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Just looking at some tutorial videos, I'm noticing somewhat of a trend... but it wasn't spelled out explicitly, so I'd like to verify if this theory of mine is correct... (forgive my horribly un-exact notation...)

Given $m\in\mathbb{N}$ vectors in $\mathbb{R}^n$, the span of those vectors is $$a_1\vec{v}_1+a_2\vec{v}_2+a_3\vec{v}_3+a_4\vec{v}_4+\cdots+a_m\vec{v}_m$$ If the null space of those vectors is a span of just $p\in\mathbb{N}$ of the $m$ vectors, does that mean that the column space is a span of the other $m-p$ vectors?

Basically, is it true that the column space "added" to the null space gives all of the original vectors?

Mirrana
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2 Answers2

27

You're close. What you are touching on is the fundamental theorem of linear algebra, which establishes that there are what is known as the $4$ fundamental subspaces of an $m\times{n}$ matrix.

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The row space and the nullspace together span the domain of the linear transformation: $\mathbb{R}^n$. Their intersection only contains $1$ element: the $n$ component $0$ vector.

Similarly, the column space and the left nullspace together span the co-domain of the linear transformation: $\mathbb{R}^m$. Their intersection only contains $1$ element: the $m$ component $0$ vector.

Furthermore, the rank of the matrix is the dimension of both the column space and the row space. The dimension of the nullspace is $n-r$, and the dimension of the left nullspace is $m-r$.

kimchi lover
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rurouniwallace
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6

Your wording is a little unusual: the null space of those vectors. The null space is usually defined for a linear function, not for a set of vectors. You can, nonetheless, put those vectors together and make up a linear function. More specifically, define $T:\mathbb R^m \to \mathbb R^n$ by $$ T(\vec a) = a_1\vec v_1 + a_2\vec v_2 + \ldots + a_m\vec v_m. $$ where $a_i$ are components of $\vec a$. Or in matrix-vector form, this is $$ T\begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_m \end{pmatrix} = \begin{pmatrix} | & | & \dots & | \\ \vec v_1 & \vec v_2 & \ldots & \vec v_m \\ | & | & \ldots & | \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_m \end{pmatrix} = \sum_{i=1}^m a_i\vec v_i. $$ Then you can talk about the null space of $T$. The space spanned by $\vec v_i$ is the image of $T$, which is exactly the column space of the matrix $(\vec v_1\ \ \vec v_2\ \ \ldots\ \ \vec v_m)$. Now, the rank-nullity theorem coincides with what you stated.

A little more explanation: The vector $\vec a \in \mathbb R^m$ indicates a choice of linear combination for a fixed set of vectors $\vec v_i$. $\vec a$ will belong to the null space of $T$ if $T(\vec a) = 0$, i.e., $\vec a$ corresponds to a linear combination that is zero. If $\vec v_i$ are linearly independent, the only linear combination that is zero is the zero linear combination, which is when all coefficients are zero. Then the null space of $T$ has only one member: $\vec a = 0$. In a sense, the dimension of the null space of $T$ measures how dependent (or redundant) $\vec v_i$ are.

Tunococ
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