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# Geometric Definition

The definition of a tangent space is the space of all vectors tangent to a surface at some point.

# Tangent Spaces of Images

MathJax TeX Test Page If you've heard of domain/range, an image is just like a range. It is the set of all c s.t. $f(x) = c$.

It turns out the span of the Jacobian is the tangent space to an image. In every direction $f_x$, the partial represents the instantaneous rate of change, so you can create a tangent line at that point using the partial. Therefore, in every direction, you have a tangent line. The tangent space is the span of all of these lines. $$\text{Span}\left(\begin{bmatrix}\frac{\partial f}{\partial x_1} & \frac{\partial f}{\partial x_2} & ... & \frac{\partial f}{\partial x_n} \end{bmatrix}\right)$$ Special case: $$f: \mathbb{R}^n \to \mathbb{R}$$ If this is the case, you can write $w = f(x,y,z) \to f(x,y,z) - w = 0$. In this case, it's the same type of problem as a level set.

# Example

MathJax TeX Test Page Let's think of a curve: $$f(t) = < t, t^2>$$ The tangent space is $\text{Span}\left(\begin{bmatrix}\frac{\partial f_1}{\partial t} \\ \frac{\partial f_2}{\partial t}\end{bmatrix}\right) = \text{Span}\left(\begin{bmatrix}1 \\ 2t\end{bmatrix}\right)$. So, it's a line with that slope. That agrees with what we expect for $y = x^2$.

Now, let's imagine a unit sphere: $$f(\theta, \phi) = < \cos(\theta)\sin(\phi), \sin(\theta)\sin(\phi), \cos(\phi)>$$ You can find how I got this here. Now, this is our Jacobian: $$\text{Span}\left(\begin{bmatrix}\frac{\partial f_1}{\partial \theta} & \frac{\partial f_1}{\partial \theta}\\ \frac{\partial f_2}{\partial \theta} & \frac{\partial f_2}{\partial \phi} \\ \frac{\partial f_3}{\partial \phi} & \frac{\partial f_3}{\partial \phi} \end{bmatrix}\right)$$ $$\text{Span}\left(\begin{bmatrix}-\sin(\theta)\sin(\phi) & \cos(\theta)\cos(\phi)\\ \cos(\theta)\sin(\phi) & \sin(\theta)\cos(\phi) \\ 0 & -\sin(\phi) \end{bmatrix}\right)$$ What answer do we expect? We expect to see a plane. The span of two vectors, by definition, is a plane.

# Tangent Spaces to Graphs

MathJax TeX Test Page A graph is pretty similar to what we normally think of graphs, they equal $$\Gamma(x) = \{\begin{bmatrix}x\\f(x)\end{bmatrix} \mid x \in \mathbb{R}^n\}$$ Think about graphing $y = x^2$, you graph the x and y points together: $$\begin{bmatrix}0 \\ 0\end{bmatrix}, \begin{bmatrix}3 \\ 9\end{bmatrix}, \begin{bmatrix}-2 \\ 4\end{bmatrix} \dots$$ The tangent space equals the graph of the derivative, meaning it's $$\{\begin{bmatrix}x\\Df(p)(x)\end{bmatrix} \mid x \in \mathbb{R}^n\}$$ The Jacobian for $\begin{bmatrix}x^2\end{bmatrix}$ is very simple: $\begin{bmatrix}2x\end{bmatrix}$. Therefore, the graph of the derivative is $x$ and the linear approximation at that point, which is $2x$ times $x$ = $2x^2$. Why is this? Imagine we had an x-value: $3$, the Jacobian would be $[6]$. We're graphing $\begin{bmatrix}x \\ Jf(p)(x)\end{bmatrix}$ = $\begin{bmatrix}x \\ 6x\end{bmatrix}$. Now, this is what we actually get in the general case: $$\begin{bmatrix}x \\ 2x^2\end{bmatrix} = \text{Span}\left(\begin{bmatrix}1 \\ 2x\end{bmatrix}\right)$$

David Witten