How to find matrices that satisfy subspaces conditions?

Question

I'm having a hard time trying to learn some linear algebra wizardry. Can someone explain to me how to solve this problem:

P4.18 Find 2x3 matrices A,B, and C, that satisfy the conditions.

P4.18 Find $2 \times 3$ matrices $A,B,C$ that satisfy the conditions.

1. $R(A) = \operatorname{span}((1,2,3))$ and $C(A) = \operatorname{span}((1,2)^T)$,
2. $N(B) = \operatorname{span}((1,1,1)^T)$,
3. $N(C^T) = \operatorname{span}((1,3))$.

Note: when I write $R(A)$, $N(B)$, $N(C^T)$, I mean rowspace of A, Null-space of B and left-null-space of C.

Thank you considering answering this question!! The problem is from the book No Bullshit Linear Algebra by Ivan Savov, on page 255. The suggested solution is:

a)

1, 2, 3
2, 4, 6

b)

1, -1, 0
1, 0, -1

c)

3, 6, 9
-1, -2, -3

Other answers are possible.

Answer 1

I’ll give this a shot:

Let’s say $A$ looks like $\begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix}$, then $R(A)= K_1\cdot \begin{bmatrix} a_{11} & a_{12} & a_{13}\end{bmatrix} + K_2\cdot \begin{bmatrix} a_{21} & a_{22} & a_{23} \end{bmatrix}=K\cdot\begin{bmatrix} 1&2&3\end{bmatrix} = span\left\{\begin{bmatrix}1&2&3\end{bmatrix}\right\}$

for some constants $K,K_1,K_2$

and since there is only one vector in the span, we have a basis for the row-space of $A$ and the rows of $A$ must be linearly dependent and hence multiples of each other:

$k\cdot\begin{bmatrix} a_{11} & a_{12} & a_{13}\end{bmatrix}= \begin{bmatrix} a_{21} & a_{22} & a_{23}\end{bmatrix}$

for some constant $k$. Now every column vector of $A$ looks like $a_{1j}\cdot\begin{bmatrix}1 \\k\end{bmatrix}$ for $j=1,2,3$.

$C(A)= a_{11}\cdot\begin{bmatrix}1 \\k\end{bmatrix}+ a_{12}\cdot\begin{bmatrix}1 \\k\end{bmatrix}+ a_{13}\cdot\begin{bmatrix}1 \\k\end{bmatrix}=\mathcal{K}\cdot\begin{bmatrix}1 \\k\end{bmatrix}=span\left\{\begin{bmatrix}1 \\k\end{bmatrix}\right\} $

and the second clue gives us that $C(A)= span\left\{\begin{bmatrix}1 \\2\end{bmatrix}\right\}$ and $k=2$.

$A=\begin{bmatrix} 1&2&3\\2&4&6\end{bmatrix}$ seems to work fine (letting $[a_{11} \: a_{12} \: a_{13} ]= [1\:2\:3]$).

On the other hand, the null-space is the solution space of the homogeneous system(essentially add a column of zeros, and solve the system)

The solution space of $\left[\begin{array}{ccc|c}b_{11}&b_{12}&b_{13}&0\\b_{21}&b_{22}&b_{23}&0\end{array}\right]$ is said to look like $t\cdot \begin{bmatrix}1\\1\\1\end{bmatrix}$.

An easy $2\times 3$ matrix which would have this could be

$B=\begin{bmatrix}1&0&-1\\0&1&-1\end{bmatrix}$ which is similar to the answer provided only the rows are flipped.

To see why, one can imagine solving the system of linear equations parametrically:

$\left\{\begin{array}{c}x-z=0\\y-z=0\end{array}\right.$ one can bring $z$ over to the right and let $z=t$

$\left\{\begin{array}{c}x=z=t\\y=z=t\end{array}\right.$ so the general solution looks like $\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}t\\t\\t\end{bmatrix}=t\cdot\begin{bmatrix}1\\1\\1\end{bmatrix}$.

Similarly for $C=\begin{bmatrix}3&6&9\\-1&-2&-3\end{bmatrix}$ is an acceptable answer because $C^T = \begin{bmatrix}3&-1\\6&-2\\9&-3\end{bmatrix}$ and the system of equations

$\left\{\begin{array}{c}3x-y=0\\6x-2y=0\\9x-3y=0\end{array}\right.\Longrightarrow \left\{\begin{array}{c}3x=y\\6x=2y\\9x=3y\end{array}\right. \Longrightarrow 3x=y$

the general solution has form $\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\\3x\end{bmatrix}=x\cdot\begin{bmatrix}1\\3\end{bmatrix}=span\left\{\begin{bmatrix}1\\3\end{bmatrix}\right\}$

I think there a $^T$ missing in #3.

Hope this helps!