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Linear Algebra
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1. Vectors
2. Matrices
3. Linear equations
4. Matrix determinant
5. Vector space
6. Special matrices
7. Eigenvalues and Eigenvectors
8. Orthogonality
9. Matrix decomposition
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# Comprehensive Guide on Column Space in Linear Algebra

schedule Mar 5, 2023
Last updated
local_offer
Linear Algebra
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Definition.

# Column space

Let $\boldsymbol{A}$ be any $m\times{n}$ matrix:

\begin{align*} \boldsymbol{A}= \begin{pmatrix} \vert&\vert&\vert&\vert\\ \boldsymbol{a}_1&\boldsymbol{a}_2&\cdots&\boldsymbol{a}_n\\ \vert&\vert&\vert&\vert\\ \end{pmatrix} \end{align*}

The column space or range of an $m\times{n}$ matrix $\boldsymbol{A}$, denoted by $\mathrm{col}(\boldsymbol{A})$, is the span of its column vectors, that is:

\begin{align*} \mathrm{col}(\boldsymbol{A})&= \mathrm{span} (\boldsymbol{a}_1,\boldsymbol{a}_2,\cdots,\boldsymbol{a}_n) \end{align*}

Note that span is defined as the set of all the vectors that can be constructed using a linear combination of $\boldsymbol{a}_1$, $\boldsymbol{a}_2$, $\cdots$, $\boldsymbol{a}_n$, that is:

$$\mathrm{col}(\boldsymbol{A}) =\{ c_1\boldsymbol{a}_1+c_2\boldsymbol{a}_2+\cdots+c_n\boldsymbol{a}_n \;|\; c_1,c_2,\cdots,c_n\in\mathbb{R} \}$$
Example.

## Finding the column space of a matrix (1)

Consider the following matrix:

$$\boldsymbol{A}= \begin{pmatrix} 2&4\\ 5&3 \end{pmatrix}$$

Find the column space of $\boldsymbol{A}$.

Solution. By definitionlink, the column space of a matrix is the span of its column vectors. The column space of $\boldsymbol{A}$ is:

$$\mathrm{col}(\boldsymbol{A})=\mathrm{span} \left( \begin{pmatrix}2\\5\end{pmatrix},\; \begin{pmatrix}4\\3\end{pmatrix} \right)$$

Notice how the two vectors are linearly independent. By theoremlink, we know that two linearly independent vectors in $\mathbb{R}^2$ span $\mathbb{R}^2$, which means that the column space of $\boldsymbol{A}$ is the entire $\mathbb{R}^2$.

Example.

## Finding the column space of a matrix (2)

Consider the following matrix:

$$\boldsymbol{A}= \begin{pmatrix} 2&3\\ 2&3 \end{pmatrix}$$

Find the column space of $\boldsymbol{A}$.

Solution. The column space of $\boldsymbol{A}$ is the span of its column vectors:

$$\mathrm{col}(\boldsymbol{A})= \mathrm{span} \left( \begin{pmatrix}2\\2\end{pmatrix},\; \begin{pmatrix}3\\3\end{pmatrix} \right)$$

By the plus-minus theoremlink, since the column vectors of $\boldsymbol{A}$ are linearly dependent, we can remove one vector and still preserve the same span:

$$\mathrm{col}(\boldsymbol{A})= \mathrm{span} \left( \begin{pmatrix}2\\2\end{pmatrix} \right)$$

This means that the column space of $\boldsymbol{A}$ is a line traced out by the first column vector. Visually, the column space of $\boldsymbol{A}$ looks like follows:

For instance, the following vector is a particular element of the column space of $\boldsymbol{A}$:

$$\begin{pmatrix} 5\\5 \end{pmatrix}\in \mathrm{col}(\boldsymbol{A})$$
Theorem.

# The column space of a matrix is a subspace

The column space of an $m\times{n}$ matrix is a subspace of $\mathbb{R}^m$.

Proof. The column space of an $m\times{n}$ matrix $\boldsymbol{A}$ is defined as the span of its column vectors. Therefore, the column space is a subset of $\mathbb{R}^m$. To show that the column space is a subspace, we must show that the column space of $\boldsymbol{A}$ is closed under addition and scalar multiplication.

Let $\boldsymbol{v}$ and $\boldsymbol{w}$ be any two elements in the column space of $\boldsymbol{A}$. This means that there exist some vectors $\boldsymbol{x}$ and $\boldsymbol{y}$ such that:

\begin{align*} \boldsymbol{A}\boldsymbol{x}&=\boldsymbol{v}\\ \boldsymbol{A}\boldsymbol{y}&=\boldsymbol{w}\\ \end{align*}

Now, let's check if vector $\boldsymbol{v}+\boldsymbol{w}$ is contained in the column space of $\boldsymbol{A}$ like so:

\begin{align*} \boldsymbol{v}+\boldsymbol{w} &=\boldsymbol{Ax}+\boldsymbol{Ay}\\ &=\boldsymbol{A}(\boldsymbol{x}+\boldsymbol{y})\\ \end{align*}

Since $\boldsymbol{A}(\boldsymbol{x}+\boldsymbol{y})$ results in $\boldsymbol{v}+\boldsymbol{w}$, we conclude that $\boldsymbol{v}+\boldsymbol{w}$ is contained in the column space of $\boldsymbol{A}$. This means that the column space of $\boldsymbol{A}$ is closed under addition.

Next, suppose vector $\boldsymbol{v}$ is an element of the column space of $\boldsymbol{A}$. This means that there exists some vector $\boldsymbol{x}$ such that $\boldsymbol{Ax}=\boldsymbol{v}$. Now, consider a scalar $k$. Let's check if $k\boldsymbol{v}$ is included in the column space of $\boldsymbol{A}$ like so:

\begin{align*} k\boldsymbol{v} &=k(\boldsymbol{Ax})\\ &=\boldsymbol{A}(k\boldsymbol{x}) \end{align*}

This means that the column space of $\boldsymbol{A}$ is closed under scalar multiplication. Therefore, by definition, the column space of $\boldsymbol{A}$ is a subspace of $\mathbb{R}^m$. This completes the proof.

Theorem.

# Relationship between consistent linear system and column space

A system of linear equations $\boldsymbol{Ax}=\boldsymbol{b}$ is consistentlink if and only if $\boldsymbol{b}$ is contained in the column space of $\boldsymbol{A}$.

Proof. By definitionlink, the column space of an $m\times{n}$ matrix $\boldsymbol{A}$ is the span of its column vectors, that is, the set of all linear combinations of the column vectors:

$$\mathrm{col}(\boldsymbol{A}) =\{ x_1\boldsymbol{a}_1+x_2\boldsymbol{a}_2+\cdots+x_n\boldsymbol{a}_n \;|\; x_1,x_2,\cdots,x_n\in\mathbb{R} \}$$

Also, by definitionlink, if the system of linear equations $\boldsymbol{Ax}=\boldsymbol{b}$ is consistent, then there exists at least one solution $\boldsymbol{x}$ such that the product $\boldsymbol{Ax}$ yields $\boldsymbol{b}$. We know from theoremlink that $\boldsymbol{Ax}$ can be written as a linear combination of the column vectors of $\boldsymbol{A}$, that is:

\label{eq:ZIFT38Fn4iG3GO9QloM} \begin{aligned}[b] \boldsymbol{Ax}=\boldsymbol{b} \;\;\Longleftrightarrow\;\; \begin{pmatrix} \vert&\vert&\vert&\vert\\ \boldsymbol{a}_1&\boldsymbol{a}_2&\cdots&\boldsymbol{a}_n\\ \vert&\vert&\vert&\vert\\ \end{pmatrix} \begin{pmatrix} x_1\\x_2\\\vdots\\x_n \end{pmatrix}=\boldsymbol{b} \;\;\Longleftrightarrow\;\; x_1\boldsymbol{a}_1+x_2\boldsymbol{a}_2+\cdots+x_n\boldsymbol{a}_n=\boldsymbol{b} \end{aligned}

Again, given that $\boldsymbol{Ax}=\boldsymbol{b}$ is consistent, we know that there exist scalars $x_1$, $x_2$, $\cdots$, $x_n$ that make the above equality hold. Since $\boldsymbol{b}$ can be expressed as a linear combination of the column vectors of $\boldsymbol{A}$, we have that $\boldsymbol{b}$ belongs to the column space of $\boldsymbol{A}$.

Let's now prove the converse, that is, if $\boldsymbol{b}$ is contained in the column space of $\boldsymbol{A}$, then the system $\boldsymbol{Ax}=\boldsymbol{b}$ is consistent. By definitionlink, since $\boldsymbol{b}$ belongs to the column space of $\boldsymbol{A}$, we know that $\boldsymbol{b}$ can be expressed as a linear combination of the column vectors of $\boldsymbol{A}$, that is:

$$$$\label{eq:C4O1JExiiANzJPJw14a} x_1\boldsymbol{a}_1+x_2\boldsymbol{a}_2+\cdots+x_n\boldsymbol{a}_n=\boldsymbol{b}$$$$

Using the same logic as \eqref{eq:ZIFT38Fn4iG3GO9QloM}, this can be converted into the linear system $\boldsymbol{Ax}=\boldsymbol{b}$. Because we know there exist $x_1$, $x_2$, $\cdots$, $x_n$ that make \eqref{eq:C4O1JExiiANzJPJw14a} hold, we conclude that $\boldsymbol{Ax}=\boldsymbol{b}$ is consistent. This completes the proof.

Theorem.

# Pivot columns of a matrix span the column space of the matrix

The columns in matrix $\boldsymbol{A}$ corresponding to the pivot columnslink in the reduced row echelon form of $\boldsymbol{A}$ span the column space of $\boldsymbol{A}$.

Proof. Suppose $\boldsymbol{A}$ is an $m\times{n}$ matrix. The column space of $\boldsymbol{A}$ is defined as the span of its column vectors:

$$\mathrm{col}(\boldsymbol{A})=\mathrm{span}(\boldsymbol{a}_1,\boldsymbol{a}_2 ,\cdots,\boldsymbol{a}_n )$$

We know from the plus/minus theoremlink that we can remove linearly dependent vectors from the spanning set while preserving the span. From theoremlink, we know that columns in $\boldsymbol{A}$ corresponding to the non-pivot columns can be expressed as a linear combination of the columns in $\boldsymbol{A}$ corresponding to pivot columns. In other words, non-pivot columns are linearly dependent on pivot columns.

Therefore, we can remove all columns in $\boldsymbol{A}$ corresponding to the non-pivot columns from the spanning set while preserving the span. This means that only the columns in $\boldsymbol{A}$ corresponding to the pivot columns are sufficient to span the column space of $\boldsymbol{A}$.

This completes the proof.

Theorem.

# Pivot columns of a matrix form a basis for the column space of the matrix

The columns in matrix $\boldsymbol{A}$ corresponding to the pivot columns of the reduced row echelon formlink of $\boldsymbol{A}$ form a basislink for the column space of $\boldsymbol{A}$.

Proof. From theoremlink, we know that the columns in $\boldsymbol{A}$ corresponding to the pivot columns in $\mathrm{rref}(\boldsymbol{A})$ span the column space of $\boldsymbol{A}$. From theoremlink, we know that the columns in $\boldsymbol{A}$ corresponding to the pivot columns in $\mathrm{rref}(\boldsymbol{A})$ are linearly independent. Therefore, by definitionlink of basis, the columns in $\boldsymbol{A}$ corresponding to the pivot columns in $\mathrm{rref}(\boldsymbol{A})$ form a basis for the column space of $\boldsymbol{A}$. This completes the proof.

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