is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

Tangent Spaces

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.


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: $$\begin{bmatrix}x \\ 2x^2\end{bmatrix} = \text{Span}\left(\begin{bmatrix}1 \\ 2x\end{bmatrix}\right)$$


MathJax TeX Test Page Let $f(x,y) = \begin{bmatrix}x^2 + y^2 + 1\\2xy \end{bmatrix}$. Find the tangent space of its graph if $(x,y) = (2,3)$. The graph of f(x,y) equals $$\begin{bmatrix}x\\y\\x^2 + y^2 + 1\\2xy \end{bmatrix}$$ The Jacobian of f(x,y) equals $$\begin{bmatrix}2x & 2y \\2y & 2x\end{bmatrix} \to \begin{bmatrix}2(2) & 2(3) \\2(3) & 2(2)\end{bmatrix} = \begin{bmatrix}4 & 6 \\6 & 4\end{bmatrix}$$ The graph of the derivative equals $$\begin{bmatrix}x\\y\\Jf(p)\begin{bmatrix}x\\y\end{bmatrix}\end{bmatrix}$$ $$\begin{bmatrix}x \\ y\\ 4x + 6y\\6x + 4y\end{bmatrix}$$ $$= x\begin{bmatrix}1 \\0 \\4\\6\end{bmatrix} + y\begin{bmatrix}0\\1\\6\\4\end{bmatrix}$$ Because x and y can be anything, the tangent space of this graph is a plane (span of 2 vectors). However, this tangent plane is linear, meaning it goes through the origin. However, when $(x,y) = (0,0)$, $f(x,y) = \begin{bmatrix}1\\0\end{bmatrix}$. We want the tangent space to touch at the point (2,3, 14, 12), so we just add that point! $$\begin{bmatrix}x+2\\ y+3\\ 4x + 6y + 14\\6x + 4y + 12\end{bmatrix}

Tangent Spaces to Level Curves

In the post about gradients, we talked about gradients were orthogonal to level curves. So, how do we combine our understanding of tangent spaces: the span of the derivative (Jacobian) and the span of the graph of the derivative.

MathJax TeX Test Page The tangent space of a level curve is the kernel of the derivative. This means $Df(p) \cdot v = 0$. If that's the case, then $Df(p)^{T}v = 0 \to \nabla f \cdot v = 0$ Therefore, $\nabla f$ is the normal vector to the tangent space.


MathJax TeX Test Page Find the tangent space to $x^2 + y^2 - z^2 = 0$ at the point $(3,4,5)$. The gradient is $\begin{bmatrix}2x\\2y\\-2z\end{bmatrix} = \begin{bmatrix}6\\8\\-10\end{bmatrix}$ The kernel of that means $\begin{bmatrix}6\\8\\-10\end{bmatrix}\cdot v = 0$, so the tangent space is just a plane with that normal vector. $$6x + 8y - 10z = 0$$
David Witten