Construct $A = \mathbf{c_1{r_1}^T + c_2{r_2}^T}$ whose column space has basis $(1, 2, 4), (2, 2, 1)$ and whose row space has basis $(1, 0), (1, 1)$.
Answer: $A = \begin{bmatrix}
\mathbf{c_1} & \mathbf{c_2} \\
\end{bmatrix}
\begin{bmatrix}
\mathbf{{r_1}^T} \\
\mathbf{{r_2}^T} \\
\end{bmatrix} =
\begin{bmatrix}
1 & 2 \\
2 & 2 \\
4 & 1 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
1 & 1 \\
\end{bmatrix}
$
has coln space spanned by $\mathbf{c_i}$ and row space by $\mathbf{r_i}. \quad \square$P124: The column space of $A = C(A)$ consists of all linear combinations of the columns.
The combinations are all possible vectors $\mathbf{Ax}$.
P171: The row space is the column space of $A^T: C(A^T)$.
$\Large{{1-2.}}$ How and why does the construction in the answer work? How was it conceived? Possibly because the definitions for column and row spaces are given for matrix $\times$ vector, and not multiplication of matrices? I recognise though that vectors are matrices.
$\Large{{3.}}$ In general, suppose $C(B)$ has basis $\{\mathbf{c_i}\}_{1 \le i \le s}$ and $C(B^T)$ has basis $\{\mathbf{r_i}\}_{1 \le i \le s}$.
Would $B_{m \times n} = \begin{bmatrix}
\vert & & \vert \\
\mathbf{c_1} & \cdots & \mathbf{c_s} \\
\vert & & \vert \\
\end{bmatrix}_{m \times s}
\begin{bmatrix}
— & \mathbf{{r_1}^T} & — \\
& … & \\
— & \mathbf{{r_s}^T} & — \\
\end{bmatrix}_{s \times n}
= \sum\limits_{1 \le i \le s}\mathbf{c_s{r_s}^T} $ ? How and why?
Best Answer
1-2) This construction works because the columns are on the left and the rows are on the right. I imagine it was conceived by thinking about how matrix multiplication on the left and right produces a vector. On a more subtle note, vectors are only matrices when you are working in some basis. More generally, vectors do not a priori have a default matrix representation.
3) I you multiply $$B_{m \times n} = \begin{bmatrix} \vert & & \vert \\ \mathbf{c_1} & \cdots & \mathbf{c_s} \\ \vert & & \vert \\ \end{bmatrix}_{m \times s} \begin{bmatrix} -- & \mathbf{{r_1}^T} & -- \\ & ... & \\ -- & \mathbf{{r_s}^T} & -- \\ \end{bmatrix}_{s \times n} = \sum\limits_{1 \le i \le s}\mathbf{c_s{r_s}^T} $$ by a column vector on the right, the column times $\begin{bmatrix} -- & \mathbf{{r_1}^T} & -- \\ & ... & \\ -- & \mathbf{{r_s}^T} & -- \\ \end{bmatrix}_{s \times n}$ gives a new column. Multiplying this column by $\begin{bmatrix} \vert & & \vert \\ \mathbf{c_1} & \cdots & \mathbf{c_s} \\ \vert & & \vert \\ \end{bmatrix}_{m \times s}$ gives you an element of the column space. If you multiply a row vector on the left of $B_{m\times n}$ it similarly gives you an element of the row space.