# Thursday, March 28, 2013

Maintainer: admin

## 1The square root of a diagonalizable matrix¶

e.g. $P^{-1}AP=\begin{pmatrix}0 & 0 \\ 0 & 2\end{pmatrix}$

Set $B = \begin{pmatrix}0 & 0 \\ 0 & \sqrt{2}\end{pmatrix}$, then $B^2 = \begin{pmatrix}0 & 0 \\ 0 & 2\end{pmatrix} = P^{-1}AP$

Look at $(PBP^{-1})^2 = (PBP^{-1})(PBP^{-1})=PB(P^{-1}P)BP^{-1} = PB^2P^{-1} = P(P^{-1} AP)P^{-1} = A$.

e.g. $P^{-1}AP = \begin{pmatrix}4 & 0 \\ 0 & 9\end{pmatrix}$. Try one of the four possible $B = \begin{pmatrix}\pm 2 & 0 \\ 0 & \pm 3\end{pmatrix}$:

$$B_1 = \begin{pmatrix}- 2 & 0 \\ 0 & -3\end{pmatrix} \qquad B_2 = \begin{pmatrix}- 2 & 0 \\ 0 & 3\end{pmatrix} \qquad B_3 = \begin{pmatrix} 2 & 0 \\ 0 & -3\end{pmatrix} \qquad B_4 = \begin{pmatrix}2 & 0 \\ 0 & 3\end{pmatrix}$$

So $PB_jP^{-1}$ is a square root of $A$.

e.g. square root of $a+bi \qquad a,b \in \mathbb{R}$

\begin{aligned} (c+di)^2 = a+bi \\ (c^2 - d^2) + i(2cd) = a+bi\end{aligned}

Need $\begin{cases} c^2 - d^2 = a \\ 2cd = b \end{cases}$ so for $b\neq 0, d= \frac{b}{2c}$

Subtitute into a:

\begin{aligned} a &= c^2 - d^2 = c^2 - \frac{b^2}{4c^2} \\ ac^2 &= c^4 - \frac{b^2}{4} \qquad \text{quadratic in }\mathbb{C}^2 \\ c^4 - ac^2 - \frac{b^2}{4} &= 0 \\ c^2 &= \frac{-(-1) \pm \sqrt{(-a)^2 - 4(\frac{-b^2}{4})}}{2} \\ c^2 &= \frac{+a \pm \sqrt{a^2 + b^2}}{2} \\ c &= \pm \sqrt{\frac{a + \sqrt{a^2 + b^2}}{2}}\end{aligned}

Definition A real matrix is skew symmetric iff $A^T = -A$

$$\bar{(iA)^T} = \bar{i} A^T = -i(-A) = iA$$

$iA$ is hermitian so $iA$ has real eigenvalues, thus $A$ has purely complex eigenvalues $(0+ib)$.

e.g. eigenvalues of $\begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}$

$$\begin{vmatrix}\lambda - 0 & 1 \\ -1 & \lambda - 0 \end{vmatrix} = \lambda^2 + 1 \qquad \lambda = \pm i$$

$$\begin{pmatrix}i & -1 \\ 1 & i\end{pmatrix}\begin{pmatrix}\alpha \\ \beta \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}$$

$$\begin{cases}i\alpha - \beta = 0 \\ \alpha + i\beta = 0\end{cases} \beta = i \alpha$$

$\begin{pmatrix}1 \\ i\end{pmatrix}$ would do for $\lambda_1 = i$

—-

$$\begin{pmatrix}-i & -1 \\ 1 & -i\end{pmatrix}\begin{pmatrix}\gamma \\ \delta \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}$$

$$\begin{cases}-i\gamma - \delta = 0 \\ \gamma - i\delta = 0\end{cases} \gamma = i \delta$$

$\begin{pmatrix}i \\ 1\end{pmatrix}$ would do for $\lambda_2 = -i$

Recall that $<\begin{pmatrix}1 \\ i\end{pmatrix},\begin{pmatrix}i \\ 1\end{pmatrix}> = (1,i)\begin{pmatrix}-i \\ 1\end{pmatrix} = -i + i = 0 \therefore \mathbf{v_1} \perp \mathbf{v_2}$etc.

## 2Quadratic form over $\mathbb{R}$¶

e.g. $q(\mathbf{v}) = 2x^2 + 2xy + 2y^2 = (x,y) \begin{pmatrix}2 & 1 \\ 1 & 2\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}$

$A = A^T = \begin{pmatrix}2 & 1 \\ 1 & 2\end{pmatrix}$ real eigenvalues

Change of basis $\begin{pmatrix}x \\ y\end{pmatrix} P \begin{bmatrix}X \\ Y\end{bmatrix} \qquad (x,y) = [X,Y]P^T$

$q(\mathbf{v}) = (x,y)A\begin{pmatrix}x \\ y\end{pmatrix} = [X,Y]P^T AP\begin{bmatrix}X \\ Y\end{bmatrix}$

If we can find an orthogonal change of basis $(P^T = P^{-1})$ then $P$ ...

Eigenvalues of $A = \begin{pmatrix}2 & 1 \\ 1 & 2\end{pmatrix}$

\begin{aligned} 0 &= \begin{vmatrix}\lambda I - A \end{vmatrix} = \begin{vmatrix} \lambda -2 & -1 \\ -1 & \lambda -2\end{vmatrix}= (\lambda -2)^2 - 1 \\ 0 &= (\lambda -2)^2 - 1^2 = ((\lambda -2)-1)((\lambda -2)+1) \\ 0 &= (\lambda -3)(\lambda -1)\end{aligned}

$\begin{cases}\lambda_1 = 3 \lambda_2 = 1\end{cases}$ eigenvalues

$3X^2 + Y^2$ is $q$ in the new system,

Eigenvalue $\lambda_1 = 3$

$\begin{pmatrix} 3-2 & -1 \\ -1 & 3-2 \end{pmatrix} \begin{pmatrix}\alpha \\ \beta\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix} \begin{cases} \alpha - \beta = 0 \\ -\alpha + \beta = 0 \end{cases} so \alpha = \beta \begin{pmatrix}1 \\ 1\end{pmatrix}$ will do

Note $<\begin{pmatrix}1 \\ 1\end{pmatrix},\begin{pmatrix}-1 \\ 1\end{pmatrix}> = 0$ orthogonal basis of eigenvectors

$\|\begin{pmatrix}1 \\ 1\end{pmatrix}\|^2 = 2 =\|\begin{pmatrix}-1 \\ 1\end{pmatrix}\|^2$

Corresponding orthonormal basis:

$\{\begin{pmatrix}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{pmatrix}, \begin{pmatrix}-\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{pmatrix}\}$

But another can be

$\{\begin{pmatrix}\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\end{pmatrix}, \begin{pmatrix}\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\end{pmatrix}\}$ and two more candidates...

We may take $P = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{pmatrix} = (P^{-1})^T$

Note $det P = \pm 1$ (rotation)

Claim $\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{pmatrix} \begin{pmatrix} 2 & 1 \\ 1 & 2\end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{pmatrix} = P^{-1}AP$

$\begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{pmatrix} \begin{bmatrix} X \\ Y \end{bmatrix}=\begin{bmatrix} \frac{X-Y}{\sqrt{2}} \\ \frac{X+Y}{\sqrt{2}} \end{bmatrix}$

$2x^2 + 2xy + 2y^2 = 2 (\frac{X-Y}{\sqrt{2}})^2 + 2(\frac{X-Y}{\sqrt{2}})(\frac{X+Y}{\sqrt{2}})+2(\frac{X+Y}{\sqrt{2}}) = [X^2 - 2XY + Y^2] + [X^2 - Y^2] + [X^2 + 2XY + Y^2] = 3X^2+Y^2$ as predicted.

e.g. n = 3

$A= \begin{pmatrix} 1&1&1 \\ 1&1&1 \\ 1&1&1 \end{pmatrix} = A^T$ (real eigenvalues)

$0 = |A -\lambda I| = \begin{vmatrix} 1-\lambda&1&1 \\ 1&1-\lambda&1 \\ 1&1&1-\lambda \end{vmatrix} = \begin{vmatrix} 1-\lambda&1&1 \\ 0&-\lambda&\lambda \\ 1&1&1-\lambda \end{vmatrix} = \lambda\begin{vmatrix} 1-\lambda&1&1 \\ 0&-1&1 \\ 1&1&1-\lambda \end{vmatrix} = \lambda [(1-\lambda) \begin{vmatrix}-1 & 1 \\ 1 & 1 -\lambda \end{vmatrix} - 0 \begin{vmatrix}- & - \\ - & -\end{vmatrix} + 1 \begin{vmatrix}1 & 1 \\ -1 & 1 \end{vmatrix}] = \lambda [(1-\lambda)(1-(1-\lambda) - 1) + 2]$ ... more work etc.

You end up with two distinct eigenvalues: $\begin{cases} \lambda_1 = 3 \\ \lambda_2 = 0 \\ \lambda_3 =0 \end{cases}$

Case $\lambda_2 = \lambda_3 = 0$

$\begin{pmatrix}1&1&1 \\ 1&1&1 \\ 1&1&1 \end{pmatrix} \begin{pmatrix}a \\ b\\ c\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}$

$\{\begin{pmatrix}0 \\ 1 \\ -1\end{pmatrix}, \begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix} \}$ basis eigenspace for $\lambda_2 = \lambda_3 = 0$

Basis so far $\{\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}, \begin{pmatrix}0 \\ 1 \\ -1\end{pmatrix}, \begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix} \}$. To normalise the basis, apply G-S.