**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.