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.

Let $n$ be an odd positive integer.
Determine whether there exists an $n \times n$ real matrix $A$ such that
\[A^2+I=O,\] where $I$ is the $n \times n$ identity matrix and $O$ is the $n \times n$ zero matrix.

If such a matrix $A$ exists, find an example. If not, prove that there is no such $A$.

How about when $n$ is an even positive number?

Hint.

The key technique to solve this problem very easily is determinant.
Recall the following properties of the determinant.
Let $A, B$ be $n\times n$ matrices and $c$ be a scalar Then we have

• $\det(AB)=\det(A)\det(B)$
• $\det(cA)=c^n\det(A)$.

Solution.

When $n$ is odd.

When $n$ is odd, we prove that there is no $n \times n$ real matrix $A$ such that $A^2+I=O$.
Seeking a contradiction, assume that we have $A$ such that $A^2+I=O$.
Since we have $A^2=-I$, we have
\[\det(A^2)=\det(-I).\]

Using the properties of determinant, we obtain
\[\det(A)^2=(-1)^n\det(I)=-1\] because $n$ is odd and $\det(I)=1$.

Since $A$ is a real matrix, the determinant of $A$ is also real.
Thus, $\det(A)^2=-1$ is impossible. Hence there is no such $A$.

When $n$ is even.

On the other hand, if $n$ is even there is $A$ such that $A^2+I=O$.
For example, consider
\[A=\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}.\] Then a direct computation shows that $A^2=-I$, hence $A^2+I=O$.

Comment.

Recall that the imaginary number $i$ is the number whose square is $-1$.
Similarly, we found above that the square of the matrix $A=\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}$ is $-I$.

But when $n$ is odd, there is no such matrix $A$ as we showed.

The post Is there an Odd Matrix Whose Square is $-I$? first appeared on Problems in Mathematics.