Dimension of the column space and null space does not match my expectations?

Question

I'm looking at the last part of Question 31 in the link which states:

Section 3.6 Problem 31 : $\mathbf{M}$ is the space of 3 by 3 matrices. Multiply each matrix $X$ in $\mathbf{M}$ by $$ \begin{bmatrix} 1 & 0 & -1\\ -1 & 1 & 0 \\ 0 & -1 & 1 \\ \end{bmatrix}. \text{Notice}: A \begin{bmatrix} 1\\ 1\\ 1\\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0\\ \end{bmatrix} $$

(We are skipping the previous parts of this question)

$(a)$ find the "nullspace" of that operation $AX$ and $(b)$ find the "column space". Why do the dimensions add to $(n-r)+r=9$?

I see that for part a you can solve for the nullspace of A. It says you're solving for the nullspace of the operation AX but because AX is a linear combination of A, it's my understanding that it doesn't make a difference.

However, for part b I'm pretty lost. Solving for the column space of B definitely will not get $\dim(C(B))=6$ or $\dim(N(B))=3$. Where are these dimensions coming from?

Answer 1

Essentially what they are saying is that any $3\times3$ matrix $B$ such that $AX=B$ must have the prescribed form. There are nine elements in your matrix, and the first two rows can be freely chosen, hence the "column space" being 6-dimensional. The last row is then uniquely determined, and so your "null space" is 3-dimensional.

These spaces are not referring to $B$, but rather to the $9\times9$ matrix that you get by equating the nine elements of $B$ with the nine elements of $AX$.

Answer 2

The space of $3\times 3$ real matrices is a 9-dimensional vector space over the reals. The null space of your mapping $X\mapsto AX$ is the subspace of all $3\times 3$ matrices $X$ such that $AX$ is the $3\times 3$ zero matrix.

The hint tells you that $A$ applied to a scalar multiple of the vector $(1,1,1)$ is the zero vector; in fact, these are the only vectors that $A$ maps to zero. Using the fact that the columns of the matrix $AX$ are just $A$ times the respective columns of $X$, we conclude that for $AX$ to be the zero matrix, the columns of $X$ must be scalar multiples of $(1,1,1)$. But it doesn't matter which scalar multiple, so you have three degrees of freedom: one multiple for the first column, one for the second, and one for the third. This shows the nullspace of the mapping $X\mapsto AX$ is 3-dimensional. It follows by the rank-nullify theorem that the column space is 6-dimensional.

(Note carefully that the nullspace of $A$ construed as a mapping from vectors to vectors is not the same thing as the nullspace of $A$ construed as a mapping from matrices to matrices. In the former case $A$ sends a $3\times 1$ vector to a $3\times 1$ vector. In the latter case, $A$ sends a $3\times 3$ matrix to a $3\times 3$ matrix. These are entirely different objects. The former is a linear endomorphism on a $3$-dimensional vector space. The latter is a linear endomorphism on a $9$-dimensional vector space.)

Answer 3

People often misapprehend what a vector is: a vector is, quite simply, any element of a vector space.

This sort of begs the question: what is a vector space?

The simple (and sort of inaccurate) answer is: a structure where we can:

$1)$ Add stuff (vectors) together.

$2)$ Scale (intuitively, this means "stretch" or "shrink") vectors by a field element (this scaling is called "scalar multiplication", and the field elements are thus called "scalars").

The more correct (but more formal, and somewhat mystifying) answer is, a vector space is any set, together with an associated field, that satisfies a certain set of axioms (these are usually listed as $10$ to $12$ "rules", such as "associativity and commutativity of vector addition"). The field itself, has to satisfy another set of axioms (the "field axioms", which are usually just accepted as "the rules of arithmetic", when one is talking about well-known fields, such as the rational or real numbers). For example one vector space axiom is:

For any two scalars $a,b$ and any vector $v$:

$(a+b)v = ab + bv$ (note that the "$+$" means two different things on each side of the equals sign-the one on the left is the sum of two scalars, the one on the right is the sum of two vectors).

I won't list the vector space axioms here, they can be found in almost any worthwhile text on linear algebra.

So, it turns out that any $3 \times 3$ matrix:

$\begin{bmatrix}a&b&c\\d&e&f\\g&h&j\end{bmatrix}$ can be written as the linear combination of nine basis matrices, like so:

$\begin{bmatrix}a&b&c\\d&e&f\\g&h&j\end{bmatrix} = a\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}+b\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}+c\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}\\ +d\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}+e\begin{bmatrix}0&0&0\\0&1&0\\0&0&0\end{bmatrix}+f\begin{bmatrix}0&0&0\\0&0&1\\0&0&0\end{bmatrix}\\ +g\begin{bmatrix}0&0&0\\0&0&0\\1&0&0\end{bmatrix}+h\begin{bmatrix}0&0&0\\0&0&0\\0&1&0\end{bmatrix}+j\begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix}.$

If we want to find the nullspace of the linear transformation:

$L_A: \mathbf{M} \to \mathbf{M}$ which takes $X$ to the matrix $AX$, and we write:

$A = \begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}$ and:

$X = \begin{bmatrix}x_{11}&x_{12}&x_{13}\\x_{21}&x_{22}&x_{23}\\x_{31}&x_{32}&x_{33}\end{bmatrix}$

This involves solving the $9$ linear equations (in the $9$ unknowns $x_{ij}$):

$a_{11}x_{11} + a_{12}x_{21} + a_{13}x_{31} = 0\\ a_{21}x_{11} + a_{22}x_{21} + a_{23}x_{31} = 0\\ a_{31}x_{11} + a_{32}x_{21} + a_{33}x_{31} = 0$

---

$a_{11}x_{12} + a_{12}x_{22} + a_{13}x_{32} = 0\\ a_{21}x_{12} + a_{22}x_{22} + a_{23}x_{32} = 0\\ a_{31}x_{12} + a_{32}x_{22} + a_{33}x_{32} = 0$

---

$a_{11}x_{13} + a_{12}x_{23} + a_{13}x_{33} = 0\\ a_{21}x_{13} + a_{22}x_{23} + a_{23}x_{33} = 0\\ a_{31}x_{13} + a_{32}x_{23} + a_{33}x_{33} = 0.$

Note how I have grouped the nine equations. If we write $X_i = \begin{bmatrix}x_{1i}\\x_{2i}\\x_{3i}\end{bmatrix}$ for $i = 1,2,3$

we see that each group of three represents the condition $AX_i = \begin{bmatrix}0\\0\\0\end{bmatrix}$,

that is to say, each $X_i$ is in the nullsapce of the matrix $A$ (not $L_A$).

Now this is true if and only if $X_i = c_i\begin{bmatrix}1\\1\\1\end{bmatrix}$ for some scalars $c_i$ (they might conceivably be different for different $i$), that is to say, it is necessary and sufficient to merely specify the first row of $X$, since the subsequent row entries must match the matching column entry above. It takes $3$ scalars to do so, so the dimension of the nullspace of $L_A$ is $3$, it has the basis:

$\left\{\begin{bmatrix}1&0&0\\1&0&0\\1&0&0\end{bmatrix}, \begin{bmatrix}0&1&0\\0&1&0\\0&1&0\end{bmatrix}, \begin{bmatrix}0&0&1\\0&0&1\\0&0&1\end{bmatrix}\right\}$

The computation of the range of $L_A$ is done in essentially the same way. We see that the columns of $AX$ are the $3$-vectors $AX_i$, so, for example, with $i = 1$, we have:

$x_{11} - x_{31} = b_{11}\\ -x_{11} + x_{21} = b_{21}\\ -x_{21} + x_{31} = b_{31}$

so that $b_{11} + b_{21} + b_{31} = 0$.

So specifying two elements of each column of $AX$ determines the third, which means we need $6$ scalars to completely determine $AX$, that is: the range of $L_A$ has dimension $6$.