# Quiz 3. Condition that Vectors are Linearly Dependent/ Orthogonal Vectors are Linearly Independent

## Problem 281

(a) For what value(s) of $a$ is the following set $S$ linearly dependent?
$S=\left \{\,\begin{bmatrix} 1 \\ 2 \\ 3 \\ a \end{bmatrix}, \begin{bmatrix} a \\ 0 \\ -1 \\ 2 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ a^2 \\ 7 \end{bmatrix}, \begin{bmatrix} 1 \\ a \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ -2 \\ 3 \\ a^3 \end{bmatrix} \, \right\}.$

(b) Let $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ be a set of nonzero vectors in $\R^m$ such that the dot product
$\mathbf{v}_i\cdot \mathbf{v}_j=0$ when $i\neq j$.
Prove that the set is linearly independent.

## Solution.

### (a) Linearly dependent vectors

Since the set $S$ consists of five $4$-dimensional vectors, it is linearly dependent regardless of the value of $a$. Thus, for any value of $a$ the set $S$ is linearly dependent.

### (b) Perpendicular nonzero vectors are linearly independent

Suppose that we have the linear combination
$c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3=\mathbf{0} \tag{*}$ for some scalars $c_1, c_2, c_3$.
To show that the set is linearly independent, we must show that $c_1=c_2=c_3=0$.

Taking the dot product with $\mathbf{v}_1$, we have
\begin{align*}
\mathbf{v}_1\cdot (c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3)=\mathbf{v}_1\cdot \mathbf{0}=\mathbf{0}.
\end{align*}
Distributing $\mathbf{v}_1$, we have
\begin{align*}
c_1\mathbf{v}_1\cdot\mathbf{v}_1+c_2\mathbf{v}_1\cdot\mathbf{v}_2+c_3\mathbf{v}_1\cdot\mathbf{v}_3= \mathbf{0}.
\end{align*}

Since the dot products $\mathbf{v}_1\cdot\mathbf{v}_2=0, \mathbf{v}_1\cdot\mathbf{v}_3=0$, we have
$c_1||\mathbf{v}_1||^2=c_1\mathbf{v}_1\cdot\mathbf{v}_1=0.$ Since the vector $\mathbf{v}_1\neq \mathbf{0}$, the length $||\mathbf{v}||\neq 0$.
Thus, we must have $c_1=0$.

We repeat the above argument with $\mathbf{v}_1$ replaced by $\mathbf{v}_2$ as follows.
Since $c_1=0$, the (*) becomes
$c_2\mathbf{v}_2+c_3\mathbf{v}_3=\mathbf{0}.$ We have
\begin{align*}
\mathbf{0}&=\mathbf{v}_2\cdot \mathbf{0}=\mathbf{v}_2\cdot (c_2\mathbf{v}_2+c_3\mathbf{v}_3)\\
&=c_2\mathbf{v}_2\cdot\mathbf{v}_2+c_3\mathbf{v}_2\cdot\mathbf{v}_3.
\end{align*}
Since $\mathbf{v}_2\cdot\mathbf{v}_3=0$, we have
$\mathbf{0}=c_2\mathbf{v}_2\cdot\mathbf{v}_2=c_2||\mathbf{v}_2||^2.$ Since $\mathbf{x}\neq \mathbf{0}$, we have $||\mathbf{v}_2||\neq 0$, and thus $c_2=0$.

Hence (*) is now just $c_3\mathbf{v}_3=\mathbf{0}$, and since $\mathbf{x}_3\neq \mathbf{0}$, we conclude that $c_3=0$. Therefore, we have shown that $c_1=c_2=c_3=0$, and thus the set $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is linearly independent.

## Comment.

These are Quiz 3 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.

### List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions.

### 3 Responses

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