What is the meaning of (A - λI)v = 0?

1. A and λI act identically on v. 2. when you apply A to v, the result doesn't change direction. it's just scaled by λ.

Define a singular matrix

1. not invertible 2. determinant = 0, meaning it collapses some direction to 0

why must A - λI be singular for an eigenvalue?

Because ofr a nonzero v to satisfy (A - λI)v = 0, that matrix must have a nontrivial null space and thus a determinant of 0.

What is the bisection method used for?

Finding zeroes of a continuous function by repeatedly halving an interval where the function changes sign

When does the bisection method not work?

if f(0) and f(1) have the same sign the algorithm may never start, which will always happen if there are an even amount of eigenvalues within the interval

How many bisection steps for n-digit accuracy?

log_2(10^n) steps

Why are eigenvalues the ones that make A - λI singular?

Because only for those λ's does Av = λv have a nonzero solution v

Why does symmetry of a matrix A matter computationally?

It guarantees real eigenvalues and allows orthogonal tridiagonalization.

what happens to non-eigenvectors when you apply a matrix (linear transformation)?

they usually change direction.

Why do symmetric matrices always have real eigenvalues?

Because for symmetric A, you can show v^T Av is always real, and the quadratic form guarantees real stretching, not rotation.

What does the eigenvalue tell you geometrically?

The scale factor by which the transformation stretches or compresses space in that eigenvector's direction

What does it mean geometrically if det(A - λI) = 0?

It means at that λ, the transformation A−λI flattens space in at least one direction. that direction is the eigenvector.

What happens if a matrix has fewer than n distinct eigenvalues?

Some directions share the same scaling factor; the matrix isn't fully diagonalizable if it lacks enough independent eigenvectors.

Why not just compute the characteristic polynomial and find its roots directly?

It's numerically unstable for large matrices.

Why do we use the sign of the determinant in the bisection method?

Because sign changes indicate where the determinate(A−λI) crosses zero and where A−λI becomes singular

What does it mean to find the critical points?

Set the partial derivatives to 0 and then solve for them

Determinant trace test criteria

If D > 0 and T > 0 local minimum If D > 0 and T < 0 local maximum If D < 0 saddle

Determinant trace test steps to get D and T

Get fxx, fxy and fyy and then D = ac - b^2 and T = a + c