Linear Space of Matrices

Sep 30, 2020
7 min read
Mar 11, 2022 15:59 UTC
The column space, row space and rank of a matrix and their properties.
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In the previous two sections (vector space and geometrical considerations) we’ve been focusing on vector operations and properties. Now we’re ready to come back to matrices. Associated with any matrix is a very important characteristic called the rank. There are several (consistent) ways of defining the rank. The most fundamental of these is in terms of the dimension of a linear space.


Recall a matrix $\boldsymbol{A}: m \times n$ has $\boldsymbol{a}_1, \cdots, \boldsymbol{a}_n \in \mathbb{R}^m$ columns and $\boldsymbol{\alpha}_1^\prime, \cdots, \boldsymbol{\alpha}_m^\prime \in \left( \mathbb{R}^n \right)^\prime$ rows. The column space of $\boldsymbol{A}$ contains all linear combinations of the columns:

$$ \begin{aligned} \mathcal{C}(\boldsymbol{A}) &= \mathcal{L}(\boldsymbol{a}_1, \cdots, \boldsymbol{a}_n) \\ &= \left\{ \sum_{j=1}^n c_j \boldsymbol{a}_j, \quad c_j \in \mathbb{R} \right\} \\ &= \{ \boldsymbol{Ac}, \quad \boldsymbol{c} \in \mathbb{R}^n \}, \quad \boldsymbol{c} = (c_1, \cdots, c_n)^\prime \end{aligned} $$

This means for any $\boldsymbol{y} \in \mathcal{C}(\boldsymbol{A})$, there exists $\boldsymbol{x} \in \mathbb{R}^n$ such that $\boldsymbol{y} = \boldsymbol{Ax}$.

Similarly, the row space of $\boldsymbol{A}$ is

$$ \begin{aligned} \mathcal{R}(\boldsymbol{A}) &= \mathcal{L}(\boldsymbol{\alpha}_1^\prime, \cdots, \boldsymbol{\alpha}_m^\prime) \\ &= \left\{ \sum_{i=1}^m c_i \boldsymbol{\alpha}_j^\prime, \quad c_i \in \mathbb{R} \right\} \\ &= \{ \boldsymbol{c}^\prime \boldsymbol{A}, \quad \boldsymbol{c} \in \mathbb{R}^m \} \end{aligned} $$

Note that

$$ \begin{gathered} \boldsymbol{y} \in \mathcal{C}(\boldsymbol{A}) \Longleftrightarrow \boldsymbol{y}^\prime \in \mathcal{R}(\boldsymbol{A}^\prime) \\ \boldsymbol{x}^\prime \in \mathcal{R}(\boldsymbol{A}) \Longleftrightarrow \boldsymbol{x} \in \mathcal{C}(\boldsymbol{A}^\prime) \end{gathered} $$


  1. $dim(\mathcal{C}(\boldsymbol{A})) \leq \min(n, m)$. The dimension of the column space of $\boldsymbol{A}$ is called the rank of $\boldsymbol{A}$, which is written as $rank(\boldsymbol{A}) \triangleq r(\boldsymbol{A})$.

  2. $dim(\mathcal{R}(\boldsymbol{A})) = dim(\mathcal{C}(\boldsymbol{A})) = r(\boldsymbol{A})$, which implies the number of LIN columns is equal to the number of LIN rows.

    Proof: Show that the row rank is equal to the column rank. Let $r$ denote the row rank, that is we have $r$ linearly independent rows. This means $\mathcal{R}(\boldsymbol{A})$ has a basis $\{\boldsymbol{x}_1^\prime, \cdots, \boldsymbol{x}_r^\prime\}$. We will show that $\{\boldsymbol{Ax}_1, \cdots, \boldsymbol{Ax}_r\}$ is linearly independent.

    We show this because each vector $\boldsymbol{Ax}_i \in \mathcal{C}(\boldsymbol{A})$. If we can show the set is linearly independent, then the dimension of the column space has to be at least $r$. To show this, we set

    $$ \sum_{i=1}^r c_i \boldsymbol{Ax}_i = \boldsymbol{0} $$

    where the $c_i$’s are scalars. If we take $\boldsymbol{A}$ outside,

    $$ \boldsymbol{A} \left( \sum_{i=1}^r c_i \boldsymbol{x}_i \right) = \boldsymbol{0} $$

    Let $\boldsymbol{v} = \sum_{i=1}^r c_i \boldsymbol{x}_i$. If $\boldsymbol{Av} = \boldsymbol{0}$, then $\boldsymbol{v}^\prime$ has to be orthogonal to each row of $\boldsymbol{A}$, which means $\boldsymbol{v}^\prime \perp \mathcal{R}(\boldsymbol{A})$. Note that $\boldsymbol{v}^\prime$ is a linear combination of the rows of $\boldsymbol{A}$, so $\boldsymbol{v}^\prime \in \mathcal{R}(\boldsymbol{A})$. The only vector that belongs to a subspace and is orthogonal to that subspace is the zero vector.

    If $\boldsymbol{v} = \boldsymbol{0}$, then $c_i = 0$ for all $i$ because the $\boldsymbol{x}_i$’s are linearly independent. Thus, the $\boldsymbol{Ax}_i$’s are linearly independent. The column space of $\boldsymbol{A}$ has a set of $r$ linearly independent vectors, so $dim(\mathcal{C}(\boldsymbol{A})) \geq r$.

    We can repeat the above argument with $c \triangleq dim(\mathcal{C}(\boldsymbol{A}))$, and show that $dim(\mathcal{R}(\boldsymbol{A})) \geq c$. By showing both sides of the inequality, we arrive at the conclusion that $r=c$, i.e. the dimension of the column space is the same as the dimension of the row space, and that’s what we call the rank of $\boldsymbol{A}$.

  3. If $r(\boldsymbol{A}) = m$, then we say $\boldsymbol{A}$ has full row rank1. If $r(\boldsymbol{A}) = n$, then we say $\boldsymbol{A}$ has full column rank. Either case, we say $\boldsymbol{A}$ has full rank. If a matrix is not of full rank, it is rank-deficient.

  4. If $r(\boldsymbol{A}) = r$, then there exists matrix $\boldsymbol{B}: m \times r$, $r(\boldsymbol{B}) = r$ (full column rank) and matrix $\boldsymbol{T}: r \times n$, $r(\boldsymbol{T}) = r$ (full row rank), such that $\boldsymbol{A} = \boldsymbol{BT}$.

    Any matrix can be written as a product of two full rank matrices. $\mathcal{C}(\boldsymbol{A})$ has $r$ basis vectors, and we can put them into an $m \times r$ matrix $\boldsymbol{B}$. Each column of $\boldsymbol{A}$, $\boldsymbol{a}_i$, is in $\mathcal{C}(\boldsymbol{A})$, so $\boldsymbol{a}_i$ can be expressed as $\boldsymbol{a}_i = \boldsymbol{By}_i$ where the $\boldsymbol{y}_i$’s are $r \times 1$ vectors, and we can put them together into an $r \times n$ matrix $\boldsymbol{T}$.

  5. $r(\boldsymbol{AB}) \leq \min(r(\boldsymbol{A}), r(\boldsymbol{B}))$ for any $\boldsymbol{A}$, $\boldsymbol{B}$ matrices.

    Proof: We first show that $r(\boldsymbol{AB}) \leq r(\boldsymbol{A})$ from the fact that $\mathcal{C}(\boldsymbol{AB}) \subset \mathcal{C}(\boldsymbol{A})$. Note that

    $$ \boldsymbol{AB} = [\boldsymbol{Ab}_1, \cdots, \boldsymbol{Ab}_q] $$

    where $\boldsymbol{B}$ has $q$ columns. Each column of $\boldsymbol{AB}$ has a form of $\boldsymbol{Ab}_i$, which is in $\mathcal{C}(\boldsymbol{A})$. In other words, any vector that lives in the column space of $\boldsymbol{AB}$ is in the column space of $\boldsymbol{A}$.

    Similarly, we can show $r(\boldsymbol{AB}) \leq r(\boldsymbol{B})$ from the fact that $\mathcal{R}(\boldsymbol{AB}) \subset \mathcal{R}(\boldsymbol{B})$2.

  6. If an $m \times n$ matrix $\boldsymbol{A}$ with rank $r$ is written as $\boldsymbol{A} = \boldsymbol{BT}$ with $\boldsymbol{B}: m \times r$ and $\boldsymbol{T}: r \times n$, then $r(\boldsymbol{B}) = r(\boldsymbol{T}) = r$.

    This means both $\boldsymbol{B}$ and $\boldsymbol{T}$ have full rank, specifically $\boldsymbol{B}$ has full column rank and $\boldsymbol{T}$ has full row rank. This is true because

    $$ r = r(\boldsymbol{A}) = r(\boldsymbol{BT}) \leq r(\boldsymbol{B}) \leq r \Rightarrow r(\boldsymbol{B}) = r $$

  7. The null space of $\boldsymbol{A}: m \times n$ is the collection of $n \times 1$ vectors that such that

    $$ N(\boldsymbol{A}) = \{\boldsymbol{x} \mid \boldsymbol{Ax} = \boldsymbol{0}\} $$

    The null space is a subspace in $\mathbb{R}^n$. If $\boldsymbol{x} \in N(\boldsymbol{A})$, $\boldsymbol{x}$ is orthogonal to each row of $\boldsymbol{A}$, thus

    $$ N(A) = \mathcal{R}(A)^\perp $$

  8. A square matrix $\boldsymbol{A}: n \times n$ with rank $n$ is called non-singular. If $r(\boldsymbol{A}) < n$, it is singular.

  9. For matrices $\boldsymbol{A}: m \times n$ and $\boldsymbol{B}: m \times q$, the horizontal concatenation of $\boldsymbol{A}$ and $\boldsymbol{B}$ is an $m \times (n+q)$ matrix denoted $[\boldsymbol{A}, \boldsymbol{B}]$. Since $\mathcal{C}(\boldsymbol{A}) \subset \mathcal{C}([\boldsymbol{A}, \boldsymbol{B}])$, we have $r(\boldsymbol{A}) \leq r([\boldsymbol{A}, \boldsymbol{B}])$.

  10. $r([\boldsymbol{A}, \boldsymbol{B}]) \leq r(\boldsymbol{A}) + r(\boldsymbol{B})$. Think in terms of linearly independent columns.

  11. $\mathcal{C}(\boldsymbol{A} + \boldsymbol{B}) \subset \mathcal{C}([\boldsymbol{A}, \boldsymbol{B}])$. The reason is

    $$ \begin{gathered} \mathcal{C}(\boldsymbol{A} + \boldsymbol{B}) = \{ (\boldsymbol{A} + \boldsymbol{B})\boldsymbol{x} = \boldsymbol{Ax} + \boldsymbol{Bx} \} \\ \mathcal{C}([\boldsymbol{A}, \boldsymbol{B}]) = \left\{ [\boldsymbol{A}, \boldsymbol{B}]\begin{pmatrix} \boldsymbol{y}_1 \\ \boldsymbol{y}_2 \end{pmatrix} = \boldsymbol{Ay}_1 + \boldsymbol{By}_2 \right\} \end{gathered} $$

    From this we can see that

    $$ r(\boldsymbol{A} + \boldsymbol{B}) \leq r([\boldsymbol{A}, \boldsymbol{B}]) \leq r(\boldsymbol{A}) + r(\boldsymbol{B}) $$

  12. $\mathcal{C}(\boldsymbol{A}^\prime \boldsymbol{A}) = \mathcal{C}(\boldsymbol{A}^\prime)$, and $\mathcal{C}(\boldsymbol{AA}^\prime) = \mathcal{C}(\boldsymbol{A})$. Similarly for the row spaces,

    $$ \mathcal{R}(\boldsymbol{A}^\prime \boldsymbol{A}) = \mathcal{R}(\boldsymbol{A}), \quad \mathcal{R}(\boldsymbol{AA}^\prime) = \mathcal{R}(\boldsymbol{A}^\prime) $$

    and the ranks:

    $$ r(\boldsymbol{A}^\prime \boldsymbol{A}) = r(\boldsymbol{A}) = r(\boldsymbol{AA}^\prime) = r(\boldsymbol{A}^\prime) $$


Recall that in Projection to a subspace we showed that $P(\boldsymbol{x} \mid \mathcal{V})$ has the shortest distance to $\boldsymbol{x}$ among all $\boldsymbol{y} \in \mathcal{V}$, i.e. for any $\boldsymbol{y} \in \mathcal{V}$,

$$ \|\boldsymbol{y} - \boldsymbol{x}\|^2 \geq \| P(\boldsymbol{x} \mid \mathcal{V}) - \boldsymbol{x} \|^2 $$

The statistical application for this is that suppose we have data $\boldsymbol{y} = (y_1, \cdots, y_n)^\prime$, and our goal is to explain or predict $\boldsymbol{y}$ using another variable $\boldsymbol{x} = (x_1, \cdots, x_n)^\prime$.

In simple linear regression (SLR), we find $\beta_0$ and $\beta_1$ such that $\beta_0 + \beta_1 \boldsymbol{x}$ best explains $\boldsymbol{y}$. In other words, we find the vector in the subspace

$$ \mathcal{V} = \{ \beta_0 \boldsymbol{1}_n + \beta_1 \boldsymbol{x} \mid \beta_0, \beta_1 \in \mathbb{R} \} $$

that is closest to $\boldsymbol{y}$.

Projection of $\boldsymbol{y}$ onto the subspace spanned by $\{\boldsymbol{1}, \boldsymbol{x}\|$.

So how do we find the projection $P(\boldsymbol{y} \mid \mathcal{V})$? $\{ \boldsymbol{1}_n, \boldsymbol{x} \}$ is a basis.

By definition, we have

$$ \begin{gathered} P(\boldsymbol{y} \mid \mathcal{V}) = \beta_0 \boldsymbol{1}_n + \beta_1 \boldsymbol{x} \\ \begin{cases} \boldsymbol{y} - (\beta_0 \boldsymbol{1}_n + \beta_1 \boldsymbol{x}) \cdot \boldsymbol{1}_n = 0 \\ \boldsymbol{y} - (\beta_0 \boldsymbol{1}_n + \beta_1 \boldsymbol{x}) \cdot \boldsymbol{x} = 0 \end{cases} \Rightarrow \begin{cases} \boldsymbol{y}^\prime \boldsymbol{1}_n - \beta_0 n - \beta_1 \boldsymbol{x}^\prime \boldsymbol{1}_n = 0 \\ \boldsymbol{y}^\prime \boldsymbol{x} - \beta_0 \boldsymbol{x}^\prime \boldsymbol{1}_n - \beta_1 \boldsymbol{x}^\prime \boldsymbol{x} = 0 \end{cases} \\ \hat\beta_0 = \bar{y} - \beta_1 \bar{x} \\ \hat\beta_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum(x_i - \bar{x})^2} = \frac{S_{xy}}{S_{xx}} \end{gathered} $$

What we have here is

$$ \hat{y} = \hat\beta_0 + \hat\beta_1 \boldsymbol{x} $$

which is the projection of $\boldsymbol{y}$ onto $\mathcal{V}$. To ensure that $\boldsymbol{1}_n$ and $\boldsymbol{x}$ are orthogonal, we can use Gram-Schmidt orthogonalization:

$$ \boldsymbol{1}_n, \boldsymbol{x} \xrightarrow{G-S} \boldsymbol{1}_n, \boldsymbol{x} - \bar{x}\boldsymbol{1}_n = \begin{pmatrix} x_1 - \bar{x} \\ \vdots \\ x_n - \bar{x} \end{pmatrix} $$

  1. It’s more likely a short, wide matrix. Similarly, full column rank matrices are likely to be narrow and tall. ↩︎

  2. $\mathcal{C}(\boldsymbol{AB})$ and $\mathcal{C}(\boldsymbol{B})$ can’t even be compared usless $m = n$. However their dimensions could be discussed. ↩︎

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