[Linear Algebra] 13. Eigenvectors and eigenvalues

Eigenvectors and eigenvalues

Most vectors knocked off their span(=line) during transformation. But some special vectors remain on their line after transformation. It means that transformation just stretches or squishes vectors. Moreover, any other vector in that line also stretch or squish by a same scalar. These special vector are called the "eigenvectors" of that transformation. And each eigenvector has "eigenvalue", which is factor that stretches or squishes during transformation.

$$
A\mathbf{\vec{v}} = \lambda\mathbf{\vec{v}}
$$

 

We could understand transformation by using "matrix". But finding "eigenvectors" and "eigenvalues" can be the better way to help understanding "what linear transformation actually do".

---

To get "eigenvectors" and "eigenvalues", re-organize formula:

$$
A\mathbf{\vec{v}} = (\lambda I)\mathbf{\vec{v}}
\longrightarrow
(A - \lambda I)\mathbf{\vec{v}} = \mathit{0}
$$

 

There are two type of solution. 1). This always be true if $\mathbf{\vec{v}}$ is zero vector. 2). If $\mathbf{\vec{v}}$ is non-zero vector, solution only exists when transformation $A - \lambda I$ squishes into lower dimension. So the goal is to find a value of $\lambda$ that makes this determinant $0$.

Eigenbasis

A set of basis vector, which are also eigenvectors, is called "eigenbasis".

 

If linear transformation is described by diagonal matrix, way to interpret this matrix is that all the basis vector are eigenvectors and diagonal entries of this matrix are eigenvalues.

 

It's very easy to compute if we multiply diagonal matrix bunch of times:

$$
\begin{bmatrix}
a & 0 \\
0 & b \\
\end{bmatrix}^{n}
\begin{bmatrix} x \\ y \end{bmatrix}
=
\begin{bmatrix}
a^n & 0 \\
0 & b^n \\
\end{bmatrix}
\begin{bmatrix} x \\ y \end{bmatrix}
$$

 

So it's lucky to have eigenbasis. If you have to compute some transformation multiple times and have eigenbasis, 1). translate $\begin{bmatrix} x_{standard} \\ y_{standard} \end{bmatrix}$ to $\begin{bmatrix} x_{new} \\ y_{new} \end{bmatrix}$ by $V^{-1}$, 2). compute transformation multiple times by $V^{-1}AV=\Lambda$ and 3) translate $\begin{bmatrix} x_{new} \\ y_{new} \end{bmatrix}$ to $\begin{bmatrix} x_{standard} \\ y_{standard} \end{bmatrix}$ by $V$. Reasoning for doing this is that $V^{-1}AV=\Lambda$ is diagonal. Because new coordinate system was established by eigenbasis. In other words, all basis vector of new coordinate system are eigenvectors.