Question

For a matrix 𝑀, let Rowspace(𝑀) denote the linear span of the rows of 𝑀 and
Colspace(𝑀) denote the linear span of the columns of 𝑀. Which of the
following hold(s) for all 𝐴, 𝐵, 𝐶 ∈ ${𝑀}_{10}$ (ℝ) satisfying 𝐴 = 𝐵𝐶 ?

1. Rowspace(𝐴) ⊆ Rowspace(𝐵)
2. Rowspace(𝐴) ⊆ Rowspace(𝐶)
3. Colspace(𝐴) ⊆ Colspace(𝐵)
4. Colspace(𝐴) ⊆ Colspace(𝐶)

Answer:
Option-2. Rowspace(𝐴) ⊆ Rowspace(𝐶)
Option-3. Colspace(𝐴) ⊆ Colspace(𝐵)

Solution :-

We know that is $A$ is square matrix of order $n$ with entries from $ℝ$ then, $Rowspace\left(A\right)=Colspace\left(A^T\right)=\{A^{T}x:x\in {ℝ}^{n\times 1}\}$ and $Colspace\left(A\right)=\{Ax:x\in {ℝ}^{n\times 1}\}\mathrm{.}$

Also, $Rowspace\left(A\right)$ and $Colspace\left(A\right)$ is subspace of ${ℝ}^n\mathrm{.}$

Therefore, if 𝐴, 𝐵, 𝐶 ∈ ${𝑀}_{10}$ (ℝ) then,

$Rowspace\left(A\right)=Colspace\left(A^T\right)=\{A^{T}x:x\in {ℝ}^{10\times 1}\}$ , which is a subspace of ${ℝ}^{10}$ and $Colspace\left(A\right)=\{Ax:x\in {ℝ}^{10\times 1}\}$ , which is a subspace of ${ℝ}^{10}\mathrm{.}$

$Rowspace\left(B\right)=Colspace\left(B^T\right)=\{B^{T}x:x\in {ℝ}^{10\times 1}\}$ , which is a subspace of ${ℝ}^{10}$ and $Colspace\left(B\right)=\{Bx:x\in {ℝ}^{10\times 1}\}$ , which is a subspace of ${ℝ}^{10}\mathrm{.}$

$Rowspace\left(C\right)=Colspace\left(C^T\right)=\{C^{T}x:x\in {ℝ}^{10\times 1}\}$ , which is a subspace of ${ℝ}^{10}$ and $Colspace\left(C\right)=\{Cx:x\in {ℝ}^{10\times 1}\}$ , which is a subspace of ${ℝ}^{10}\mathrm{.}$

Now, we want to check if we take a general vector from Row Space of A and use the fact 𝐴 = 𝐵𝐶 then that general vector where to go.

So, if $y\in Rowspace\left(A\right)$ then $y$ is of this form $y=A^{T}x$ where, $x\in {ℝ}^{10\times 1}\mathrm{.}$

Here, we can see that we took a general vector $y$ from the Row Space of A then that vector $y$ is also in Row Space of C, if 𝐴 = 𝐵𝐶 then.

Now, we want to check if we take a general vector from Column Space of A and use the fact 𝐴 = 𝐵𝐶 then that general vector where to go.

So, if $y\in Colspace\left(A\right)$ then $y$ is of this form $y=Ax$ where, $x\in {ℝ}^{10\times 1}\mathrm{.}$

Here, we can see that we took a general vector $y$ from the Column Space of A then that vector $y$ is also in Column Space of B, if 𝐴 = 𝐵𝐶 then.