# Question on the fundamental subspaces of an $m \times n$ matrix for when the domain is a subspace of $R^n$

If I have some $$m \times n$$ matrix A, typically a transformation described by $$A$$ is $$T: R^n \to R^m$$ and $$T(x) = Ax$$. If I choose to restrict the domain to be a subspace of $$R^n$$ (called $$Q$$) instead of the whole space, I'm trying to figure out how this changes the column space, row space, null space and left null space if at all.

Based on the definition of column space on Wikipedia:

In linear algebra, the column space (also called the range or image) of a matrix A is the span (set of all possible linear combinations) of its column vectors. The column space of a matrix is the image or range of the corresponding matrix transformation.

1. Here it states that the column space is all linear combinations of the columns. This seems to imply that the column space remains the same even if the domain is restricted to $$Q$$? But if that's the case the column space would no longer be the range of the transformation?

Based on the definition of null space:

the kernel (also known as null space or nullspace) of a linear map L : V → W between two vector spaces V and W, is the set of all elements v of V for which L(v) = 0, where 0 denotes the zero vector in W.

1. Does this mean that if I restrict the domain space to $$Q$$ instead of $$R^n$$, the null space of the transformation must be restricted to vectors in $$Q$$ that gets mapped to the zero vector in the range of the transformation?

2. Do the left null space and row space remain unchanged?

3. If I restrict the domain to a line in $$R^3$$ and $$A$$ is $$3x3$$ with linearly independent columns, is there some way to make sense of the rank nullity theorem as domain space has dim = 1, null space has dim = 0, rank of the matrix is 3?

I'll modify your notation slightly. Let $$A$$ be an $$m \times n$$ matrix with entries in the field $$\Bbb{R}$$. To each such matrix, you can define an associated linear transformation (called the left multiplication by $$A$$ transformation )$$L_A: \Bbb{R}^n \to \Bbb{R}^m$$, defined by $$L_A(x) = A \cdot x$$. Also, let $$Q$$ be a given subspace of $$\Bbb{R}^n$$

I feel that most of your confusion is because you haven't been careful enough in reading the definitions, and you haven't been specific enough in your choice of words.

Question $$1$$:

First, you write

Here it states that the column space is all linear combinations of the columns.

Let's be careful with the choice of words here. What the article actually says is the column space of a given matrix $$A$$ is the span of its columns (i.e set of all linear combinations of columns of $$A$$). In terms of the connection between matrices and linear transformations, it says \begin{align} \text{col}(A) = \text{image}(L_A). \end{align} Next, you write

This seems to imply that the column space remains the same even if the domain is restricted to $$Q$$?

You are right. The column space of a matrix has been defined without any reference to linear transformations. I think your confusion stems from the fact you weren't specific enough with your wording (column space of what? domain of what?).

When you restrict the domain of $$L_A$$ to $$Q$$, you actually get a different linear transformation; you're getting the map $$L_A\big|_Q : Q \to \Bbb{R}^m$$. So, even though the maps $$L_A$$ and $$L_A \big|_Q$$ follow the same "rule", they are strictly speaking different functions (because their domains are different). Obviously in general, \begin{align} \text{image}(L_A) \neq \text{image}\big( L_A \big|_Q \big) \end{align}

So, the correct and specific statement is (in general) "the column space of the given matrix $$A$$ equals the image of $$L_A$$, but in general is not equal to the image of $$L_A \big|_Q$$." Or as an equation: \begin{align} \text{col}(A) = \text{image}(L_A) \neq \text{image} \big( L_A \big|_Q \big) \end{align} (where once again I stress that the last $$\neq$$ sign could actually be $$=$$ in specific cases). So, I hope this serves as a good reminder to be specific about what you're actually talking about.

Question $$2$$:

1. Does this mean that if I restrict the domain space to $$Q$$ instead of $$R^n$$, the null space of the transformation must be restricted to vectors in $$Q$$ that gets mapped to the zero vector in the range of the transformation?

The answer is yes, because from an abstract perspective, if $$V,W$$ are vector spaces over a field $$F$$, and $$T:V \to W$$ is linear, we define the null space of $$T$$ to be \begin{align} \text{null}(T) := \{x \in V : \, T(x) = 0\} \end{align} It is elements of the domain which get mapped to the zero vector of the target space. So, in your particular case, when you restrict the domain of $$L_A$$ to $$Q$$ and ask about its nullspace, you're actually asking what is the null space of the linear transformation $$L_A \big|_Q : Q \to \Bbb{R}^m$$. You can actually prove (it's very easy) that \begin{align} \text{null}\big( L_A \big|_Q \big) = \text{null}(L_A) \cap Q \end{align}

Question $$3$$:

I've actually never heard the term "left null space", but the answer to your question is yes. The reason is because the concept of "row space" and "left null space" are defined for matrices, NOT for linear transformations, so it doesn't matter what you do to a linear transformation. Once again, your confusion here probably stems from not being specific enough in your choice of words.

Question $$4$$:

The answer here is "yes, you can make sense of the rank-nullity theorem", but once again you need to be specific with your wording (domain of what? domain space of what? null space of what?). You wrote

1. If I restrict the domain to a line in $$R^3$$ and $$A$$ is $$3x3$$ with linearly independent columns, is there some way to make sense of the rank nullity theorem as domain space has dim = 1, null space has dim = 0, rank of the matrix is 3?

So, in this particular case, $$m=n=3$$, and $$Q$$ is a certain $$1$$-dimensional subspace of $$\Bbb{R}^3$$, and $$A$$ is a matrix with full rank. The hypothesis can be equivalently phrased as saying that $$L_A: \Bbb{R}^3 \to \Bbb{R}^3$$ has an image of dimension $$3$$ (in this particular case, it means $$\text{image}(L_A) = \Bbb{R}^3$$). Now, we have two linear transformations in the game, the first is $$L_A$$ and the second is $$L_A\big|_Q$$, and we can apply the rank-nullity theorem to each of them. Doing so yields two equations: \begin{align} \begin{cases} \dim \text{null}(L_A) + \dim \text{image}(L_A) &= \dim \text{domain}(L_A) \\ \dim \text{null}(L_A\big|_Q) + \dim \text{image}(L_A \big|_Q) &= \dim \text{domain}(L_A \big|_Q) \end{cases} \end{align}

So, if you use the second of these equations and plug in the numbers you know, you can conclude that the image of the restricted map $$L_A\big|_Q$$ is a $$1$$-dimensional subspace of the target space $$\Bbb{R}^3$$. In this sense, there is no contradiction to the rank-nullity theorem; you just need to be careful about which linear map you're applying it to.

• Is $Col(A) = image(L_A)$ in this case becuase the domain of $L_A$ is the entire $R^n$? So of the subspaces I listed, only the null space relate to the linear transformation (LT), the rest I should view as some subspace of $R^n$ or $R^m$ indep. of the LT? WRT the rank-nullity theorem, my confusion was from substituting Rank(A) = 3 where $\dim \text{image}(L_A \big|_Q)$ sits. So I should really be using dim of the image of the actual LT instead of the rank of its matrix / dim of the column space; however if the domain is $R^n$, all of these equal? Commented Aug 7, 2019 at 23:48
• first question: yes. Next, no, both the null space and image are subspaces which depend on the linear transformation. As for rank nullity, we have that $\text{rank}(A) = \dim \text{image}(L_A) = 3$. So, substituting this into the first equation I wrote implies that $\dim \text{null}(L_A) = 0$. Now, since $\text{null}(L_A|_Q) = \text{null}(L_A) \cap Q$, this implies the null space of $L_A|_Q$ is also the trivial subspace $\{0\}$; hence has dimension zero. Substituting this into the second R-N equation I wrote implies that $\dim \text{image}(L_A|_Q) = 1$ Commented Aug 8, 2019 at 0:55
• Note that the matrix $A$ corresponds to the linear map $L_A : \Bbb{R}^n \to \Bbb{R}^m$. The matrix $A$ has (roughly speaking in your context) nothing to do with $L_A|_Q$. So, I suggest you keep this in mind: if you have some property of a matrix $A$, then you can equivalently phrase it in terms of the linear map $L_A$. That's it. If for some reason you restrict the domain of $L_A$, then that new map has almost nothing to do with $A$ (in reality, there is some connection, but I suggest you focus on the basics) Commented Aug 8, 2019 at 0:57
• Based on this article, is it correct to say that a real $m \times n$ matrix $A$ always correspond to some linear map with domain $R^n$ (and why the Wikipedia article in my OP states $Col(A)$ and $image(L_A)$ as equivalent), so in my case where the linear map that is some subspace of $R^n$, I shouldn't be using a matrix to represent this linear map? Commented Aug 8, 2019 at 19:18
• To each $m \times n$ matrix $A$, we can associate a unique linear transformation with domain $\Bbb{R}^n$ and target $\Bbb{R}^m$ (for example $L_A$) (and vice-versa). Also, if you have a linear map $L_A|_Q:Q \to \Bbb{R}^m$, you CAN associate it to a matrix; but the matrix depends on the choice of basis for $Q$. So, it is very likely that the matrix representation of $L_A|_Q$ and the initial matrix $A$ are very different. Commented Aug 8, 2019 at 21:45