The Timpson Lyceum

1. (2 pts) Determine whether the columns of $\begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}$ are linearly independent.

Solution

1. Neither column is a scalar multiple of the other: $(2,5) = k(1,3)$ would force $k = 2$ from the first entry yet $k = \tfrac{5}{3}$ from the second, which is impossible. By the two-vector criterion the columns are linearly independent. (Equivalently, row reduce $\begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix} \xrightarrow{R_2 - 3R_1} \begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix}$ : a pivot in every column, so $c_1\mathbf{a}_1 + c_2\mathbf{a}_2 = \mathbf{0}$ has only the trivial solution.)

2. (3 pts) Determine whether $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$ , $\mathbf{v}_2 = \begin{bmatrix} 2 \\ 4 \\ 3 \end{bmatrix}$ , $\mathbf{v}_3 = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}$ are linearly independent. If they are dependent, give an explicit dependence relation.

Solution

2. Solve $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3 = \mathbf{0}$ by reducing $[\mathbf{v}_1\;\mathbf{v}_2\;\mathbf{v}_3]$ :

$\begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 3 & 0 \end{bmatrix} \xrightarrow[R_3 - R_1]{R_2 - 2R_1} \begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 1 & -1 \end{bmatrix}.$

Only two pivots (columns 1 and 2), so $c_3$ is free — a nontrivial solution exists, and the set is linearly dependent. Taking $c_3 = 1$ : from $c_2 - c_3 = 0$ , $c_2 = 1$ ; from $c_1 + 2c_2 + c_3 = 0$ , $c_1 = -3$ . Dependence relation:

$-3\mathbf{v}_1 + \mathbf{v}_2 + \mathbf{v}_3 = \mathbf{0}, \quad\text{i.e.}\quad \mathbf{v}_2 + \mathbf{v}_3 = 3\mathbf{v}_1.$

3. (3 pts) Determine whether $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$ , $\mathbf{v}_2 = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}$ , $\mathbf{v}_3 = \begin{bmatrix} 0 \\ 1 \\ 3 \end{bmatrix}$ are linearly independent.

Solution

3. Reduce $[\mathbf{v}_1\;\mathbf{v}_2\;\mathbf{v}_3]$ :

$\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 3 \end{bmatrix} \xrightarrow{R_3 - R_1} \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 3 \end{bmatrix} \xrightarrow{R_3 + 2R_2} \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 5 \end{bmatrix}.$

There is a pivot in every column, so $c_1 = c_2 = c_3 = 0$ is the only solution. The vectors are linearly independent.

4. (3 pts) Find all values of $h$ for which the vectors $\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix}$ , $\begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix}$ , $\begin{bmatrix} h \\ 2 \\ -3 \end{bmatrix}$ are linearly dependent.

Solution

4. Row reduce $[\mathbf{v}_1\;\mathbf{v}_2\;\mathbf{v}_3]$ , keeping $h$ symbolic:

$\begin{bmatrix} 1 & 3 & h \\ -1 & 0 & 2 \\ 2 & 1 & -3 \end{bmatrix} \xrightarrow[R_3 - 2R_1]{R_2 + R_1} \begin{bmatrix} 1 & 3 & h \\ 0 & 3 & h+2 \\ 0 & -5 & -3-2h \end{bmatrix} \xrightarrow{R_3 \,\to\, 3R_3 + 5R_2} \begin{bmatrix} 1 & 3 & h \\ 0 & 3 & h+2 \\ 0 & 0 & 1 - h \end{bmatrix}.$

The set is linearly dependent exactly when column 3 has no pivot, i.e. when $1 - h = 0$ . So the vectors are linearly dependent iff $h = 1$ (and independent for every other value of $h$ ).

5. (3 pts) State whether each of the following is true or false, and justify each answer.

(a) A set of two vectors is linearly dependent if and only if one is a scalar multiple of the other.

(b) Any set of four vectors in $\mathbb{R}^3$ is linearly dependent.

(c) If $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is linearly dependent, then each of the three vectors is a linear combination of the other two.

Solution

5.

(a) True. If $\mathbf{v}_2 = k\mathbf{v}_1$ then $k\mathbf{v}_1 - \mathbf{v}_2 = \mathbf{0}$ is a nontrivial relation; conversely a nontrivial relation $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 = \mathbf{0}$ with, say, $c_2 \neq 0$ gives $\mathbf{v}_2 = -\tfrac{c_1}{c_2}\mathbf{v}_1$ . (The zero-vector edge cases are included: $\mathbf{0} = 0\cdot\mathbf{v}$ .)

(b) True. Four vectors in $\mathbb{R}^3$ are more vectors than entries; by Theorem 8 any such set is linearly dependent.

(c) False. Dependence only guarantees some vector is a combination of the rest. Counterexample: $\mathbf{v}_1 = (1,0)$ , $\mathbf{v}_2 = (0,1)$ , $\mathbf{v}_3 = \mathbf{0}$ . The set is dependent, but $\mathbf{v}_1$ is not a combination of $\mathbf{v}_2$ and $\mathbf{v}_3$ .

6. (3 pts) Without any calculation, decide whether each set is linearly independent or dependent, and name the theorem or fact that justifies your answer.

(a) $\left\{\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}, \begin{bmatrix} 5 \\ 6 \\ 7 \\ 8 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}\right\}$

(b) A set of five vectors in $\mathbb{R}^4$ .

(c) $\left\{\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\right\}$

Solution

6.

(a) Dependent. The set contains the zero vector; a set containing $\mathbf{0}$ is always linearly dependent (take coefficient $1$ on $\mathbf{0}$ and $0$ elsewhere).

(b) Dependent. Five vectors in $\mathbb{R}^4$ is more vectors than entries; by Theorem 8 the set is dependent.

(c) Independent. Two vectors, neither a scalar multiple of the other (they point along different axes), so by the two-vector criterion the set is independent.

7. (3 pts) Prove that $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is linearly dependent if and only if at least one of the vectors is a linear combination of the other two.

Solution

7. ( $\Leftarrow$ ) Suppose one vector, say $\mathbf{v}_3 = a\mathbf{v}_1 + b\mathbf{v}_2$ . Then $a\mathbf{v}_1 + b\mathbf{v}_2 - \mathbf{v}_3 = \mathbf{0}$ is a linear dependence with a nonzero coefficient ( $-1$ on $\mathbf{v}_3$ ), so the set is dependent.

( $\Rightarrow$ ) Suppose the set is dependent: there are scalars $c_1, c_2, c_3$ , not all zero, with $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3 = \mathbf{0}$ . Pick an index $k$ with $c_k \neq 0$ ; then $\mathbf{v}_k = -\tfrac{1}{c_k}\sum_{j \neq k} c_j\mathbf{v}_j$ , expressing $\mathbf{v}_k$ as a linear combination of the other two. $\blacksquare$

8. (4 pts ••) Suppose $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is linearly independent. Prove that $\{\mathbf{v}_1 + \mathbf{v}_2, \; \mathbf{v}_2 + \mathbf{v}_3, \; \mathbf{v}_1 + \mathbf{v}_3\}$ is also linearly independent.

Solution

8. Suppose $a(\mathbf{v}_1 + \mathbf{v}_2) + b(\mathbf{v}_2 + \mathbf{v}_3) + c(\mathbf{v}_1 + \mathbf{v}_3) = \mathbf{0}$ . Grouping by $\mathbf{v}_i$ :

$(a + c)\mathbf{v}_1 + (a + b)\mathbf{v}_2 + (b + c)\mathbf{v}_3 = \mathbf{0}.$

Since $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is linearly independent, all coefficients vanish:

$a + c = 0, \quad a + b = 0, \quad b + c = 0.$

Subtracting the second from the first gives $c - b = 0$ , so $b = c$ ; the third then gives $2b = 0$ , so $b = c = 0$ , and $a + c = 0$ gives $a = 0$ . The only solution is $a = b = c = 0$ , so the set $\{\mathbf{v}_1 + \mathbf{v}_2, \mathbf{v}_2 + \mathbf{v}_3, \mathbf{v}_1 + \mathbf{v}_3\}$ is linearly independent. $\blacksquare$