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− | === $C(A)$ === | + | === The Column Space $C(A)$ === |

Let $A$ be $m \times n$ matrix: | Let $A$ be $m \times n$ matrix: | ||

* $A = \left[ \mathop{a_1}\limits_|^| \, \mathop{a_2}\limits_|^| \ \cdots \ \mathop{a_n}\limits_|^| \right]$ | * $A = \left[ \mathop{a_1}\limits_|^| \, \mathop{a_2}\limits_|^| \ \cdots \ \mathop{a_n}\limits_|^| \right]$ | ||

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* $C(A)$ is in $\mathbb R^r$ space where $r \leqslant n$ is the $A$'s [[Rank (Matrix)|Rank]] | * $C(A)$ is in $\mathbb R^r$ space where $r \leqslant n$ is the $A$'s [[Rank (Matrix)|Rank]] | ||

* so the dimensionality is at most the number of columns, and at least the rank of the matrix | * so the dimensionality is at most the number of columns, and at least the rank of the matrix | ||

+ | |||

+ | |||

+ | <img width="50%" src="http://alexeygrigorev.com/projects/imsem-ws14-lina/img-svg/diagram1.svg" /> | ||

+ | |||

+ | |||

+ | It's also called the ''range'' of $A$: | ||

+ | * $\text{ran}(A) = \{ \mathbf y \in \mathbb R^m \ : \ \mathbf y = A \mathbf x \ \forall \mathbf x \in \mathbb R^n \}$ | ||

+ | * if we think about [[Linear Transformation]] $T_{A}$ formed by $A$, this is what it does to an $n$-dimensional vector $\mathbf x$: it brings it to an $m$-dinesional vector $\mathbf y$ | ||

+ | * all such vectors $\mathbf y$ form the column space of $A$ | ||

+ | |||

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For example, the following can solve it: | For example, the following can solve it: | ||

* $\mathbf 0_4$, because $\mathbf x = \mathbf 0_3$ will solve it | * $\mathbf 0_4$, because $\mathbf x = \mathbf 0_3$ will solve it | ||

− | * | + | * <math>\begin{bmatrix} |

1 \\ | 1 \\ | ||

2 \\ | 2 \\ | ||

3 \\ | 3 \\ | ||

4 \\ | 4 \\ | ||

− | \end{bmatrix} | + | \end{bmatrix}</math> - one of the columns, so <math>\mathbf x = \begin{bmatrix} |

1 \\ | 1 \\ | ||

0 \\ | 0 \\ | ||

0 \\ | 0 \\ | ||

− | \end{bmatrix} | + | \end{bmatrix}</math> |

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== Sources == | == Sources == | ||

* [[Linear Algebra MIT 18.06 (OCW)]] | * [[Linear Algebra MIT 18.06 (OCW)]] | ||

+ | * [[Matrix Computations (book)]] | ||

[[Category:Linear Algebra]] | [[Category:Linear Algebra]] |

A column space $C(A)$ of a matrix $A$ is a subspace formed by columns of $A$

- it's one of is one of the Four Fundamental Subspaces of $A$

Let $A$ be $m \times n$ matrix:

- $A = \left[ \mathop{a_1}\limits_|^| \, \mathop{a_2}\limits_|^| \ \cdots \ \mathop{a_n}\limits_|^| \right]$
- the columns of $A$ form a subspace - a hyperplane through the origin
- $C(A)$ is in $\mathbb R^r$ space where $r \leqslant n$ is the $A$'s Rank
- so the dimensionality is at most the number of columns, and at least the rank of the matrix

It's also called the *range* of $A$:

- $\text{ran}(A) = \{ \mathbf y \in \mathbb R^m \ : \ \mathbf y = A \mathbf x \ \forall \mathbf x \in \mathbb R^n \}$
- if we think about Linear Transformation $T_{A}$ formed by $A$, this is what it does to an $n$-dimensional vector $\mathbf x$: it brings it to an $m$-dinesional vector $\mathbf y$
- all such vectors $\mathbf y$ form the column space of $A$

It's a subspace:

- if we take any vectors from $C(A)$, the linear combination will still be $C(A)$ (by definition)

Suppose we have a matrix $A \in \mathbb R^{3 \times 2}$

$A = \begin{bmatrix} 1 & 3 \\ 2 & 3 \\ 4 & 1 \\ \end{bmatrix}$

Subspace from columns - $C(A)$ - the **Column Space** of $A$:

- we cannot just take the two columns and call it a subspace:
- it also must include all linear combinations of these columns
- these linear combinations of two vectors form a plane - a subspace $\mathbb R^2$ in the space $\mathbb R^3$
- since we include all possible combinations, we're guaranteed to have a subspace
- $v_1$ and $v_2$ are 1st and 2nd columns of $A$ - they form a plane through the origin

Column Space $C(A)$ of $A$ is important: the system $A \mathbf x = \mathbf b$ has the solution only when $\mathbf b \in C(A)$

For example:

$A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 1 & 4 \\ 4 & 1 & 5 \\ \end{bmatrix}$.

- There are 3 columns and they are 4-dim vectors
- so $C(A)$ is a subspace $\mathbb R^4$
- but how big it is? is it the entire $\mathbb R^4$? No - we have only 3 vectors, so it's at most $\mathbb R^3$

Since there are only 3 columns, in $A \mathbf x = \mathbf b$

- $\mathbf x \in \mathbb R^3$ and $\mathbf b \in \mathbb R^4$
- does it always have a solution? no: we have 4 equations and 3 unknowns
- there are many $\mathbf b$'s that can't solve the system
- but there are some that can: they are linear combinations of the columns - so those $\mathbf b$ that are from $C(A)$

For example, the following can solve it:

- $\mathbf 0_4$, because $\mathbf x = \mathbf 0_3$ will solve it
- [math]\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ \end{bmatrix}[/math] - one of the columns, so [math]\mathbf x = \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix}[/math]

if $\mathbf b \not \in C(A)$ there's no way to solve the system

What's the dimension of $C(A)$?

- $\text{dim } C(A) = r$ where $r$ -rank of $A$
- the easiest way to determine it - is calculate the number of pivot columns during Gaussian Elimination
- in this case, $\text{dim } C(A) = 2$ because the rank is 2 (the 3rd column is a linear combination of 1st and 2nd)