# Fundamental Subspaces

The **fundamental subspaces** are four vector spaces defined by a given $m \times n$ matrix $A$ (and its transpose): the column space and nullspace (or kernel) of $A$, the column space of $A^T$ $($also called the *row space* of $A),$ and the nullspace of $A^T$ $($also called the *left nullspace* of $A).$

The fundamental subspaces are useful for a number of linear algebra applications, including analyzing the rank of a matrix. The subspaces are also closely related by the *fundamental theorem of linear algebra*.

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## Column Space

The **column space** of a matrix $A$ is the vector space formed by the columns of $A$, essentially meaning all linear combinations of the columns of $A$. Equivalently, the column space consists of all matrices $Ax$ for some vector $x$.

For this reason, the column space is also known as the **image** of $A$ $\big($denoted $\text{im}(A)\big),$ as it is the result when $A$ is viewed as a linear transformation of the vector space $\mathbb{R}^m$ $(\text{where } m$ is the number of rows of $A).$ In particular, the image of $A$ is necessarily a subspace of $\mathbb{R}^m$, hence the term "fundamental subspace."

For example, consider the matrix

$A = \begin{pmatrix} 1 & 2 & 3 & 3 \\ 2 & 0 & 6 & 2 \\ 3 & 4 & 9 & 7\end{pmatrix}.$

Then the column vectors are

$\begin{pmatrix}1\\2\\3\end{pmatrix}, \begin{pmatrix}2\\0\\4\end{pmatrix}, \begin{pmatrix}3\\6\\9\end{pmatrix},\begin{pmatrix}3\\2\\7\end{pmatrix},$

which form a vector space that has a basis of $\left\{\begin{pmatrix}1\\2\\3\end{pmatrix}, \begin{pmatrix}2\\0\\4\end{pmatrix}\right\}$. As there are 2 vectors in this basis, the dimension of the column space is 2; hence, the rank of $A$ is 2.

## Nullspace

The **nullspace** or **kernel** of a matrix $A$ $\big($denoted $\text{ker}(A)\big)$ is the set of all vectors $x$ for which

$Ax = 0.$

If $x$ and $y$ are in the nullspace, then $c_1x+c_2y$ is also in the nullspace as

$A(c_1x + c_2y) = c_1(Ax) + c_2(Ay) = 0 + 0 = 0,$

so the nullspace is in fact a vector space. When $A$ is viewed as a linear transformation, the nullspace is the subspace of $\mathbb{R}^n$ that is sent to 0 under the map $A$, hence the term "fundamental subspace."

For example, consider the matrix

$A = \begin{pmatrix} 1 & 2 & 3 & 3 \\ 2 & 0 & 6 & 2 \\ 3 & 4 & 9 & 7\end{pmatrix}.$

The nullspace consists of all vectors $x$ such that $Ax = 0$. This defines a system of linear equations that can be solved to give the family of solutions

$\begin{pmatrix} -x-3y\\-x\\y\\x\end{pmatrix}, x,y \in \mathbb{R},$

which defines a vector space with basis $\left\{\begin{pmatrix}-1\\-1\\0\\1\end{pmatrix}, \begin{pmatrix}-3\\0\\1\\0\end{pmatrix}\right\}$. As there are two vectors in this basis, the dimension of the nullspace is 2.

## Row Space and Left Nullspace

The **row space** and **left nullspace** are defined as the column space and nullspace of $A^T$, the transpose of $A$, respectively. In this way, they are subspaces of $\mathbb{R}^n$.

## Fundamental Theorem of Linear Algebra

The **fundamental theorem of linear algebra** relates all four of the fundamental subspaces in a number of different ways. There are main parts to the theorem:

**Part 1:**

The first part of the fundamental theorem of linear algebra relates the dimensions of the four fundamental subspaces:

The column and row spaces of an $m \times n$ matrix $A$

bothhave dimension $r$, the rank of the matrix. The nullspace has dimension $n-r$, and the left nullspace has dimension $m-r$.

This is illustrated by the example in previous sections: the dimension of the column space of

$A = \begin{pmatrix} 1 & 2 & 3 & 3 \\ 2 & 0 & 6 & 2 \\ 3 & 4 & 9 & 7\end{pmatrix}$

is 2, and the dimension of the nullspace of $A$ is $n - r = 4 - 2 = 2$.

This first part of the fundamental theorem of linear algebra is sometimes referred to by name as the rank-nullity theorem.

**Part 2:**

The second part of the fundamental theorem of linear algebra relates the fundamental subspaces more directly:

The nullspace and row space are orthogonal. The left nullspace and the column space are also orthogonal.

In other words, if $v$ is in the nullspace of $A$ and $w$ is in the row space of $A$, the dot product $v \cdot w$ is 0. This is true because any vector in the nullspace is orthogonal to each row vector by definition, so it is also orthogonal to any linear combination of them.

**Part 3:**

The final part of the fundamental theorem of linear algebra constructs an orthonormal basis, and demonstrates a **singular value decomposition**: any matrix $M$ can be written in the form $U\sum V^T$, where

- $U$ is an $m \times m$ unitary matrix;
- $\sum$ is an $m \times n$ matrix with nonnegative values on the diagonal;
- $V$ is an $n \times n$ unitary matrix.

This part of the fundamental theorem allows one to immediately find a basis of the subspace in question.

This can be summarized in the following table:

Subspace | Subspace of$\hspace{10mm}$ | Symbol$\hspace{10mm}$ | Dimension$\hspace{10mm}$ | Basis |

column space | $\mathbb{R}^m$ | $\text{im}(A)$ | $r$ = rank | First $r$ columns of $U$ |

nullspace (kernel) | $\mathbb{R}^n$ | $\text{ker}(A)$ | $n - r$ | Last $n-r$ columns of $V$ |

row space | $\mathbb{R}^n$ | $\text{im}(A^T)$ | $r$ | First $r$ columns of $V$ |

left nullspace (kernel)$\hspace{10mm}$ | $\mathbb{R}^m$ | $\text{ker}(A^T)$ | $m - r$ | Last $m - r$ columns of $U$ |

Or it can also be summarized in the following image, where the arrows represent the results of multiplication:

## See Also

**Cite as:**Fundamental Subspaces.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/fundamental-subspaces/