# Null Space The set of vectors that are transformed to the origin are referred to as the **null space** or **kernel** of the matrix. ![](Inverse_matrices_column_space_and_null_space__Chapter_7_Esse%20(1).gif) ### Linear Equations Perspective Consider the following visualization. We have a matrix that represents a linear system of equations, where each row in the matrix *defines* a hyperplane. The intersection if these hyperplanes represents the null space in this case (since the intersection represents the set of all $(x, y, z)$ tuples that are mapped to the $0$ vector ): ![](Dear_linear_algebra_students_This_is_what_matrices_and_matri.gif) As we add two of the equations in the system, while scaling one, we see that the null space is *preserved*. We can say that it is invariant to this kind of operation. This is very useful when trying to practically solve one of these systems (i.e. solve for a set of $(x, y, z)$ tuples that are mapped to the zero vector) since even though the plane rotates it does not change our solution set, but it makes the equations a bit cleaner to deal with. In the above example (see more [here](https://youtu.be/4csuTO7UTMo?t=251)), we see that if we do the same thing with equation 1 and 3, they actually become the same plan! Now, let us consider a vector perspective. If we treat each row in our matrix as a vector, we can think of the first row as saying the dot product with $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ yields $0$. What this means is that the first row (as a vector) is perpendicular to the null space. ![](Dear%20linear%20algebra%20students,%20This%20is%20what%20matrices%20(and%20matrix%20manipulation)%20really%20look%20like%205-51%20screenshot.png) To be clear, the reason this is true is because our vector $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ represents a solution to our system, specifically one that is mapped to zero. Any vector mapped to $0$ by a linear transformation is part of the null space. And thanks to our dot product interpretation, we also know that this solution has a dot product of $0$ with our first row vector, hence they are perpendicular. See more [here](https://youtu.be/4csuTO7UTMo?t=327). --- Date: 20211229 Links to: [Linear Algebra MOC](Linear%20Algebra%20MOC.md) Tags: References: * [3b1b video](https://youtu.be/uQhTuRlWMxw?t=639) * [Matrix Manipulation Visualizations](https://www.youtube.com/watch?v=4csuTO7UTMo)