Intuition behind Eigenvalue solution matrix

Question

I've been watching the excellent course by 3Blue1Brown on Linear Algebra which is oriented towards giving students intuition into Linear Algebra concepts.

I am trying to find an intuitive way to understand the matrix we use to calculate eigenvalues. Specifically, I am trying to get an intuition for the matrix shown below (screen snapshot from 3Blue1Brown course).

I understand the derivation of this matrix calculation. I also understand that the determinant of this matrix is zero because the matrix transformation is "transforming" into a lower dimension because many (eigen) vectors are being collapsed/transformed into a single span in the new "transformed" vector space.

However, I'm trying to get an intuition as to what does it mean to subtract a scalar from the diagonal of a matrix.

That is, can anyone give me a geometric or algebraic intuition as to what it means to subtract off a scalar value from the diagonal of a matrix?

Answer 1

The scalars are being subtracted from the diagonal because those scalars times the identity matrix are being subtracted from the other matrix:

$\begin{bmatrix} 3 & 1 & 4\\ 1 & 5 & 9\\ 2 & 6 & 5 \end{bmatrix}$-λ$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$

One way to visualize this would be to imagine the transformation of the first matrix, imagine the transformation of the scalar matrix, and then imagine subtracting the vectors from the second transformation from the first.

That's why this method finds eigenvalues. In both cases the eigenvectors are basically just being scaled by the same amount, so when they're subtracted from one another they produce the zero vector.

Answer 2

Instead of trying to make sense of “subtracting a bunch of scalars from the diagonal of a matrix,” back up a step: this matrix is the result of subtracting a multiple of the identity matrix from $A$. If you want to think of it in terms of the linear transformations that these matrices represent, you apply $A$ and then subtract the result of a uniform scaling by a factor of $\lambda$. Any vectors that are only scaled correspondingly by $A$ end up getting mapped to $0$.

This is just a different way to find these vectors. Ultimately, you’re comparing the action of the transformation represented by the matrix $A$ with uniform scaling by a factor of $\lambda$, looking for any vectors on which these transformations have the same effect. That’s the content of the eigenvector equation $A\mathbf v=\lambda\mathbf v$.