Fix the row vector $\mathbf{b} = \begin{bmatrix} -1 & 3 & -1 \end{bmatrix}$, and let $\R^3$ be the vector space of $3 \times 1$ column vectors. Define
\[W = \{ \mathbf{v} \in \R^3 \mid \mathbf{b} \mathbf{v} = 0 \}.\] Prove that $W$ is a vector subspace of $\R^3$.

Proof.

We verify the subspace criteria: the zero vector $\mathbf{0}$ of $\R^3$ is in $W$, and $W$ is closed under addition and scalar multiplication.

First, the zero element in $\R^3$ is $\mathbf{0}$, the $3 \times 1$ column vector whose entries are all $0$. Then clearly $\mathbf{b} \mathbf{0} = 0$, and so $\mathbf{0} \in W$.

Next, suppose $\mathbf{v} , \mathbf{w} \in W$, and $c \in \mathbb{R}$. Then $\mathbf{b} \mathbf{v} = \mathbf{b} \mathbf{w} = 0$, and so
\[\mathbf{b} ( \mathbf{v} + \mathbf{w} ) = \mathbf{b} \mathbf{v} + \mathbf{b} \mathbf{w} = 0.\] Thus, $\mathbf{v} + \mathbf{w} \in W$.

Because, again, $\mathbf{b} \mathbf{v} = \mathbf{0}$, we have
\[\mathbf{b} ( c \mathbf{v} ) = c \mathbf{b} \mathbf{v} = c \mathbf{0} = \mathbf{0}.\] Thus $c \mathbf{v} \in W$. These three criteria show that $W$ is a vector subspace of $\R^3$.

Comment.

We can generalize the problem with an arbitrary $1\times 3$ row vector $\mathbf{b}$.

The proof is almost identical.
(Look at the proof. We didn’t use components of the row vector $\mathbf{b} = \begin{bmatrix} -1 & 3 & -1 \end{bmatrix}$.)

Note that vectors $\mathbf{u}, \mathbf{v}\in \R^3$ is said to be perpendicular if
\[\mathbf{u}\cdot \mathbf{v}=\mathbf{u}^{\trans}\mathbf{v}=0.\]

Thus, the result of the problem says that for a fixed vector $\mathbf{u}\in \R^3$, the set of vectors $\mathbf{v}$ that are perpendicular to $\mathbf{u}$ is a subspace in $\R^3$.
(Note that we appy the problem to $\mathbf{b}=\mathbf{u}^{\trans}$.)

The post The Set of Vectors Perpendicular to a Given Vector is a Subspace first appeared on Problems in Mathematics.

Let
\[\mathbf{v}=\begin{bmatrix}
a \\
b \\
c
\end{bmatrix}, \qquad \mathbf{v}_1=\begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}, \qquad \mathbf{v}_2=\begin{bmatrix}
2 \\
-1 \\
2
\end{bmatrix}.\] Find the necessary and sufficient condition so that the vector $\mathbf{v}$ is a linear combination of the vectors $\mathbf{v}_1, \mathbf{v}_2$.

 

We give two solutions.

Solution 1. (Use the range)

The question is equivalent to finding the condition so that the vector $\mathbf{v}$ is in the range of the matrix
\[A=[\mathbf{v}_1, \mathbf{v}_2]=\begin{bmatrix}
1 & 2 \\
2 & -1 \\
0 &2
\end{bmatrix}.\] The vector $\mathbf{v}$ is in the range $\calR(A)$ if and only if the system $A\mathbf{x}=\mathbf{v}$ is consistent.

We reduce the augmented matrix of the system by elementary row operations as follows.
\begin{align*}
[A\mid \mathbf{v}]=\left[\begin{array}{rr|r}
1 & 2 & a \\
2 &-1 &b \\
0 & 2 & c
\end{array}\right] \xrightarrow{R_2-2R_1}
\left[\begin{array}{rr|r}
1 & 2 & a \\
0 &-5 &b-2a \\
0 & 2 & c
\end{array}\right]\\[6pt] \xrightarrow{R_2+3R_3}
\left[\begin{array}{rr|r}
1 & 2 & a \\
0 &1 &b-2a+3c \\
0 & 2 & c
\end{array}\right]\\
\xrightarrow{R_3-2R_2}
\left[\begin{array}{rr|r}
1 & 2 & a \\
0 &1 &b-2a+3c \\
0 & 0 & 4a-2b-5c
\end{array}\right].
\end{align*}
The last matrix is in echelon form and the system is consistent if and only if $4a-2b-5c=0$.
Therefore, the condition that $\mathbf{v}$ be a linear combination of $\mathbf{v}_1, \mathbf{v}_2$ is $4a-2b-5c=0$.

Solution 2. (Use the cross product)

Note that the vectors $\mathbf{v}_1, \mathbf{v}_2$ spans a plane $P$ in $\R^3$.
Thus, the vector $\mathbf{v}$ is a linear combination of $\mathbf{v}_1, \mathbf{v}_2$ if and only if $\mathbf{v}$ lies on the plane $P$.

The cross product
\[\mathbf{v}_1\times \mathbf{v}_2=\begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}\times \begin{bmatrix}
2 \\
-1 \\
2
\end{bmatrix}=
\begin{bmatrix}
\begin{vmatrix}
2 & -1\\
0& 2
\end{vmatrix} \\[10pt] – \begin{vmatrix}
1 & 2\\
0& 2
\end{vmatrix} \\[10pt] \begin{vmatrix}
1 & 2\\
2& -1
\end{vmatrix}
\end{bmatrix}=\begin{bmatrix}
4 \\
-2 \\
-5
\end{bmatrix}\] is perpendicular to the plane $P$.

Therefore, the vector $\mathbf{v}$ is on the plane if and only if the dot (inner) product
\[\mathbf{v}\cdot (\mathbf{v}_1\times \mathbf{v}_2)=0.\] Namely,
\[\begin{bmatrix}
a \\
b \\
c
\end{bmatrix}\cdot \begin{bmatrix}
4 \\
-2 \\
-5
\end{bmatrix}=4a-2b-5c=0,\] and we obtained the same condition as in Solution 1.

The post Find a Condition that a Vector be a Linear Combination first appeared on Problems in Mathematics.

(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.

• Quiz 1. Gauss-Jordan elimination / homogeneous system.
• Quiz 2. The vector form for the general solution / Transpose matrices.
• Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent
• Quiz 4. Inverse matrix/ Nonsingular matrix satisfying a relation
• Quiz 5. Example and non-example of subspaces in 3-dimensional space
• Quiz 6. Determine vectors in null space, range / Find a basis of null space
• Quiz 7. Find a basis of the range, rank, and nullity of a matrix
• Quiz 8. Determine subsets are subspaces: functions taking integer values / set of skew-symmetric matrices
• Quiz 9. Find a basis of the subspace spanned by four matrices
• Quiz 10. Find orthogonal basis / Find value of linear transformation
• Quiz 11. Find eigenvalues and eigenvectors/ Properties of determinants
• Quiz 12. Find eigenvalues and their algebraic and geometric multiplicities
• Quiz 13 (Part 1). Diagonalize a matrix.
• Quiz 13 (Part 2). Find eigenvalues and eigenvectors of a special matrix

The post Quiz 3. Condition that Vectors are Linearly Dependent/ Orthogonal Vectors are Linearly Independent first appeared on Problems in Mathematics.

Let $\mathbf{a}$ and $\mathbf{b}$ be fixed vectors in $\R^3$, and let $W$ be the subset of $\R^3$ defined by
\[W=\{\mathbf{x}\in \R^3 \mid \mathbf{a}^{\trans} \mathbf{x}=0 \text{ and } \mathbf{b}^{\trans} \mathbf{x}=0\}.\]

Prove that the subset $W$ is a subspace of $\R^3$.
 

Proof.

We prove the following criteria for the subset $W$ to be a subspace of $\R^3$.

For (a), note that $\mathbf{a}^{\trans} \mathbf{0}=0$ and $\mathbf{b}^{\trans} \mathbf{0}=0$. Thus the zero vector $\mathbf{0}\in \R^3$ is in $W$.

To check (b), let $\mathbf{x}, \mathbf{y} \in W$. Then we have the following relations.
\[\mathbf{a}^{\trans} \mathbf{x}=0 \text{ and } \mathbf{b}^{\trans} \mathbf{x}=0,
\text{ and }
\mathbf{a}^{\trans} \mathbf{y}=0 \text{ and } \mathbf{b}^{\trans} \mathbf{y}=0.\tag{*}\] To show that $\mathbf{x}+\mathbf{y} \in W$, we need to show that
\[\mathbf{a}^{\trans}(\mathbf{x}+\mathbf{y})=0 \text{ and } \mathbf{b}^{\trans}(\mathbf{x}+\mathbf{y})=0.\]

We first compute
\begin{align*}
\mathbf{a}^{\trans}(\mathbf{x}+\mathbf{y}) &=\mathbf{a}^{\trans}\mathbf{x}+\mathbf{a}^{\trans}\mathbf{y}\\
&= 0+0=0
\end{align*}
by the relations (*).
Similarly, we have
\begin{align*}
\mathbf{b}^{\trans}(\mathbf{x}+\mathbf{y}) &=\mathbf{b}^{\trans}\mathbf{x}+\mathbf{b}^{\trans}\mathbf{y}\\
&= 0+0=0
\end{align*}
by the relations (*).

Thus the vector $\mathbf{x}+\mathbf{y}$ satisfies the defining relations for $W$, hence $\mathbf{x}+\mathbf{y} \in W$.

Finally, to prove (c), let $\mathbf{x} \in W$ and let $c\in \R$.
Since $\mathbf{x} \in W$, we have
\[\mathbf{a}^{\trans} \mathbf{x}=0 \text{ and } \mathbf{b}^{\trans} \mathbf{x}=0.\] Multiplying by the scalar $c$ from the left, we obtain
\[\mathbf{a}^{\trans} (\mathbf{cx})=0 \text{ and } \mathbf{b}^{\trans} (\mathbf{cx})=0.\] (Note that since $c$ is a scalar, we can switch the order of the product of $c$ and $\mathbf{a}^{\trans}$. Same for $c$ and $\mathbf{b}^{\trans}$.)

These equalities proves that the vector $c\mathbf{x}$ satisfies the defining relation for $W$. Thus $c\mathbf{x} \in W$.

Therefore the criteria (a)-(c) are all met, and we conclude that $W$ is a subspace of $\R^3$.

The post Subset of Vectors Perpendicular to Two Vectors is a Subspace first appeared on Problems in Mathematics.