# Recitation 8A

Definition: A subspace is a subset $S$ of $\mathbb{R}^n$ satisfying (1) the zero vector is in $S$; (2) if $\mathbf{u}$ and $\mathbf{v}$ are in $S$, then $\mathbf{u}+\mathbf{v}$ is in $S$; (3) if $\mathbf{u}$ is in $S$, then $c\mathbf{u}$ is in $S$ for all $c\in\mathbb{R}$.

Problem 1: Why the following sets are not subspace of $\mathbb{R}^2$? $$S_1=\{x\geq 0, y\geq 0\}, S_2=\{xy \leq 0\}, S_3=\{|x|\leq 1\}, S_4=\{x+y\leq 0\}.$$

Solution: Pick $(1,1)$ from $S_1$. But $(-1,-1)=(-1)(1,1)$ is not in $S_1$. This violates property (3). Pick $(1,0),(0,1)$ from $S_2$. But $(1,1)=(1,0)+(0,1)$ is not in $S_2$. This violates property (2). Pick $(1,0)$ from $S_3$. But $(2,0)=2(1,0)$ is not in $S_3$. This violates property (3). Pick $(-1,-1)$ from $S_4$. But $(1,1)=(-1)(-1,-1)$ is not in $S_4$. This violates property (3).

Definition: The column space of a matrix $M$ is the subspace generated by the column vectors of $M$. The null space of $M$ is the solution set of $M\mathbf{x}=\mathbf{0}$.

Problem 2: Construct bases for the column space and the null space of $A$, where $A=\begin{bmatrix}4&5&9&-2\\6&5&1&12\\3&4&8&-3\end{bmatrix}$ whose echelon form is given by $B=\begin{bmatrix}1&2&6&-5\\0&1&5&-6\\0&0&0&0\end{bmatrix}$.

Solution: The pivot columns of $A$, $\begin{bmatrix}4\\6\\3\end{bmatrix}, \begin{bmatrix}5\\5\\4\end{bmatrix}$ form a basis of the column space of $A$. To get the null space, we need to solve the linear system $A\mathbf{x}=\mathbf{0}$. Using the echelon form of $A$, we get $x_2=-5x_3+6x_4, x_1=-2x_2-6x_3+5x_4=-2(-5x_3+6x_4)-6x_3+5x_4=4x_3-7x_4$. In its parametric vector form, the solution set can be written as $\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}4\\-5\\1\\0\end{bmatrix}x_3+\begin{bmatrix}4\\6\\0\\1\end{bmatrix}x_4$. Therefore $\begin{bmatrix}4\\-5\\1\\0\end{bmatrix}$ and $\begin{bmatrix}4\\6\\0\\1\end{bmatrix}$ form a basis of the null space.

Problem 3: Find $\mathcal{B}$-coordinate vector of $\mathbf{x}$, where $\mathcal{B}$ is a subspace with basis $\mathbf{b_1}=\begin{bmatrix}1\\4\\-3\end{bmatrix}, \mathbf{b_2}=\begin{bmatrix}-2\\-7\\5\end{bmatrix}$, and $\mathbf{x}=\begin{bmatrix}2\\9\\-7\end{bmatrix}$.

Solution: Suppose $(x_1,x_2)$ is the $\mathcal{B}$-coordinate vector of $\mathbf{x}$. Then $x_1\mathbf{b_1}+x_2\mathbf{b_2}=\mathbf{x}$. Solve the linear system and get $(x_1,x_2)=(4,-1)$.

Problem 4: Is $\lambda=2$ an eigenvalue of $A=\begin{bmatrix}3 & 2 \\ 3 & 8\end{bmatrix}$?

Solution: Since the determinant of $A-\lambda I=\begin{bmatrix}1&2\\3&6\end{bmatrix}$ is zero, $A-\lambda I$ is not invertible. Therefore $2$ is indeed an eigenvalue of $A$.