$\newcommand{\identity}{\mathrm{id}} \newcommand{\notdivide}{{\not{\mid}}} \newcommand{\notsubset}{\not\subset} \newcommand{\lcm}{\operatorname{lcm}} \newcommand{\gf}{\operatorname{GF}} \newcommand{\inn}{\operatorname{Inn}} \newcommand{\aut}{\operatorname{Aut}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\cis}{\operatorname{cis}} \newcommand{\chr}{\operatorname{char}} \newcommand{\Null}{\operatorname{Null}} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&}$

## Section4.3Week 9 (Jul 23-27)

Last week we were introduced to the concept of eigenvectors: vectors which, when a linear transformation acts upon them, are transformed simply into (scalar) multiples of their old selves. These are vectors that, for a given matrix $A\text{,}$ satisfy the eigenvalue-eigenvector equation (4.2.1).

So, how do we find such vectors? That is the task this week, and it can involve some subtlety. Solving $A{\bf v} = \lambda {\bf v}$ is typically a two-phase process, and one that begins after subtracting $\lambda{\bf v}$ from both sides of the equation to obtain the version that we can use our linear solving techniques to investigate:

\begin{align} \bigl( A - \lambda I \bigr) {\bf v} \amp=\amp {\bf 0}.\tag{4.3.1} \end{align}

Then, we insist that (4.3.1) have nontrivial solutions ${\bf v}\text{,}$ in other words, something besides the zero vector must satisfy this equation in order for the equation to be interesting. (After all, (4.2.1) is always satisfied by ${\bf v} = {\bf 0}$ regardless of the value of $\lambda\text{.}$) But the only way this can happen is if there exists a nontrivial linear combination of the columns of $(A-\lambda I)$ which equals zero. In other words, we want for the columns of $(A-\lambda I)$ to not be linearly independent. This means that we want $(A-\lambda I)$ not to be an invertible matrix! And that kicks off our two-step process.

1. Find the eigenvalues $\lambda$ by insisting that $(A-\lambda I)$ be not invertible. This means that the determinant of this matrix -- which depends on $\lambda$ -- must equal zero. Solving this characteristic equation will determine the eigenvalues: \begin{align} \det (A-\lambda I) \amp=\amp 0\tag{4.3.2} \end{align}
2. For each eigenvalue $\lambda\text{,}$ find its eigenvector(s) by finding the parametric vector solution of (4.3.1). As long as your eigenvalues $\lambda$ are correct, there will always be at least one free variable! So you'll want to express your answer as a basis of solutions.

To do this week:

NOTE: Portfolio 4 will be combined with Portfolio 3. Continue to keep copies of your best work on each learning standard to prepare for submission of a cumulative portfolio (containing all of Chapters 1-4) upon Portfolio 4's due date.

By Tuesday 7/24:

1. Read and annotate 4.2: Finding eigenvalues and eigenvectors https://via.hypothes.is/http://merganser.math.gvsu.edu/david/linear.algebra/ula/ula/sec-eigen-find.html

By Wednesday 7/25:

1. Watch for Lecture 9 here.
2. Submit Quiz 8R via Blackboard.
3. Submit Quiz 9 via Blackboard.

By Friday 7/27:

1. Submit Quiz 9R via Blackboard.