# Recitation 3

Problem 1: Let $\mathbf{u}=\begin{bmatrix}3\\2\\-4\end{bmatrix}$, $\mathbf{v}=\begin{bmatrix}-6\\1\\7\end{bmatrix}$, $\mathbf{w}=\begin{bmatrix}0\\-5\\2\end{bmatrix}$ and $\mathbf{z}=\begin{bmatrix}3\\7\\-5\end{bmatrix}$.

1. Are the sets $\{\mathbf{u}, \mathbf{v}\}, \{\mathbf{u},\mathbf{w}\}, \{\mathbf{u},\mathbf{z}\}, \{\mathbf{v},\mathbf{w}\}, \{\mathbf{v},\mathbf{z}\}$ and $\{\mathbf{w},\mathbf{z}\}$ each linearly independent? Why or why not?
2. Does the answer to Problem 1 imply that $\{\mathbf{u}, \mathbf{v}, \mathbf{w}, \mathbf{z}\}$ is linearly independent?
3. To determine if $\{\mathbf{u}, \mathbf{v}, \mathbf{w}, \mathbf{z}\}$ is linearly dependent, is it wise to check if, say $\mathbf{w}$ is a linear combination of $\mathbf{u}, \mathbf{v}$ and $\mathbf{z}$?
4. Is $\{\mathbf{u}, \mathbf{v}, \mathbf{w}, \mathbf{z}\}$ linearly dependent?

Solution:

1. All the sets are linearly independent because none of the vectors is a multiple of another.
2. The answer to part 1 doesn’t imply that $\{\mathbf{u}, \mathbf{v}, \mathbf{w}, \mathbf{z}\}$ is linearly independent. In general, pairwise linear independency does not guarantee the global linear independency.
3. It is not wise to check if $\mathbf{w}$ is a linear combination of $\mathbf{u}, \mathbf{v}$ and $\mathbf{z}$ to determine if $\{\mathbf{u}, \mathbf{v}, \mathbf{w}, \mathbf{z}\}$ is linearly dependent. However, if $\mathbf{w}$ happens to be a linear combination of $\mathbf{u}, \mathbf{v}$ and $\mathbf{z}$, we can conclude that $\{\mathbf{u}, \mathbf{v}, \mathbf{w}, \mathbf{z}\}$ is linearly dependent. On the other hand, it is not necessarily true that $\mathbf{w}$ being a linear combination of $\mathbf{u}, \mathbf{v}$ and $\mathbf{z}$,a linear combination of $\mathbf{u}, \mathbf{v}$ and $\mathbf{z}$ implies that $\{\mathbf{u}, \mathbf{v}, \mathbf{w}, \mathbf{z}\}$ is linearly dependent.
4. The set $\{\mathbf{u}, \mathbf{v}, \mathbf{w}, \mathbf{z}\}$ is linearly dependent because the number of vectors is greater than the dimension.

Problem 2: For what values of $h$ is $\mathbf{v_3}$ in $\mathrm{span}(\mathbf{v_1}, \mathbf{v_2})$ and for what values of $h$ is $\{\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3}\}$ linearly dependent? Justify each answer. $$\mathbf{v_1}=\begin{bmatrix}1\\-3\\2\end{bmatrix}, \mathbf{v_2}=\begin{bmatrix}-3\\9\\-6\end{bmatrix}, \mathbf{v_3}=\begin{bmatrix}5\\-7\\h\end{bmatrix}.$$

Solution: Consider the vector equation, $x_1\mathbf{v_1}+x_2\mathbf{v_2}=\mathbf{v_3}$. The correspondent augmented matrix is $$\begin{bmatrix}1 & -3 & 5\\-3 & 9 & -7\\2 & -6 & h\end{bmatrix}.$$ After a row reduction, we obtain $$\begin{bmatrix}1 & -3 & 5\\0 & 0 & 8\\2 & -6 & h\end{bmatrix},$$ in which the second row implies a contradiction. Hence the vector equation has no solutions. Therefore $\mathbf{v_3}$ is not in the span of $\mathbf{v_1}$ and $\mathbf{v_2}$ for all values of $h$.

On the other hand, consider the vector equation $x_1\mathbf{v_1}+x_2\mathbf{v_2}+x_3\mathbf{v_3}=0$. The correspondent augmented matrix is $$\begin{bmatrix}1 & -3 & 5 & 0\\-3 & 9 & -7 & 0\\2 & -6 & h & 0\end{bmatrix}.$$ After three row reductions, we obatin $$\begin{bmatrix}1 & -3 & 5 & 0\\0 & 0 & 8 & 0\\0 & 0 & 0 & 0\end{bmatrix},$$ in which the second column is not a pivot column. Hence the vector equation has infinite solutions. Therefore $\{\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3}\}$ is linearly dependent for all values of $h$.

Problem 3: Describe the possible echelon forms of a $2\times 2$ matrix with linearly dependent columns.

Solution: The echelon form of a $2\times 2$ matrix $A$ can be written as $$\begin{bmatrix}a & b\\0 & d\end{bmatrix}$$. Because $A$ has linearly dependent columns, the augmented matrix $$\begin{bmatrix}A & \mathbf{0}\end{bmatrix}$$ has infinitely many solutions. Since $A$ is equivalent to its echelon form, the augmented matrix $$\begin{bmatrix}A & \mathbf{0}\end{bmatrix}$$ is equivalent to the augmented matrix $$\begin{bmatrix}a & b & 0\\0 & d & 0\end{bmatrix}$$. Therefore, the augmented matrix $$\begin{bmatrix}a & b & 0\\0 & d & 0\end{bmatrix}$$ has infinite solutions. Either $a=0$ or $d=0$ can guarantee this. Hence the possible echelon forms are $$\begin{bmatrix}0 & b\\0 & d\end{bmatrix}, \begin{bmatrix}a & b\\0 & 0\end{bmatrix}.$$

Problem 4: Suppose an $m\times n$ matrix $A$ has $n$ pivot columns. Explain why for each $\mathbf{b}$ in $\mathbb{R}^m$ the equation $A\mathbf{x}=\mathbf{b}$ has at most one solution.

Solution: Assume for sake of contradiction that $A\mathbf{x}=\mathbf{b}$ has two different solutions, say $\mathbf{x_1}$ and $\mathbf{x_2}$. Then $A\mathbf{x_1}=\mathbf{b}$ and $A\mathbf{x_2}=\mathbf{b}$ imply that $A\mathbf{x_0}=\mathbf{0}$, where $\mathbf{x_0} = \mathbf{x_1} - \mathbf{x_2} \neq \mathbf{0}$. Since $A$ has $n$ pivot columns, i.e., all its columns are pivot columns, the augmented matrix $$\begin{bmatrix}A & \mathbf{0}\end{bmatrix}$$ has a unique solution, that is the zero solution. However $\mathbf{x_0}$ is not a zero vector and satisfies $A\mathbf{x_0}=\mathbf{0}$. This is a contradiction.