Learning Resources
- $\begin{bmatrix}\hat{i}_x & \hat{j}_x \\ \hat{i}_y & \hat{j}_y\end{bmatrix}
\begin{bmatrix}x \\ y\end{bmatrix}
= f\left( \begin{bmatrix}x \\ y\end{bmatrix} \right)$, where $\begin{bmatrix}\hat{i}_x \\ \hat{i}_y\end{bmatrix}$ is where the basis vector $\hat{i}$ lands after the transformation, and $\begin{bmatrix}\hat{j}_x \\ \hat{j}_y\end{bmatrix}$ is where the basis vector $\hat{j}$ lands.
- $det\left(\begin{bmatrix}v_x & u_x \\ v_y & u_y \end{bmatrix}\right)
=
\text{Area of parallelogram spanned by } \vec{v} \text{ and } \vec{u}$
- More accurately, the determinant is the increase in scale of the area (from the area of parallelogram formed by basis vectors $\hat{i}$ and $\hat{j}$). Since the basis vectors are unit vectors, that area is always 1.
- Matrix $A$ with $n$ columns will have a Rank $r$ (dimension of the span of all possible output vectors transformed by $A$) + nullity $X$, where $X=n-r$.
- All possible vectors in $n$-dimension space must either end up in the rank space or the nullspace.
- https://www.youtube.com/watch?v=DA-iQkSSjKw
- Duality: $\begin{bmatrix}
a \\
b
\end{bmatrix}
\cdot
\begin{bmatrix}
c \\
d
\end{bmatrix}
=
\begin{bmatrix}
a & b
\end{bmatrix}
\begin{bmatrix}
c \\
d
\end{bmatrix}
= \text{A scalar value}$
- The Dot Product applies Linear Transformation $\begin{bmatrix}a & b\end{bmatrix}$ to an input vector $\begin{bmatrix}c \\ d\end{bmatrix}$ and outputs a scalar value (single number on the 1D number line).
- If you flip this transformation vector $\begin{bmatrix}a & b\end{bmatrix}$ on its head to become $\begin{bmatrix}
a \\
b
\end{bmatrix}$, you get the dual vector of this transformation in the original (pre-transformation) plane.
- $\vec{i}=\text{input vector, and } \vec{d}=\text{dual vector} \\ \vec{i} \cdot \vec{d} = X \\
\text{where } X= \text{the number on the number line (1D scalar) which the input vector lands on after the transformation}$
- We derive the Cross Product formula using the determinant. But why is the determinant present in this formula?
- $\det(A)$ = Volume of parallelepiped spanned by the column vectors of $A$, when $A$ has 3 columns (3D).
- Volume = Area $\times$ Height.
- This applies even to parallelepipeds. Imagine a slanted stack of books of equal area, and pushing them so that they are stacked upright. Always V = Base Area $\times$ Height.
- The diagram to the right is a variant of the Cross Product formula using determinants.
-
What vector $\vec{p}$, when dot-producted with an input vector $\vec{u} =
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}$, will yield the volume of the parallelepiped spanned by that input vector and two fixed vectors $\vec{v}$ and $\vec{w}$?
-
Recall that the dot product takes the projection of one vector onto another (so that we only keep the components of both vectors which are in the direction of only one of the vectors) and multiple the magnitudes together.
-
To find the volume of the parallelepiped, $\det(A)$, we should take the area of the parallelogram and multiply it by the height.
- $\det(A) = \text{Height of parallelogram} \cdot \text{Area of parallelogram}$.
- Let’s define a new vector $\vec{c}$, which is a vector which is perpendicular to the parallelogram, and $|\vec{c}|=1$.
- $\det(A) = (\vec{c} \cdot \vec{u} ) \cdot \text{Area of parallelogram}$
- Note that $(\vec{c} \cdot \vec{u})=\text{Height of parallelepiped}$.
- We are taking only the component of $\vec{u}$ which is perpendicular to the parallelogram, by definition of the dot product. This is because $\vec{c}$ is a unit vector which is perpendicular to the parallelogram.
-
However, if instead of $|\vec{c}|=1$, we set $|\vec{c}| = \text{Area of parallelogram}$.
-
We now have $\det(A) = \vec{c} \cdot \vec{u}$, because the magnitude multiplication step of taking the dot product will include the length of $\vec{c}$, which is the area of the parallelogram.
-
$\vec{v}\times\vec{w}=\vec{c}$, where $|\vec{c}|=\text{Area of parallelogram}$, and $\vec{c}$ is perpendicular to the parallelogram.
