Intuition behind subtraction of diagonal matrices build from eigenvalues

Question

Suppose we have a matrix A: \begin{array}{ccc} 3 & 1 & 1 \\ 3 & 2 & 3 \\ 3 & 2 & 1 \\ \end{array}

It's eigenvalues are: {6, -1, 1}

If I make a diagonal matrix out of any of them for example: \begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \\ \end{array}

And subtract it from A the rank of the resulting matrix is one less than the rank of matrix A. Why is that? What role does the eigenvalue play here? I'd be very thankful for any intuition.

Answer 1

Hint: Because $det(A-\lambda I) = 0$ by the definition of the eigenvalue. Also, as there three distinct eigenvalues for $A$ there two distinct eigenvalues for $B = A-6I$ too. $\{-5, -7\}$ are eigenvalues of $B$ as $B+5I = A-I$ and $B+7I = A+I$.

Answer 2

Remember: $Av= \lambda v$, then you can rewrite this as: $(A - \lambda I)v = 0$, which means that $det(A-\lambda I) = 0$ (assuming $v$ is not 0, $v$ is the eigenvector corresponding to $\lambda$).