Subspace Spanned By Cosine and Sine Functions

Kyoto University Exam problems and solutions in mathematrics

Problem 435

Let $\calF[0, 2\pi]$ be the vector space of all real valued functions defined on the interval $[0, 2\pi]$.
Define the map $f:\R^2 \to \calF[0, 2\pi]$ by
\[\left(\, f\left(\, \begin{bmatrix}
\alpha \\
\beta
\end{bmatrix} \,\right) \,\right)(x):=\alpha \cos x + \beta \sin x.\] We put
\[V:=\im f=\{\alpha \cos x + \beta \sin x \in \calF[0, 2\pi] \mid \alpha, \beta \in \R\}.\]

(a) Prove that the map $f$ is a linear transformation.

(b) Prove that the set $\{\cos x, \sin x\}$ is a basis of the vector space $V$.

(c) Prove that the kernel is trivial, that is, $\ker f=\{\mathbf{0}\}$.
(This yields an isomorphism of $\R^2$ and $V$.)

(d) Define a map $g:V \to V$ by
\[g(\alpha \cos x + \beta \sin x):=\frac{d}{dx}(\alpha \cos x+ \beta \sin x)=\beta \cos x -\alpha \sin x.\] Prove that the map $g$ is a linear transformation.

(e) Find the matrix representation of the linear transformation $g$ with respect to the basis $\{\cos x, \sin x\}$.

(Kyoto University, Linear Algebra exam problem)

 
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Terminology (image=range, kernel=null space)

We remark that the image is also called the range, and the kernel is also called the null space.

Proof.

(a) Prove that the map $f$ is a linear transformation.

To prove that $f:\R^2 \to \calF[0, 2\pi]$ is a linear transformation, we check the following two properties:

  1. \[f\left(\, \begin{bmatrix}
    \alpha \\
    \beta
    \end{bmatrix}+\begin{bmatrix}
    \alpha’ \\
    \beta’
    \end{bmatrix} \,\right)=f\left(\, \begin{bmatrix}
    \alpha \\
    \beta
    \end{bmatrix}\,\right)+f\left(\, \begin{bmatrix}
    \alpha’ \\
    \beta’
    \end{bmatrix} \,\right)\] for any vectors $\begin{bmatrix}
    \alpha\\
    \beta
    \end{bmatrix}, \begin{bmatrix}
    \alpha’ \\
    \beta’
    \end{bmatrix} \in \R^2$.
  2. \[f\left(\, r\begin{bmatrix}
    \alpha \\
    \beta
    \end{bmatrix} \,\right)=rf\left(\, \begin{bmatrix}
    \alpha \\
    \beta
    \end{bmatrix}\,\right)\] for any vector $\begin{bmatrix}
    \alpha\\
    \beta
    \end{bmatrix} \in \R^2$ and any scalar $r\in \R$.

For any vectors $\begin{bmatrix}
\alpha\\
\beta
\end{bmatrix}, \begin{bmatrix}
\alpha’ \\
\beta’
\end{bmatrix} \in \R^2$, we have
\begin{align*}
\left(\, f\left(\, \begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}+\begin{bmatrix}
\alpha’ \\
\beta’
\end{bmatrix} \,\right) \,\right)(x)&=
\left(\, f\left(\, \begin{bmatrix}
\alpha+\alpha’ \\
\beta+\beta’
\end{bmatrix} \,\right) \,\right)(x)\\[6pt] &=(\alpha+\alpha’) \cos x + (\beta+\beta’) \sin x\\
&=(\alpha \cos x + \beta \sin x)+(\alpha’\cos x + \beta’ \sin x)\\
&=\left(\, f\left(\, \begin{bmatrix}
\alpha \\
\beta
\end{bmatrix} \,\right) \,\right)(x)+\left(\, f\left(\, \begin{bmatrix}
\alpha’ \\
\beta’
\end{bmatrix} \,\right) \,\right)(x),
\end{align*}
and hence the first property is proved.

For the second one, let $\begin{bmatrix}
\alpha\\
\beta
\end{bmatrix} \in \R^2$ and $r\in \R$.
We have
\begin{align*}
\left(\, f\left(\, r\begin{bmatrix}
\alpha \\
\beta
\end{bmatrix} \,\right) \,\right)(x)&=
\left(\, f\left(\, \begin{bmatrix}
r\alpha \\
r\beta
\end{bmatrix} \,\right) \,\right)(x)\\
&=(r\alpha) \cos x + (r\beta) \sin x\\
&=r(\alpha \cos x + \beta \sin x)\\
&=r\left(\, f\left(\, \begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}\,\right) \,\right)(x).
\end{align*}
This proves the property 2, and hence the map $f$ is a linear transformation.

(b) Prove that the set $\{\cos x, \sin x\}$ is a basis of the vector space $V$.

It is clear that the set $\{\cos x, \sin x\}$ spans the vector space $V$ by definition. Thus, it suffices to prove that $\{\cos x, \sin x\}$ is a linearly independent set.
Suppose that we have
\[\alpha \cos x + \beta \sin x=0\] for some $\alpha, \beta \in \R$.
This is an equality as vectors in $V$, that is, this equality holds for all $x\in [0, 2\pi]$.

In particular, we substitute $x=0$ and obtain $\alpha=0$.
Also, we substitute $x=\pi/2$ and obtain $\beta=0$.
It follows that $\{\cos x, \sin x\}$ is linearly independent, and hence it is a basis of $V$.

(c) Prove that the kernel $\ker f=\{\mathbf{0}\}$.

If $\begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}\in \ker f$, then we have
\[\left(\, f\left(\, \begin{bmatrix}
\alpha \\
\beta
\end{bmatrix} \,\right) \,\right)(x)=\alpha \cos x + \beta \sin x=0\] for all $x\in [0, 2\pi]$.
As in part (b) this implies that $\alpha=\beta=0$.
Hence we have
\[\begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}=\begin{bmatrix}
0 \\
0
\end{bmatrix}=\mathbf{0},\] and thus we obtain $\ker f=\{\mathbf{0}\}$.

By the isomorphism theorem, we obtain
\[\R^2=\R^2/\{0\}=\R^2/\ker f \cong \im f =V.\]

(d) Prove that the map $g$ is a linear transformation.

Let $\alpha \cos x + \beta \sin x, \alpha’ \cos x + \beta’ \sin x$ be arbitrary vectors in $V$. Then we have
\begin{align*}
&g\left(\, (\alpha \cos x + \beta \sin x)+(\alpha’ \cos x + \beta’ \sin x) \,\right)\\
&=g\left(\, (\alpha+\alpha’) \cos x + (\beta+\beta’) \sin x)\,\right)\\
&=(\beta+\beta’)\cos x-(\alpha+\alpha’) \sin x\\
&=(\beta \cos x -\alpha \sin x)+(\beta’ \cos x -\alpha’ \sin x)\\
&=g(\alpha \cos x + \beta \sin x)+g(\alpha’ \cos x + \beta’ \sin x).
\end{align*}

We also have for any vector $\alpha \cos x + \beta \sin x \in V$ and any scalar $r\in \R$
\begin{align*}
&g\left(\, r(\alpha \cos x + \beta \sin x) \,\right)\\
&=g\left(\, (r\alpha) \cos x + (r\beta) \sin x) \,\right)\\
&=(r\beta)\cos x – (r\alpha)\sin x\\
&=r(\beta\cos x – \alpha\sin x)\\
&=rg(\alpha \cos x + \beta \sin x).
\end{align*}
Therefore $g$ is a linear transformation.

(e) Find the matrix representation of the linear transformation $g$ with respect to the basis $\{\cos x, \sin x\}$.

We have
\begin{align*}
g(\cos x)=-\sin x \text{ and } g(\sin x)=\cos x.
\end{align*}
Thus, the coordinate vectors with respect to the basis $B:=\{\cos x, \sin x\}$ are
\[[g(\cos x)]_B=\begin{bmatrix}
0 \\
-1
\end{bmatrix} \text{ and } [g(\sin x)]_B=\begin{bmatrix}
1 \\
0
\end{bmatrix}.\] It follows that the matrix representation of $g$ is
\[ [\,[g(\cos x)]_B, [g(\sin x)]_B\,]=\begin{bmatrix}
0 & 1\\
-1& 0
\end{bmatrix}.\]

Alternatively, we can write
\[g[\cos x, \sin x]=[\cos x, \sin x] \begin{bmatrix}
0 & 1\\
-1& 0
\end{bmatrix}.\]


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