# Intuition behind Eigenvalue solution matrix

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?

• You transform a vector and then you subtract the original vector scaled. I don't think there is much more intuition to be found other than what you've seen in the video.
– tst
Sep 21, 2019 at 1:24

$$\begin{bmatrix} 3 & 1 & 4\\ 1 & 5 & 9\\ 2 & 6 & 5 \end{bmatrix}$$$$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$
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 $$A\mathbf v=\lambda\mathbf v$$.