Activity for Josh Hyattâ€
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #285978 |
How is the constraint $T\vec{x}\wedge\vec{y}=0$ converted into a set of linear constraints? For $m=2$, one can use the general form of the area $A$ of a parallelogram defined by two vectors $\vec{u}$ and $\vec{v}$: $A=\sqrt{|\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2}$. Substituting $A=0$, this qu... (more) |
— | over 2 years ago |
| Comment | Post #285978 |
If $m$ is 2, the direction constraint simplifies to a dot product which can be represented as a single linear expression. On the other hand, if $m$ is 3, the wedge product can be replaced with a cross product, but I don't see how that could be represented with fewer than 3 linear equations. I'd guess... (more) |
— | almost 3 years ago |
| Comment | Post #285977 |
Nice catch, thanks for pointing it out. (more) |
— | almost 3 years ago |
| Edit | Post #285977 |
Post edited: Fix dimension of vector y |
— | almost 3 years ago |
| Edit | Post #285977 | Initial revision | — | almost 3 years ago |
| Question | — |
Maximize Independent Variable of Matrix Multiplication Let $T$ be an $m\times n$ matrix with column vectors $\vec{Ti}$: $$ \vec{Ti}=\begin{bmatrix} \vec{T1} & ... & \vec{Tn} \end{bmatrix} $$ Let $\vec{x}$ be an unknown $n$-element vector: $$ \vec{x}=\begin{bmatrix} x1 \\\\ \vdots \\\\ xn \end{bmatrix} $$ Suppose ... (more) |
— | almost 3 years ago |