Class 03/22/21

Unit 3 Continued

Eigenvectors & Eigenvalues

Let A be an $N \times N$ matrix $v \neq = 0$ is an eigenvector of A if $A v = λ v$ for some scalar $λ$ , which is the eigenvalue of $v .$

Finding Eigenvectors

Suppose $v \neq = 0$ is an eigenvector of A.

A v = λ v A v - λ v = v v (A - λ I_{N}) = 0

Therefore, $A - λ I$ is singular, i.e, non-invertible.

For 2x2 Matrix

A = [a c b d]

We know A is singular if and only if $a d - b c = 0$ .

A^{- 1} = \frac{1}{a d - b c} [d - c - b a]

d e t ([a c b d]) = a d - b c

If $v$ is an eigenvector of A, with Eigen value $λ$ then det $(A - λ I) = 0$

Example

Find the eigenvalues and eigenvectors.

A = [1331] A - λ I = [1331] - [1001] = [1 - λ 3 3 1 - λ] d e t (A - λ I) = (1 - λ)^{2} - 9 λ^{2} - 2 λ + 1 = 9 (λ - 4) (λ + 2) = 0 eigenvalues: λ_{1} = 4, λ_{2} = - 2 Now find coresponding eigenvectors: We need x of (A - λ_{1} I) x = 0 Substitute in ([1331] - [4004]) [x_{1} x_{2}] = 0 ([- 3 3 3 - 3]) [x_{1} x_{2}] = 0 reduce (for 2x2 this is not neccesary) [- 3 0 30] - x_{1} + x_{2} = 0 E_{λ_{1}} = s p an ([11]) The other solution is the same, but with -2 ([1331] - (- 2) [1001]) [x_{1} x_{2}] = 0 [3333] [x_{1} x_{2}] = 0 x_{1} + x_{2} = 0 E_{λ_{2}} = s p an ([1 - 1])

Complex Numbers Example

Complex numbers have the form $a + bi$ where a and b are real numbers.

$a + bi + c + d i = (a + c) + (b + d) i \leftarrow$ Treat imaginaries like variables

All complex numbers follow the rules of algebra.

$(a + bi) (c + d i) = a c + a d i + b c i + b d (i)^{2} \leftarrow$ $i^{2}$ is a real number = $- 1$ .

Division of complex numbers

$\frac{a + bi}{c + d i} \cdot \frac{a - d i}{a - d i} = \frac{a c - a d i + b c i - b d ( i ) ^{2}}{c ^{2} - c d i + c d i - d ^{2} ( i ^{2} )} = \frac{( a c + b d ) + ( b c - a d ) i}{c ^{2} + d ^{2}}$

This shows the bottom is now real.

$\frac{1 + i}{2 - i} \cdot \frac{2 + i}{2 + i} = \frac{2 + i + 2 i - 1}{4 + 1} = \frac{1}{5} + \frac{3}{5} i$