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.

These are True or False problems.
For each of the following statements, determine if it contains a wrong information or not.

1. Let $A$ be a $5\times 3$ matrix. Then the range of $A$ is a subspace in $\R^3$.
2. The function $f(x)=x^2+1$ is not in the vector space $C[-1,1]$ because $f(0)=1\neq 0$.
3. Since we have $\sin(x+y)=\sin(x)+\sin(y)$, the function $\sin(x)$ is a linear transformation.
4. The set
   \[\left\{\, \begin{bmatrix}
   1 \\
   0 \\
   0
   \end{bmatrix}, \begin{bmatrix}
   0 \\
   1 \\
   1
   \end{bmatrix} \,\right\}\] is an orthonormal set.

(Linear Algebra Exam Problem, The Ohio State University)

Solution.

(1) True or False? Let $A$ be a $5\times 3$ matrix. Then the range of $A$ is a subspace in $\R^3$.

The answer is “False”. The definition of the range of the $5 \times 3$ matrix $A$ is
\[ \calR(A)=\{\mathbf{y}\in \R^5 \mid A\mathbf{x}=\mathbf{y} \text{ for some $\mathbf{x} \in \R^3$}\}.\] Note that to make sense the matrix product $A\mathbf{x}$, the size of the vector $\mathbf{x}$ must be $3$-dimensional because $A$ is $5\times 3$. Hence $\mathbf{y}=A\mathbf{x}$ is a $5$-dimensional vector, and thus the range $\calR(A)$ is a subspace of $\R^5$.
 

(2) True or False? The function $f(x)=x^2+1$ is not in the vector space $C[-1,1]$ because $f(0)=1\neq 0$.

The answer is “False”. The vector space $C[-1, 1]$ consists of all continuos functions defined on the interval $[-1, 1]$. Since $f(x)=x^2+1$ is a continuos function defined on $[-1, 1]$, it is in the vector space $C[-1, 1]$. The condition $f(0)=1\neq 0$ is irrelevant.
 

(3) True or False? Since we have $\sin(x+y)=\sin(x)+\sin(y)$, the function $\sin(x)$ is a linear transformation.

The answer is “False”. First of all $\sin(x+y)\neq \sin(x)+\sin(y)$. For example, let $x=y=\pi/2$. Then
\[\sin\left(\,\frac{\pi}{2}+\frac{\pi}{2}\,\right)=\sin\left(\,\pi \,\right)=0\] and
\[\sin\left(\, \frac{\pi}{2} \,\right)+\sin\left(\, \frac{\pi}{2} \,\right)=1+1=2.\] Hence $\sin(x)$ is not a linear transformation.
 

(4) True or False? The given set is an orthonormal set.

The answer is “False”. The dot product of these vectors is
\[\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}\cdot \begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix}=1\cdot 0+ 0\cdot 1 +0\cdot 1=0.\] Thus, the vectors are orthogonal. However the length of the second vector is
\[\sqrt{0^2+1^2+1^2}=\sqrt{2},\] hence it is not the unit vector.
So the set is orthogonal, but not orthonormal set.

Linear Algebra Midterm Exam 2 Problems and Solutions

• True of False Problems and Solutions (current problem): 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: 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 True or False Problems of Vector Spaces and Linear Transformations first appeared on Problems in Mathematics.

For a real number $0\leq \theta \leq \pi$, we define the real $3\times 3$ matrix $A$ by
\[A=\begin{bmatrix}
\cos\theta & -\sin\theta & 0 \\
\sin\theta &\cos\theta &0 \\
0 & 0 & 1
\end{bmatrix}.\]

(a) Find the determinant of the matrix $A$.

(b) Show that $A$ is an orthogonal matrix.

(c) Find the eigenvalues of $A$.

Solution.

(a) The determinant of the matrix $A$

By the cofactor expansion corresponding to the third row, we compute
\begin{align*}
\det(A)&=\begin{vmatrix}
\cos\theta & -\sin\theta & 0 \\
\sin\theta &\cos\theta &0 \\
0 & 0 & 1
\end{vmatrix}\\
&=0\cdot \begin{vmatrix}
-\sin \theta & 0\\
\cos \theta& 0
\end{vmatrix}-0\cdot \begin{vmatrix}
\cos \theta & 0\\
\sin \theta& 0
\end{vmatrix}+1\cdot \begin{vmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta
\end{vmatrix}\\
&=\cos^2 \theta +\sin^2 \theta\\
&=1.
\end{align*}
The last step follows from the famous trigonometry identity
\[\cos^2 \theta +\sin^2 \theta=1.\] Thus we have
\[\det(A)=1.\]

(b) The matrix $A$ is an orthogonal matrix

We give two solutions for part (b).

The first solution of (b)

The first solution computes $A^{\trans}A$ and show that it is the identity matrix $I$.
We have
\begin{align*}
A^{\trans}A&=\begin{bmatrix}
\cos\theta & \sin\theta & 0 \\
-\sin\theta &\cos\theta &0 \\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
\cos\theta & -\sin\theta & 0 \\
\sin\theta &\cos\theta &0 \\
0 & 0 & 1
\end{bmatrix}\\
&=\begin{bmatrix}
\cos^2 \theta +\sin^2\theta & 0 & 0 \\
0 &\cos^2 \theta+\sin^2 \theta &0 \\
0 & 0 & 1
\end{bmatrix}\\
&=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix}=I.
\end{align*}
Similarly, you can check that $AA^{\trans}=I$. Thus $A$ is an orthogonal matrix.

The second solution of (b)

The second proof uses the following fact: a matrix is orthogonal if and only its column vectors form an orthonormal set.
Let
\[A_1=\begin{bmatrix}
\cos \theta \\
\sin \theta \\
0
\end{bmatrix}, A_2=\begin{bmatrix}
-\sin\theta \\
\cos \theta \\
0
\end{bmatrix}, A_3=\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}\] be the column vectors of the matrix $A$. The length of these vectors are all $1$. For example, we have
\begin{align*}
||A_1||=\sqrt{(\cos\theta)^2+(\sin \theta)^2+0^2}=\sqrt{1}=1.
\end{align*}
Similarly, we have $||A_2||=||A_3||=1$.
The dot (inner) product of $A_1$ and $A_2$ is
\begin{align*}
A_1\cdot A_2=\cos \theta \cdot (-\sin \theta)+\sin \theta \cdot \cos \theta +0\cdot 0=0.
\end{align*}
Similarly, we have $A_1\cdot A_3=A_2\cdot A_3=0$.
Therefore, the column vectors $A_1, A_2, A_3$ are orthonormal vectors. Hence by the above fact, the matrix $A$ is orthogonal.

(c) The eigenvalues of $A$

We compute the characteristic polynomial $p(t)=\det(A-tI)$ as follows.
\begin{align*}
p(t)&=\det(A-tI)=\begin{vmatrix}
\cos\theta-t & -\sin\theta & 0 \\
\sin\theta &\cos\theta -t&0 \\
0 & 0 & 1-t
\end{vmatrix}\\
&=(1-t)\begin{vmatrix}
\cos \theta -t & -\sin \theta\\
\sin \theta& \cos \theta-t
\end{vmatrix} \text{ by the third row cofactor expansion}\\
&=(1-t)(\cos^2 \theta -2t \cos \theta +t^2 +\sin^2 \theta)\\
&=(1-t)(t^2-(2\cos \theta)t+1).
\end{align*}

The eigenvalues are roots of the characteristic polynomial $p(t)$, hence we solve
\[p(t)=(1-t)(t^2-(2\cos \theta)t+1)=0.\] One solution is $t=1$. The other solutions come from the quadratic polynomial in $p(t)$.
By the quadratic formula, those solutions are
\begin{align*}
t&=\cos\theta \pm\sqrt{\cos^2 \theta -1}\\
&=\cos\theta \pm \sqrt{-\sin^2 \theta}\\
&=\cos \theta \pm i \sin \theta
\end{align*}
since $\sin \theta\geq 0$ since $0 \leq \theta \leq \pi$.
Therefore the eigenvalues of the matrix $A$ are
\[1, \cos \theta \pm i \sin \theta.\]

Related Question.

The following problem treats the rotation matrix in the plane.

The solution is given in the post ↴
Rotation Matrix in the Plane and its Eigenvalues and Eigenvectors

The post Rotation Matrix in Space and its Determinant and Eigenvalues first appeared on Problems in Mathematics.