Let $\mathbb{R}^2$ be the vector space of size-2 column vectors. This vector space has an inner product defined by $ \langle \mathbf{v} , \mathbf{w} \rangle = \mathbf{v}^\trans \mathbf{w}$. A linear transformation $T : \R^2 \rightarrow \R^2$ is called an orthogonal transformation if for all $\mathbf{v} , \mathbf{w} \in \R^2$,
\[\langle T(\mathbf{v}) , T(\mathbf{w}) \rangle = \langle \mathbf{v} , \mathbf{w} \rangle.\]

For a fixed angle $\theta \in [0, 2 \pi )$ , define the matrix
\[ [T] = \begin{bmatrix} \cos (\theta) & – \sin ( \theta ) \\ \sin ( \theta ) & \cos ( \theta ) \end{bmatrix} \] and the linear transformation $T : \R^2 \rightarrow \R^2$ by
\[T( \mathbf{v} ) = [T] \mathbf{v}.\]

Prove that $T$ is an orthogonal transformation.

Solution.

Suppose we have vectors $\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}$ and $\mathbf{w} = \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} $ . Then,
\[T(\mathbf{v}) = \begin{bmatrix} \cos (\theta) & – \sin ( \theta ) \\ \sin ( \theta ) & \cos ( \theta ) \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} \cos(\theta) v_1 – \sin (\theta) v_2 \\ \sin(\theta) v_1 + \cos (\theta) v_2 \end{bmatrix},\] and
\[ T(\mathbf{w}) = \begin{bmatrix} \cos (\theta) & – \sin ( \theta ) \\ \sin ( \theta ) & \cos ( \theta ) \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} \cos(\theta) w_1 – \sin (\theta) w_2 \\ \sin(\theta) w_1 + \cos (\theta) w_2 \end{bmatrix}.\]

Then we find the inner product for these two vectors:
\begin{align*}
&\langle T(\mathbf{v} ) , T( \mathbf{w} ) \rangle \\
&= \begin{bmatrix} \cos(\theta) v_1 – \sin (\theta) v_2 & \sin(\theta) v_1 + \cos (\theta) v_2 \end{bmatrix} \begin{bmatrix} \cos(\theta) w_1 – \sin (\theta) w_2 \\ \sin(\theta) w_1 + \cos (\theta) w_2 \end{bmatrix} \\[6pt] &= \biggl( \cos(\theta) v_1 – \sin(\theta) v_2 \biggr) \biggl( \cos(\theta) w_1 – \sin ( \theta) w_2 \biggr) \\[6pt] &\qquad + \biggl( \sin (\theta) v_1 + \cos (\theta) v_2 \biggr) \biggl( \sin (\theta) w_1 + \cos(\theta) w_2 \biggr) \\[6pt] &= \cos^2(\theta) ( v_1 w_1 + v_2 w_2 ) + \sin(\theta) \cos(\theta) ( – v_1 w_2 – v_2 w_1 + v_1 w_2 + v_2 w_1 ) \\ &\qquad + \sin^2 (\theta) ( v_2 w_2 + v_1 w_1 ) \\[6pt] &= \left( \cos^2 ( \theta) + \sin^2 ( \theta ) \right) ( v_1 w_1 + v_2 w_2 ) \\
&= v_1 w_1 + v_2 w_2 \\
&= \langle \mathbf{v} , \mathbf{w} \rangle .
\end{align*}

This proves that $T$ is an orthogonal transformation. For the second-to-last equality, we used the Pythagorean identity $\sin^2 ( \theta ) + \cos^2 ( \theta ) = 1$.

The post The Rotation Matrix is an Orthogonal Transformation first appeared on Problems in Mathematics.

Let $C(\mathbb{R})$ be the vector space of real-valued functions on $\mathbb{R}$.

Consider the set of functions $W = \{ f(x) = a + b \cos(x) + c \cos(2x) \mid a, b, c \in \mathbb{R} \}$.

Prove that $W$ is a vector subspace of $C(\mathbb{R})$.

Proof.

We verify the subspace criteria: the zero vector of $C(\R)$ is in $W$, and $W$ is closed under addition and scalar multiplication.

First, the zero element of $C(\mathbb{R})$ is the zero function $\mathbf{0}$ defined by $\mathbf{0}(x) = 0$. This element lies in $W$, as $\mathbf{0}(x) = 0 + 0 \cos(x) + 0 \cos(2x)$.

Now suppose $f_1(x), f_2(x) \in W$, say $ f_1(x) = a_1 + b_1 \cos(x) + c_1 \cos(2x)$ and $f_2(x) = a_2 + b_2 \cos(x) + c_2 \cos(2x)$. Then
\[f_1(x) + f_2(x) = (a_1 + a_2) + (b_1 + b_2) \cos(x) + ( c_1 + c_2) \cos(2x)\] and so $f_1(x) + f_2(x) \in W$.

Finally, for any scalar $d \in \mathbb{R}$, we have
\[d f_1(x) = (a_1 d) + (b_1 d) \cos(x) + (c_1 d) \cos(2x),\] and so $d f_1(x) \in W$ as well.

This proves that $W$ is a subspace of $W$.

The post The Set $ \{ a + b \cos(x) + c \cos(2x) \mid a, b, c \in \mathbb{R} \}$ is a Subspace in $C(\R)$ first appeared on Problems in Mathematics.

(a) Find a function
\[g(\theta) = a \cos(\theta) + b \cos(2 \theta) + c \cos(3 \theta)\] such that $g(0) = g(\pi/2) = g(\pi) = 0$, where $a, b, c$ are constants.

(b) Find real numbers $a, b, c$ such that the function
\[g(\theta) = a \cos(\theta) + b \cos(2 \theta) + c \cos(3 \theta)\] satisfies $g(0) = 3$, $g(\pi/2) = 1$, and $g(\pi) = -5$.

Solution.

(a) Condition: $g(0) = g(\pi/2) = g(\pi) = 0$

Each condition required on $g$ can be turned into an equation involving the constants $a, b, c$. In particular, we have the system of linear equations
\begin{align*}
g(0) &= a + b + c = 0 \\[6pt] g \left( \frac{\pi}{2} \right) &= -b = 0 \\[6pt] g(\pi) &= -a + b – c = 0.
\end{align*}

To solve this system, we will use Gauss-Jordan elimination to reduce its augmented matrix.
\begin{align*}
\left[\begin{array}{rrr|r} 1 & 1 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 1 & -1 & 0 \end{array} \right] \xrightarrow[R_3 + R_2]{R_1 + R_2} \left[\begin{array}{rrr|r} 1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & -1 & 0 \end{array} \right] \xrightarrow[R_3 + R_1]{(-1) R_2} \left[\begin{array}{rrr|r} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right].
\end{align*}

The solution can now be read off: $a+c = 0$ and $b=0$.
Thus a general solution is of the form $g(\theta) = a \cos(\theta) – a \cos ( 3 \theta) $, where $a$ is any real number.

(b) Condition: $g(0) = 3$, $g(\pi/2) = 1$, and $g(\pi) = -5$

Each condition required on $g$ can be turned into an equation involving the constants $a, b, c$. In particular, we have

\begin{align*}
g(0) &= a + b + c = 3\\
g(\pi/2) &= -b = 1\\
g(\pi) &= -a + b – c = -5.
\end{align*}

To solve this system, we will use Gauss-Jordan elimination to reduce its augmented matrix.
\begin{align*}
\left[\begin{array}{rrr|r} 1 & 1 & 1 & 3 \\ 0 & -1 & 0 & 1 \\ -1 & 1 & -1 & -5 \end{array} \right] \xrightarrow[R_3 + R_2]{R_1 + R_2} \left[\begin{array}{rrr|r} 1 & 0 & 1 & 4 \\ 0 & -1 & 0 & 1 \\ -1 & 0 & -1 & -4 \end{array} \right] \xrightarrow[R_3 + R_1]{(-1) R_2} \left[\begin{array}{rrr|r} 1 & 0 & 1 & 4 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \end{array} \right] \end{align*}

The solution can now be read off: $a+c = 4$ and $b=-1$. Thus a general solution is of the form
$$ g(\theta) = a \cos(\theta) – \cos(2 \theta) + (4 – a) \cos(3 \theta) , $$
where $a$ is any real number.

The post Determine Trigonometric Functions with Given Conditions first appeared on Problems in Mathematics.

Let $C[-2\pi, 2\pi]$ be the vector space of all real-valued continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$ spanned by functions $\sin^2(x)$ and $\cos^2(x)$.

(a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$.

(b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$.

Solution.

(a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$.

By definition of the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$, the we know that $B$ is a spanning set for $W$.
Thus, it remains to show that $B$ is linearly independent set.
Suppose that
\[c_1\sin^2(x)+c_2\cos^2(x)=0.\] This equality is true for all $x\in [-2\pi, 2\pi]$.

In particular, evaluating at $x=0$, we see that $c_2=0$.
Also, plugging in $x=\pi/2$ yields $c_1=0$.

Therefore, $\sin^2(x)$ and $\cos^2(x)$ are linearly independent, that is, $B$ is linearly independent.
As $B$ is a linearly independent spanning set, it is a basis for $W$.

Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$.

Note that $\sin^2(x)-\cos^2(x)$ and $1$ are both in $W$ since both functions are linear combination of $\sin^2(x)$ and $\cos^2(x)$. Here, we used the trigonometric identity $1=\sin^2(x)+\cos^(x)$.

By part (a), we see that $\dim(W)=2$. So if we show that the functions $\sin^2(x)-\cos^2(x)$ and $1$ are linearly independent, then they form a basis for $W$.

We consider the coordinate vectors of these functions with respect to the basis $B$.
We have
\begin{align*}
[\sin^2(x)-\cos^2(x)]_B=\begin{bmatrix}
1
\\ -1
\end{bmatrix}
\text{ and }\\
[1]_B=[\sin^2(x)+\cos^2(x)]_B=\begin{bmatrix}
1\\ 1
\end{bmatrix}.
\end{align*}

Since we have
\begin{align*}
\begin{bmatrix}
1& 1 \\
-1& 1
\end{bmatrix}
\xrightarrow{R_2+R_1}
\begin{bmatrix}
1& 1 \\
0& 2
\end{bmatrix}
\xrightarrow{\frac{1}{2}R_2}
\begin{bmatrix}
1& 1 \\
0& 1
\end{bmatrix}
\xrightarrow{R_1-R_1}
\begin{bmatrix}
1& 0 \\
0& 1
\end{bmatrix},
\end{align*}
the coordinate vectors are linearly independent, and hence $\sin^2(x)-\cos^2(x)$ and $1$ are linearly independent.
We conclude that $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$.

(Another reasoning is that since the coordinate vectors form a basis for $\R^2$, $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$.)

Comment

You may directly show that $\{\sin^2(x)-\cos^2(x), 1\}$ is linearly independent just like we did for part (a).

The post Subspace Spanned by Trigonometric Functions $\sin^2(x)$ and $\cos^2(x)$ first appeared on Problems in Mathematics.

Let $C[-2\pi, 2\pi]$ be the vector space of all continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the functions \[f(x)=\sin^2(x) \text{ and } g(x)=\cos^2(x)\] in $C[-2\pi, 2\pi]$.

Prove or disprove that the functions $f(x)$ and $g(x)$ are linearly independent.

(The Ohio State University, Linear Algebra Midterm)
 

Proof.

To determine whether $f(x)$ and $g(x)$ are linearly independent or not, consider the linear combination
\[c_1f(x)+c_2g(x)=0,\] equivalently
\[c_1\sin^2(x)+c_2 \cos^2(x)=0, \tag{*}\] where $c_1, c_2$ are scalars.

If the only scalars satisfying the above equality are $c_1=0, c_2=0$, then $f(x)$ and $g(x)$ are linearly independent, otherwise they are linearly dependent.

Note that this is an equality as functions.
That is, this equality must hold for any $x$ in the interval $[-2\pi, 2\pi]$.

Let $x=0$. Then as $\sin(0)=0$ and $\cos(0)=1$, we obtain $c_2=0$ from (*).
Next, let $x=\pi/2$. Then as $\sin(\pi/2)=1$ and $\cos(\pi/2)=0$, we obtain $c_1=0$ from (*).

Therefore, we must have $c_1=c_2=0$, and hence the functions $f(x)=\sin^2(x)$ and $g(x)=\cos^2(x)$ are linearly independent.

Comment.

This is one of the midterm 2 exam problems for Linear Algebra (Math 2568) in Autumn 2017.

Here is the most common mistake.
The linear combination $c_1\sin^2(x)+c_2 \cos^2(x)$ is a function defined over the interval $[-2\pi, 2\pi]$ and we are assuming it is the zero function.

So saying that “if $c_1=1, c_2=0$, then $c_1\sin^2(x)+c_2 \cos^2(x)$ is zero at $x=0$, hence $f(x)$ and $g(x)$ are linearly independent” is totally wrong.

What you are claiming here is that the function $\sin^2(x)$ is zero at $x=0$, hence it is the zero function.
This is clearly wrong as $\sin^2(x)$ is not the zero function.

List of Midterm 2 Problems for Linear Algebra (Math 2568) in Autumn 2017

1. Vector Space of 2 by 2 Traceless Matrices
2. Find an Orthonormal Basis of the Given Two Dimensional Vector Space
3. Are the Trigonometric Functions $\sin^2(x)$ and $\cos^2(x)$ Linearly Independent?←The current problem
4. Find Bases for the Null Space, Range, and the Row Space of a $5\times 4$ Matrix
5. Matrix Representation, Rank, and Nullity of a Linear Transformation $T:\R^2\to \R^3$
6. Determine the Dimension of a Mysterious Vector Space From Coordinate Vectors
7. Find a Basis of the Subspace Spanned by Four Polynomials of Degree 3 or Less

The post Are the Trigonometric Functions $\sin^2(x)$ and $\cos^2(x)$ Linearly Independent? first appeared on Problems in Mathematics.

Let $C[-\pi, \pi]$ be the vector space of all continuous functions defined on the interval $[-\pi, \pi]$.

Show that the subset $\{\cos(x), \sin(x)\}$ in $C[-\pi, \pi]$ is linearly independent.

Proof.

Note that the zero vector in the vector space $C[-\pi, \pi]$ is the zero function
\[\theta(x):=0.\]

Let us consider a linear combination
\[a_1\cos(x)+a_2\sin(x)=\theta(x)=0 \tag{*}.\] If this linear combination has only the zero solution $a_1=a_2=0$, then the set $\{\cos(x), \sin(x)\}$ is linearly independent.

The equality (*) should be true for any values of $x\in [-\pi, \pi]$.
Setting $x=0$, we obtain from (*) that
\[a_1=0\] since $\cos(0)=1, \sin(0)=0$.

We also set $x=\pi/2$ and we obtain
\[a_2=0\] since $\cos(\pi/2)=0, \sin(\pi/2)=1$.

Therefore, we have $a_1=a_2=0$ and we conclude that the set $\{\cos(x), \sin(x)\}$ is linearly independent.

The post Cosine and Sine Functions are Linearly Independent first appeared on Problems in Mathematics.