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.

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.

Consider the $2\times 2$ matrix
\[A=\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta \end{bmatrix},\] where $\theta$ is a real number $0\leq \theta < 2\pi$.

(a) Find the characteristic polynomial of the matrix $A$.

(b) Find the eigenvalues of the matrix $A$.

(c) Determine the eigenvectors corresponding to each of the eigenvalues of $A$.

Proof.

(a) Find the characteristic polynomial of the matrix $A$.

The characteristic polynomial $p(t)$ of $A$ is computed as follows.
Let $I$ be the $2\times 2$ identity matrix.
We have
\begin{align*}
p(t)&=\det(A-tI)\\
&=\begin{vmatrix}
\cos \theta -t & -\sin \theta\\
\sin \theta& \cos \theta -t
\end{vmatrix}\\
&=(\cos \theta -t)^2+\sin^2 \theta \tag{*}\\
&=t^2-(2\cos \theta) t+\cos^2 \theta+\sin^2 \theta\\
&=t^2-(2\cos \theta) t+1
\end{align*}
by the trigonometry identity $\cos^2 \theta+\sin^2 \theta=1$.

Hence the characteristic polynomial of $A$ is
\[p(t)=t^2-(2\cos \theta) t+1.\]

(b) Find the eigenvalues of the matrix $A$.

The eigenvalues of $A$ are roots of the characteristic polynomial $p(t)$.
So let us solve
\[p(t)=t^2-(2\cos \theta) t+1=0.\] By the quadratic formula, we have
\begin{align*}
t&=\frac{2\cos \theta \pm \sqrt{(2\cos \theta)^2-4}}{2}\\[6pt] &=\cos \theta \pm \sqrt{\cos^2 \theta -1}\\
&=\cos \theta \pm \sqrt{-\sin^2 \theta}\\
&=\cos \theta \pm i \sin \theta =e^{\pm i\theta}.
\end{align*}

Hence eigenvalues of $A$ are
\[\cos \theta \pm i \sin \theta =e^{\pm i\theta}.\]

Alternatively, we could have used the equation (*).
Then we have
\begin{align*}
&p(t)=(\cos \theta -t)^2+\sin^2 \theta=0\\
&\Leftrightarrow (\cos \theta-t)^2=-\sin^2 \theta\\
&\Leftrightarrow \cos \theta -t=\pm i \sin \theta
\end{align*}
and thus, we obtain the eigenvalues
\[\cos \theta \pm i \sin \theta =e^{\pm i\theta}.\]

(c) Determine the eigenvectors

Let us first find the eigenvectors corresponding to the eigenvalue $\lambda=\cos \theta +i \sin \theta$.
We have
\begin{align*}
A-\lambda I=\begin{bmatrix}
-i \sin \theta & -\sin \theta\\
\sin \theta& -i \sin \theta
\end{bmatrix}.
\end{align*}

If $\theta=0, \pi$, then $\sin \theta=0$ and we have
\[A-\lambda I=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}\] and thus each nonzero vector of $\R^2$ is an eigenvector.

If $\lambda \neq 0, \pi$, then $\sin \theta \neq 0$.
Thus we have by elementary row operations
\begin{align*}
A-\lambda I=\begin{bmatrix}
-i \sin \theta & -\sin \theta\\
\sin \theta& -i \sin \theta
\end{bmatrix}
\xrightarrow[\frac{1}{\sin \theta} R_2]{\frac{i}{\sin \theta} R_1}
\begin{bmatrix}
1 & -i\\
1& -i
\end{bmatrix}
\xrightarrow{R_2-R_1} \begin{bmatrix}
1 & -i\\
0& 0
\end{bmatrix}.
\end{align*}
It follows that the eigenvectors for $\lambda$ are
\[\begin{bmatrix}
i \\
1
\end{bmatrix}t,\] for any nonzero scalar $t$.

The other eigenvalue is the conjugate $\bar{\lambda}$ of $\lambda$.
Since $A$ is a real matrix, the eigenvectors of $\bar{\lambda}$ are the conjugate of those of $\lambda$.
Hence the eigenvectors corresponding to $\bar{\lambda}$ is
\[\begin{bmatrix}
-i \\
1
\end{bmatrix}t,\] for any nonzero scalar $t$.

In summary, when $\theta=0, \pi$, the eigenvalues are $1, -1$, respectively, and every nonzero vector of $\R^2$ is an eigenvector.

If $\theta \neq 0, \pi$, then the eigenvectors corresponding to the eigenvalue $\cos \theta +i\sin \theta$ are
\[\begin{bmatrix}
i \\
1
\end{bmatrix}t,\] for any nonzero scalar $t$.
The eigenvectors corresponding to the eigenvalue $\cos \theta -i\sin \theta$ are
\[\begin{bmatrix}
-i \\
1
\end{bmatrix}t,\] for any nonzero scalar $t$.

Related Question.

For a similar proble, try the following.

The solution is given in the post ↴
Rotation Matrix in Space and its Determinant and Eigenvalues

The post Rotation Matrix in the Plane and its Eigenvalues and Eigenvectors first appeared on Problems in Mathematics.

Let $f(x)=\sin^2(x)$, $g(x)=\cos^2(x)$, and $h(x)=1$. These are vectors in $C[-1, 1]$.
Determine whether the set $\{f(x), \, g(x), \, h(x)\}$ is linearly dependent or linearly independent.

(The Ohio State University, Linear Algebra Midterm Exam Problem)
 

Solution.

We claim that the set is linearly dependent. To show the claim, we need to find nontrivial scalars $c_1, c_2, c_3$ such that
\[c_1 f(x)+c_2 g(x)+ c_3 h(x)=0.\]

From trigonometry, we know the identity
\[\sin^2(x)+\cos^2(x)=1.\] This implies that we have
\[\sin^2(x)+\cos^2(x)-1=0.\]

So we can choose $c_1=1, c_2=1, c_3=-1$, and thus the set is linearly dependent.

Linear Algebra Midterm Exam 2 Problems and Solutions

• True of False Problems and Solutions: True or False problems of vector spaces and linear transformations
• Problem 1 and its solution: See (7) in the post “10 examples of subsets that are not subspaces of vector spaces”
• Problem 2 and its solution (current problem): Determine whether trigonometry functions $\sin^2(x), \cos^2(x), 1$ are linearly independent or dependent
• Problem 3 and its solution: Orthonormal basis of null space and row space
• Problem 4 and its solution: Basis of span in vector space of polynomials of degree 2 or less
• Problem 5 and its solution: Determine value of linear transformation from $R^3$ to $R^2$
• Problem 6 and its solution: Rank and nullity of linear transformation from $R^3$ to $R^2$
• Problem 7 and its solution: Find matrix representation of linear transformation from $R^2$ to $R^2$
• Problem 8 and its solution: Hyperplane through origin is subspace of 4-dimensional vector space

The post Determine Whether Trigonometry Functions $\sin^2(x), \cos^2(x), 1$ are Linearly Independent or Dependent first appeared on Problems in Mathematics.