User:J D Bowen/Math725 HW10

1) We aim to show that if $M \ $ is an $m\times n \ $ matrix with linearly independent columns, then $M^HM \ $ is invertible.  Observe $M^H \ $ is a $n\times m \ $ matrix with linearly independent rows, and so $M^HM \ $ is an $n\times n \ $ matrix with $\text{rank}(M^HM)=\text{min}(\text{rank}(M),\text{rank}(M^H)) = n \ $.  So $M^HM \ $ is invertible.

2) Let $A = \begin{pmatrix} x_1 & 1 \\ \vdots & \vdots  \\ x_n & 1 \end{pmatrix}$, $x=(a,b)^t \ $, and $y=(y_1,\dots,y_n)^t \ $.

Observe that $Ax=y \ $ expands to

$ \begin{pmatrix} x_1 & 1 \\ \vdots & \vdots \\ x_n & 1 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} y_1 \\ \vdots \\ y_n \end{pmatrix} \ $

which is the system of equations $ax_j+b=y_j, \ 1\leq j \leq n \ $. Since this is the solution of the data points, the best fit line will be the best approximate solution of this matrix equation.

Observe $x=(A^HA)^{-1} A^Hy =\left({ \begin{pmatrix} x_1 & \dots & x_n \\ 1 & \dots &  1 \end{pmatrix} \begin{pmatrix} x_1 & 1 \\ \vdots & \vdots  \\ x_n & 1 \end{pmatrix}}\right)^{-1} \begin{pmatrix} x_1 & \dots & x_n \\ 1 & \dots  &  1 \end{pmatrix} \begin{pmatrix} y_1 \\ \vdots \\ y_n \end{pmatrix} $

$= \begin{pmatrix} \Sigma x_j^2 & \Sigma x_j \\ \Sigma x_j & n \end{pmatrix}^{-1} \begin{pmatrix} \Sigma x_jy_j \\ \Sigma y_j \end{pmatrix}

= \frac{1}{ (n\Sigma x_j^2)-(\Sigma x_j)^2 } \begin{pmatrix} n & -\Sigma x_j \\ -\Sigma x_j & \Sigma x_j^2 \end{pmatrix} \begin{pmatrix} \Sigma x_jy_j \\ \Sigma y_j \end{pmatrix}

= \frac{1}{ (n\Sigma x_j^2)-(\Sigma x_j)^2 } \begin{pmatrix} n\Sigma x_j y_j-(\Sigma x_j)(\Sigma y_j) \\ -(\Sigma x_j)(\Sigma x_jy_j)+(\Sigma x_j^2)(\Sigma y_j) \end{pmatrix} $

and so we have

$a=\frac{ n\Sigma x_j y_j-(\Sigma x_j)(\Sigma y_j)}{(n\Sigma x_j^2)-(\Sigma x_j)^2} \ $

as desired.

3) Given $T:V\to V \ $, we aim to show that $\lambda \ $ is an eigenvalue if and only if $\text{rank}(T-\lambda I) < \text{dim}(V) \ $.

Begin by observing that $\text{rank}(T-\lambda I) < \text{dim}(V) \ $ if and only if the columns of $\text{rank}(T-\lambda I) \ $ are not linearly independent. This is true if and only if $T-\lambda I \ $ has a non-trivial kernel, ie, there is some vector $v \neq 0 \ $ such that $(T-\lambda I)v=0 \ $. But this is true if and only if $Tv=\lambda I v = \lambda v \ $, which is equivalent to $\lambda \ $ being an eigenvalue of $T \ $.

Suppose $\lambda \ $ is an eigenvalue of $T \ $. Then we can show $\overline{\lambda} \ $ is an eigenvalue of $T^H \ $ with the following equations:

$\lambda \langle v,v, \rangle =\langle \lambda v, v \rangle = \langle Tv, v \rangle = \langle v, T^H v \rangle \ $

and

$\lambda \langle v,v, \rangle = \langle v, \overline{\lambda}v \rangle \ $

Hence

$\langle v, \overline{\lambda}v \rangle=\langle v, T^H v \rangle \ $.

and so $\overline{\lambda} \ $ is an eigenvalue of $T^H \ $. Since the conjugate of the conjugate of $\lambda \ $ is just $\lambda \ $, and $(T^H)^H = T \ $, this argument can be used to show the other direction of implication as well.

4) Suppose $U:V\to V \ $ satisfies $U^H = U^{-1} \ $, and that $\lambda \ $ is an eigenvalue.  Then $\lambda ^{-1} \ $ is the corresponding eigenvalue for $U^{1} \ $, and for some vector $v, \ Uv=\lambda v \ $.

Observe

$\lambda \langle v,v, \rangle =\langle \lambda v, v \rangle = \langle Uv, v \rangle = \langle v, U^{-1} v \rangle = \langle v, \lambda^{-1} v \rangle = \overline{\lambda^{-1}}\langle v,v \rangle \ $

We have $\lambda=\overline{\lambda^{-1}} \implies |\lambda|=1 \ $.

5) Observe that $\frac{T+T^H}{2}+\frac{T-T^H}{2} = \frac{T+T^H+T-T^H}{2} = \frac{2T}{2} = T \ $. Now, for any two vectors $v,w \ $, we have

$\langle \frac{T+T^H}{2} v, w \rangle = \frac{1}{2}\langle Tv + T^H v, w \rangle = \frac{1}{2} \left({ \langle Tv,w \rangle +\langle T^H v, w \rangle }\right) = \frac{1}{2} \left({ \langle v,T^H w \rangle + \langle v, (T^H)^H w \rangle }\right) = \frac{1}{2} \langle v, (T^H+T)w \rangle \ $

$ = \langle v, \overline{\frac{1}{2}} (T+T^H)w \rangle = \langle v, \frac{T+T^H}{2}w\rangle \ $,

and so $\frac{T+T^H}{2} \ $ is Hermitian.

Further observe

$\langle \frac{T-T^H}{2} v, w \rangle = \frac{1}{2}\langle Tv - T^H v, w \rangle = \frac{1}{2} \left({ \langle Tv,w \rangle -\langle T^H v, w \rangle }\right) = \frac{1}{2} \left({ \langle v,T^H w \rangle - \langle v, (T^H)^H w \rangle }\right) = \frac{1}{2} \langle v, (T^H-T)w \rangle \ $

$ = \langle v, -\overline{\frac{1}{2}} (T^H-T)w \rangle = \langle v, \frac{-(T^H-T)}{2}w\rangle \ $

and so $\left({ \frac{T-T^H}{2} }\right)^H = - \frac{T-T^H}{2} \ $, meaning this is anti-Hermitian.

6) I can't get octave running.