Due Oct 21 Friday @ 11:59pm
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Q1. Suppose the column space of an $m \times n$ matrix is all of $\mathbb{R}^3$. What can you say about $m$? What can you say about $n$? What can you say about the rank $r$?
Q2. Let $$ \mathbf{A}_1 = \begin{pmatrix} 1 & 3 & -2 \\ 3 & 9 & -6 \\ 2 & 6 & -4 \end{pmatrix}, \quad \mathbf{A}_2 = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}. $$
Find the matrices $\mathbf{C}_1$ and $\mathbf{C}_2$ containing independent columns of $\mathbf{A}_1$ and $\mathbf{A}_2$.
Find a rank factorization $\mathbf{A} = \mathbf{C} \mathbf{R}$ of each of $\mathbf{A}_1$ and $\mathbf{A}_2$.
Produce a basis for the column spaces of $\mathbf{A}_1$ and $\mathbf{A}_2$. What are the dimensions of those column spaces (the number of independent vectors)? What are the ranks of $\mathbf{A}_1$ and $\mathbf{A}_2$? How many independent rows in $\mathbf{A}_1$ and $\mathbf{A}_2$?
Q3. How is the null space of $\mathbf{C}$ related to the nullspaces of $\mathbf{A}$ and $\mathbf{B}$, if $$ \mathbf{C} = \begin{pmatrix} \mathbf{A} \\ \mathbf{B} \end{pmatrix}. $$
Q4. Do $\mathbf{A}^2$ and $\mathbf{A}$ always have the same null space? $\mathbf{A}$ is a square matrix.
Find a square matrix with $\text{rank}(\mathbf{A}^2) < \text{rank}(\mathbf{A})$. Confirm that $\text{rank}(\mathbf{A}'\mathbf{A}) = \text{rank}(\mathbf{A})$.
Q5. In this exercise, we show the fact $$ \text{rank}(\mathbf{A} + \mathbf{B}) \le \text{rank}(\mathbf{A}) + \text{rank}(\mathbf{B}) $$ for any two matrices $\mathbf{A}$ and $\mathbf{B}$ of same size in steps.
Show that the sum of two vector spaces $\mathcal{S}_1$ and $\mathcal{S}_2$ of same order $$ \mathcal{S}_1 + \mathcal{S}_2 = \{\mathbf{x}_1 + \mathbf{x}_2: \mathbf{x}_1 \in \mathcal{S}_1, \mathbf{x}_2 \in \mathcal{S}_2\} $$ is a vector space.
Show that $\text{dim}(\mathcal{S}_1 + \mathcal{S}_2) \le \text{dim}(\mathcal{S}_1) + \text{dim}(\mathcal{S}_2)$.
Show that $\mathcal{C}(\mathbf{A} + \mathbf{B}) \subseteq \mathcal{C}(\mathbf{A}) + \mathcal{C}(\mathbf{B})$.
Conclude that $\text{rank}(\mathbf{A} + \mathbf{B}) \le \text{rank}(\mathbf{A}) + \text{rank}(\mathbf{B})$.