For this problem, use the complex vectors
\[ \mathbf{w}_1 = \begin{bmatrix} 1 + i \\ 1 – i \\ 0 \end{bmatrix} , \, \mathbf{w}_2 = \begin{bmatrix} -i \\ 0 \\ 2 – i \end{bmatrix} , \, \mathbf{w}_3 = \begin{bmatrix} 2+i \\ 1 – 3i \\ 2i \end{bmatrix} . \]

Suppose $\mathbf{w}_4$ is another complex vector which is orthogonal to both $\mathbf{w}_2$ and $\mathbf{w}_3$, and satisfies $\mathbf{w}_1 \cdot \mathbf{w}_4 = 2i$ and $\| \mathbf{w}_4 \| = 3$.

Calculate the following expressions:

(a) $ \mathbf{w}_1 \cdot \mathbf{w}_2 $.

(b) $ \mathbf{w}_1 \cdot \mathbf{w}_3 $.

(c) $((2+i)\mathbf{w}_1 – (1+i)\mathbf{w}_2 ) \cdot \mathbf{w}_4$.

(d) $\| \mathbf{w}_1 \| , \| \mathbf{w}_2 \|$, and $\| \mathbf{w}_3 \|$.

(e) $\| 3 \mathbf{w}_4 \|$.

(f) What is the distance between $\mathbf{w}_2$ and $\mathbf{w}_3$?

Solution.

(a) $ \mathbf{w}_1 \cdot \mathbf{w}_2 $.

\[ \mathbf{w}_1 \cdot \mathbf{w}_2 = \begin{bmatrix} 1+i & 1-i & 0 \end{bmatrix} \begin{bmatrix} -i \\ 0 \\ 2-i \end{bmatrix} = (1+i)(-i) + 0 + 0 = 1 – i . \]

(b) $ \mathbf{w}_1 \cdot \mathbf{w}_3 $.

\begin{align*} \mathbf{w}_1 \cdot \mathbf{w}_3 &= \begin{bmatrix} 1+i & 1-i & 0 \end{bmatrix} \begin{bmatrix} 2+i \\ 1-3i \\ 2i \end{bmatrix} \\ &= (1+i)(2+i) + (1-i)(1-3i) + 0 \\ &= (1 + 3i) + (-2 – 4i) \\ &= -1 – i . \end{align*}

(c) $((2+i)\mathbf{w}_1 – (1+i)\mathbf{w}_2 ) \cdot \mathbf{w}_4$.

\begin{align*} ((2+i)\mathbf{w}_1 – (1+i)\mathbf{w}_2 ) \cdot \mathbf{w}_4 &= (2+i)( \mathbf{w}_1 \cdot \mathbf{w}_4) – (1+i) ( \mathbf{w}_2 \cdot \mathbf{w}_4 ) \\
&= (2+i) ( 2i ) – (1+i)(0) \\
&= -2 + 4i \end{align*}

Note that $\mathbf{w}_2 \cdot \mathbf{w}_4=0$ because these vectors are orthogonal.

(d) $\| \mathbf{w}_1 \| , \| \mathbf{w}_2 \|$, and $\| \mathbf{w}_3 \|$.

For an arbitrary complex vector $\mathbf{v}$, its length is defined to be
\[ \| \mathbf{v} \| = \sqrt{ \overline{\mathbf{v}}^\trans \mathbf{v} } . \]

Thus,
\[ \| \mathbf{w}_1 \| \, = \, \sqrt{ (1-i)(1+i) + (1+i)(1-i) + 0 } = \sqrt{ 2 + 2} = \sqrt{4} , \] \[ \| \mathbf{w}_2 \| \, = \, \sqrt{ (i)(-i) + 0 + (2+i)(2-i) } = \sqrt{1 + 5} = \sqrt{6} , \] \[ \| \mathbf{w}_3 \| \, = \, \sqrt{ (2-i)(2+i) + (1+3i)(1-3i) + (-2i)(2i) } = \sqrt{ 5 + 10 + 4} = \sqrt{19} . \]

(e) $\| 3 \mathbf{w}_4 \|$.

$ \| 3 \mathbf{w}_4 \| = 3 \| \mathbf{w}_4 \| = 3\cdot 3=9 $ .

(f) What is the distance between $\mathbf{w}_2$ and $\mathbf{w}_3$?

The distance between these vectors is given by $\| \mathbf{w}_2 – \mathbf{w}_3 \|$. First we calculate this difference:
\[ \mathbf{w}_2 – \mathbf{w}_3 \, = \, \begin{bmatrix} -i \\ 0 \\ 2 – i \end{bmatrix} – \begin{bmatrix} 2+i \\ 1 – 3i \\ 2i \end{bmatrix} \, = \, \begin{bmatrix} -2 – 2i \\ -1 + 3i \\ 2 – 3i \end{bmatrix} . \]

Now the length of the complex vector is defined to be
\begin{align*}
\| \mathbf{w}_2 – \mathbf{w}_3 \| &= \sqrt{ \left( \overline{ \mathbf{w}_2 – \mathbf{w}_3 } \right)^{\trans} \left( \mathbf{w}_2 – \mathbf{w}_3 \right) } \\[6pt] &= \sqrt{ \begin{bmatrix} -2 + 2i & -1 – 3i & 2 + 3i \end{bmatrix} \begin{bmatrix} -2 – 2i \\ -1 + 3i \\ 2 – 3i \end{bmatrix} } \\[6pt] &= \sqrt{ (-2+2i)(-2-2i) + (-1-3i)(-1+3i) + (2+3i)(2-3i) } \\[6pt] &= \sqrt{ 8 + 10 + 13 } \\[6pt] &= \sqrt{ 31} \end{align*}

The post Dot Product, Lengths, and Distances of Complex Vectors first appeared on Problems in Mathematics.

We fix a nonzero vector $\mathbf{a}$ in $\R^3$ and define a map $T:\R^3\to \R^3$ by
\[T(\mathbf{v})=\mathbf{a}\times \mathbf{v}\] for all $\mathbf{v}\in \R^3$.
Here the right-hand side is the cross product of $\mathbf{a}$ and $\mathbf{v}$.

(a) Prove that $T:\R^3\to \R^3$ is a linear transformation.

(b) Determine the eigenvalues and eigenvectors of $T$.

Proof.

(a) Prove that $T:\R^3\to \R^3$ is a linear transformation.

To prove that $T$ is a linear transformation, we need to verify the following equalities for all $\mathbf{u}, \mathbf{v}\in \R^3$ and $c\in \R$:

1. $T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})$
2. $T(c\mathbf{v})=cT(\mathbf{v})$.

These follow from the basic properties of cross products as follows.
We have
\begin{align*}
T(\mathbf{u}+\mathbf{v})&=\mathbf{a}\times (\mathbf{u}+\mathbf{v})\\
&=\mathbf{a}\times \mathbf{u}+\mathbf{a}\times \mathbf{v} &&\text{the cross product is distributive}\\
&=T(\mathbf{u})+T(\mathbf{v}).
\end{align*}
As the cross product is compatible with scalar multiplication, we also have
\begin{align*}
T(c\mathbf{v})&=\mathbf{a}\times (c\mathbf{v})=c(\mathbf{a}\times \mathbf{v}) =cT(\mathbf{v}).
\end{align*}
Therefore $T$ is a linear transformation.

(b) Determine the eigenvalues and eigenvectors of $T$.

Let $\lambda$ be an eigenvalue of the linear transformation $T$ and let $\mathbf{x}$ be an eigenvector corresponding to $\lambda$.
Then we have
\[T(\mathbf{x})=\mathbf{a}\times \mathbf{x}=\lambda \mathbf{x}.\] We take the dot product and obtain
\begin{align*}
(a\times \mathbf{x})\cdot \mathbf{x}=\lambda \mathbf{x}\cdot \mathbf{x}=\lambda \|\mathbf{x}\|^2.
\end{align*}
Note that the vector $\mathbf{a}\times \mathbf{x}$ and $\mathbf{x}$ is orthogonal, hence the left-hand side of the above equality is zero.
On the other hand, since $\mathbf{x}$ is an eigenvector, it is a nonzero vector. Hence the length $\|\mathbf{x}\|\neq 0$.
It follows that we have $0=\lambda \|\mathbf{x}\|^2$ with $\|\mathbf{x}\|\neq 0$, and thus $\lambda=0$.
Hence the eigenvalue of $T$ is $\lambda=0$.

This yields that we have $\mathbf{a}\times \mathbf{x}=0\cdot \mathbf{x}=\mathbf{0}$ for any eigenvector $\mathbf{x}$.
Note that the cross product is zero indicates that the vectors $\mathbf{a}$ and $\mathbf{x}$ are parallel.
Hence $\mathbf{x}=r\mathbf{a}$ for a nonzero $r\in \R$.

In summary, the eigenvalue of $T$ is $0$ and corresponding eigenvectors are $r\mathbf{a}$ for any nonzero real number $r\in \R$.

Comment.

Let $\{\mathbf{e}_1, \mathbf{e}_2\mathbf{e}_3\}$ be the standard basis for $\R^3$ and write $\mathbf{a}=\begin{bmatrix}
a_1 \\
a_2 \\
a_3
\end{bmatrix}$.
Then we have
\begin{align*}
T(\mathbf{e}_1)&=\begin{bmatrix}
a_1 \\
a_2 \\
a_3
\end{bmatrix}\times \begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}
=\begin{bmatrix}
0 \\
a_3 \\
-a_2
\end{bmatrix}\\[6pt] T(\mathbf{e}_2)&=\begin{bmatrix}
a_1 \\
a_2 \\
a_3
\end{bmatrix}\times \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}
=\begin{bmatrix}
-a_3 \\
0 \\
a_1
\end{bmatrix}\\[6pt] T(\mathbf{e}_3)&=\begin{bmatrix}
a_1 \\
a_2 \\
a_3
\end{bmatrix}\times \begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}
=\begin{bmatrix}
a_2 \\
-a_1 \\
0
\end{bmatrix}.
\end{align*}
It follows that the matrix representation of the linear transformation $T$ with respect to the standard basis is given by
\[A=\begin{bmatrix}
0 & -a_3 & a_2 \\
a_3 &0 &-a_1 \\
-a_2 & a_1 & 0
\end{bmatrix}.\] Note that the matrix $A$ is a real skew-symmetric matrix.
We know that each eigenvalue of a skew-symmetric matrix is either $0$ or purely imaginary.
(See the post “Eigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even“.)
Also if the degree of the matrix is odd, it has $0$ as an eigenvalue.
(See the post “The Determinant of a Skew-Symmetric Matrix is Zero“.)

The post Eigenvalues and Eigenvectors of The Cross Product Linear Transformation first appeared on Problems in Mathematics.

Let $S=\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k\}$ be a set of nonzero vectors in $\R^n$.
Suppose that $S$ is an orthogonal set.

(a) Show that $S$ is linearly independent.

(b) If $k=n$, then prove that $S$ is a basis for $\R^n$.

Proof.

(a) Show that $S$ is linearly independent.

Consider the linear combination
\[c_1\mathbf{v}_1+c_2\mathbf{v}_2+\cdots +c_k \mathbf{v}_k=\mathbf{0}.\] Our goal is to show that $c_1=c_2=\cdots=c_k=0$.

We compute the dot product of $\mathbf{v}_i$ and the above linear combination for each $i=1, 2, \dots, k$:
\begin{align*}
0&=\mathbf{v}_i\cdot \mathbf{0}\\
&=\mathbf{v}_i \cdot (c_1\mathbf{v}_1+c_2\mathbf{v}_2+\cdots +c_k \mathbf{v}_k)\\
&=c_1\mathbf{v}_i \cdot \mathbf{v}_1+c_2\mathbf{v}_i \cdot \mathbf{v}_2+\cdots +c_k \mathbf{v}_i \cdot\mathbf{v}_k.
\end{align*}

As $S$ is an orthogonal set, we have $\mathbf{v}_i\cdot \mathbf{v}_j=0$ if $i\neq j$.

Hence all terms but the $i$-th one are zero, and thus we have
\[0=c_i\mathbf{v}_i\cdot \mathbf{v}_i=c_i \|\mathbf{v}_i\|^2.\]

Since $\mathbf{v}_i$ is a nonzero vector, its length $\|\mathbf{v}_i\|$ is nonzero.
It follows that $c_i=0$.

As this computation holds for every $i=1, 2, \dots, k$, we conclude that $c_1=c_2=\cdots=c_k=0$.
Hence the set $S$ is linearly independent.

(b) If $k=n$, then prove that $S$ is a basis for $\R^n$.

Suppose that $k=n$. Then by part (a), the set $S$ consists of $n$ linearly independent vectors in the dimension $n$ vector space $\R^n$.

Thus, $S$ is also a spanning set of $\R^n$, and hence $S$ is a basis for $\R^n$.

The post Orthogonal Nonzero Vectors Are Linearly Independent first appeared on Problems in Mathematics.