A matrix is upper-triangular if the entries below the diagonal are \(0\):

\begin{equation} \mqty(\lambda_{1} & & * \\ & \ddots & \\ 0 & & \lambda_{n}) \end{equation}

properties of upper-triangular matrix

Suppose \(T \in \mathcal{L}(V)\), and \(v_1 … v_{n}\) is a basis of \(V\). Then:

1. the matrix of \(T\) w.r.t. \(v_1 … v_{n}\) is upper-triangular
2. \(Tv_{j} \in span(v_1 \dots v_{j})\) for each \(v_{j}\)
3. \(span(v_{1}, … v_{j})\) is invariant under \(T\) for each \(v_{j}\)

\(1 \implies 2\)

Recall that our matrix \(A=\mathcal{M}(T)\) is upper-triangular. So, for any \(v_{j}\) sent through \(A\), it will be multiplied to the $j$-th column vector of the matrix. Now, that $j$-th column has \(0\) for rows \(j+1 … n\), meaning that only through a linear combination of the first \(j\) vectors we can construct \(T v_{j}\). Hence, \(Tv_{j} \in span(v_1 … v_{j})\)

Gah I have to do this. Not for public consumption. California laws 2022 DL600 R7 2022.

Consequences

Not licensed

• If unlicensed person is drivnig your car, it maybe impounded for 30 days
• Hired to drive interstate commercially need to be older than 21, also need to be older than 21 to transport hazardous materials

Class C License

Driving #knw

• Two axle vehicle with a GVWL of 26,000 lbs or less
• Three axle vehicle weighing 6,000 lbs or less
• House car < 40 feet or less
• Three wheel motocycles
• Vanpool vehicle designed to carry between 10 and no more than 15 people

Towing #knw

• Single vehicle of 10,000 or less
• Vehicle weighing 4000 lbs or more unladen
  • Trailer coach under 10,000 lbs
  • Fifth wheel trailer exceeding 10,000 lbs but under 15,000 lbs, with endorsement

Mor ethings

• Class C drivers can’t tow more than one
• Motor vehile weigning under 4000 lbs cannot tow more than 6000 lbs

Getting in trouble

• Get a traffic ticket and fail to show up to court: suspend driving
• One at fault collision or one at fault traffic violation: may take action?
• Two of either at fault collision or violation conviction: no driving for 30 days unless accompanied by 25 year old adult
• Three of “”: no driving for 6 months, on probation for a year.
• Drugs or alcohol between 13-21: suspension for a year

Minor driving

Not sure if this applies

\begin{align} \min_{x}\ &c^{\top} x + d \\ s.t.\ &Gx \preceq h \\ & Ax = b \end{align}

• linear objective function
• linear constraints

single our inequality forms a half-space; the entire feasible set is denoted by a series of linear functions—-these linear equalities are each CONVEX. The resulting feasible set, then, is ALSO convex—-meaning any line within the set remains within the set. So, any local minimum is a global minimum.

This is a convex problem where all constrains and objectives are affine.

Let \(X\) denote price and \(Y\) denote volatility. The two objects obey the following process:

\begin{equation} \begin{cases} \dd{X} = \mu X \dd{t} + XY \dd{W} \\ \dd{Y} = \sigma Y \dd{B} \end{cases} \end{equation}

where, \(W\) and \(B\) are correlated Brownian motions with correlation \(\rho\) — \(E[(\dd{W})(\dd{B})] = \rho \dd{t}\).

Let’s work with \(Y\) first. We understand that \(Y\) is some continuous variable \(e^{a}\). Therefore, \(\dv{Y}{t}=ae^{a}\). Therefore, \(dY = ae^{a}dt\). Finally, then \(\frac{\dd{Y}}{Y} = \frac{ae^{a}}{e^{a}}\dd{t} = a\).

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