordan normal form) for example: $$A=\begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$

Then calculating $A-2I$ you'll see that

"If the eigenvalue has repetitiveness of $k$, then the rank decreases again by $k$"

is not true.

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