MATH 240

Wed. November 13th, 2019

Recall that knowing eigenvalues and their corresponding eigenvectors of AA gives us an idea of what happens to those vectors under repeated application of the transformation associated with AA. (They’ll just be scaled by their corresponding eigenvalue each time.)

What sort of effect does a matrix AA with no real eigenvalues have if it is "applied repeatedly" to a point x0x_0 in Rn\R^n? (x0, Ax0, A2x0, A3x0, ...x_0,\ Ax_0,\ A^2x_0,\ A^3x_0,\ ...)

Note. For convenience we will only consider 2×22\times 2 matrices.

We can essentially represent each of these transformations as a scaling and rotating of x0x_0 by constant values.

The Factor Theorem – Any polynomial of degree nn with complex coefficients has nn complex roots (counting multiplicities).
Q(λ)=k1λn+k2λn−1...+knλ0=(λ−λ1)(λ−λ2)...(λ−λn)Q(\lambda)=k_1\lambda^n+k_2\lambda^{n-1}...+k_n\lambda^0=(\lambda-\lambda_1)(\lambda-\lambda_2)...(\lambda-\lambda_n)

A 2×22\times 2 matrix AA will always have eigenvalues if you consider it as a matrix over the complex numbers. We can treat AA as a transformation from C2\cnums^2 to C2\cnums^2.

Ex.
A=(0−110)det(A−λI)=∣−λ−11−λ∣=λ2+1=0λ=i,−i\begin{aligned} A&=\begin{pmatrix} 0&-1\\1&0 \end{pmatrix}\\ \text{det}(A-\lambda I)&=\begin{vmatrix} -\lambda&-1\\1&-\lambda \end{vmatrix}\\ &=\lambda^2+1=0\\ \lambda&=i,-i \end{aligned}(0−110)(1−i)=(i1)=i(1−i)\begin{pmatrix} 0&-1\\1&0 \end{pmatrix}\begin{pmatrix} 1\\-i \end{pmatrix}=\begin{pmatrix} i\\1 \end{pmatrix}=i\begin{pmatrix} 1\\-i \end{pmatrix} Thus v1=(1−i)v_1=\begin{pmatrix} 1\\-i \end{pmatrix} is an eigenvector that corresponds to λ=i\lambda=i.(0−110)(1i)=(−i1)=−i(1i)\begin{pmatrix} 0&-1\\1&0 \end{pmatrix}\begin{pmatrix} 1\\i \end{pmatrix}=\begin{pmatrix} -i\\1 \end{pmatrix}=-i\begin{pmatrix} 1\\i \end{pmatrix} Thus v2=(1i)v_2=\begin{pmatrix} 1\\i \end{pmatrix} is an eigenvector that corresponds to λ=−i\lambda=-i.

Ex.
A=(.5−.6.751.1)det(A−λI)=λ2−1.6λ+1λ=1.6±(1.6)2−42=.8±.6i\begin{aligned} A&=\begin{pmatrix} .5&-.6\\.75&1.1 \end{pmatrix}\\ \text{det}(A-\lambda I)&=\lambda^2-1.6\lambda+1\\ \lambda&=\frac{1.6\pm \sqrt{(1.6)^2-4}}{2}\\ &=.8\pm.6i \end{aligned}

Now we look at (A−λI)x=0(A-\lambda I)x=0 and split each row into its own equation.
(−.3+.6i)x1−.6x2=0.75x1+(.3+.6i)x2=0\begin{gathered} &(-.3+.6i)x_1-.6x_2=0\\ &.75x_1+(.3+.6i)x_2=0 \end{gathered}.75x1=−(.3+.6i)x2x1=(−.4−.8i)x2\begin{aligned} .75x_1&=-(.3+.6i)x_2\\ x_1&=(-.4-.8i)x_2 \end{aligned}
One of these will be a free variable. We can choose x2=5x_2=5 to find our eigenvectors:
v1=(−2−4i5)(for λ=.8−.6i)v2=(−2+4i5)(for λ=.8+.6i)\begin{aligned} v_1&=\begin{pmatrix} -2-4i\\5 \end{pmatrix}\text{(for $\lambda=.8-.6i$)}\\ v_2&=\begin{pmatrix} -2+4i\\5 \end{pmatrix}\text{(for $\lambda=.8+.6i$)} \end{aligned} If you know that a vector vv is an eigenvector for a complex eigenvalue λ\lambda, then the conjugate of vv is an eigenvector for the conjugate of λ\lambda. (The conjugate is obtained by Re(x)−Im(x)\text{Re}(x)-\text{Im}(x))