Exponential Functions Form a Basis of a Vector Space

Problem 590

Let $C[-1, 1]$ be the vector space over $\R$ of all continuous functions defined on the interval $[-1, 1]$. Let
\[V:=\{f(x)\in C[-1,1] \mid f(x)=a e^x+b e^{2x}+c e^{3x}, a, b, c\in \R\}\] be a subset in $C[-1, 1]$.

(a) Prove that $V$ is a subspace of $C[-1, 1]$.

(b) Prove that the set $B=\{e^x, e^{2x}, e^{3x}\}$ is a basis of $V$.

(c) Prove that
\[B’=\{e^x-2e^{3x}, e^x+e^{2x}+2e^{3x}, 3e^{2x}+e^{3x}\}\] is a basis for $V$.

 
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Proof.

(a) Prove that $V$ is a subspace of $C[-1, 1]$.

Note that each function in the subset $V$ is a linear combination of the functions $e^x, e^{2x}, e^{3x}$.
Namely, we have
\[V=\Span\{e^x, e^{2x}, e^{3x}\}\] and we know that the span is always a subspace. Hence $V$ is a subspace of $C[-1,1]$.

(b) Prove that the set $B=\{e^x, e^{2x}, e^{3x}\}$ is a basis of $V$.

We noted in part (a) that $V=\Span(B)$. So it suffices to show that $B$ is linearly independent.
Consider the linear combination
\[c_1e^x+c_2 e^{2x}+c_3 e^{3x}=\theta(x),\] where $\theta(x)$ is the zero function (the zero vector in $V$).
Taking the derivative, we get
\[c_1e^x+2c_2 e^{2x}+3c_3 e^{3x}=\theta(x).\] Taking the derivative again, we obtain
\[c_1e^x+4c_2 e^{2x}+9c_3 e^{3x}=\theta(x).\]

Evaluating at $x=0$, we obtain the system of linear equations
\begin{align*}
c_1+c_2+c_3&=0\\
c_1+2c_2+3c_3&=0\\
c_1+4c_2+9c_3&=0.
\end{align*}

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We reduce the augmented matrix for this system as follows:
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 1 & 1 & 0 \\
1 &2 & 3 & 0 \\
1 & 4 & 9 & 0
\end{array} \right] \xrightarrow[R_3-R_1]{R_2-R_1}
\left[\begin{array}{rrr|r}
1 & 1 & 1 & 0 \\
0 &1 & 2 & 0 \\
0 & 3 & 8 & 0
\end{array} \right] \xrightarrow[R_3-3R_2]{R_1-R_2}\\[6pt] \left[\begin{array}{rrr|r}
1 & 0 & -1 & 0 \\
0 &1 & 2 & 0 \\
0 & 0 & 2 & 0
\end{array} \right] \xrightarrow{\frac{1}{2}R_3}
\left[\begin{array}{rrr|r}
1 & 0 & -1 & 0 \\
0 &1 & 2 & 0 \\
0 & 0 & 1 & 0
\end{array} \right] \xrightarrow[R_2-2R_2]{R_1+R_3}
\left[\begin{array}{rrr|r}
1 & 0 & 0 & 0 \\
0 &1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{array} \right].
\end{align*}
It follows that the solution of the system is $c_1=c_2=c_3=0$.
Hence the set $B$ is linearly independent, and thus $B$ is a basis for $V$.

Anotehr approach.

Alternatively, we can show that the coefficient matrix is nonsingular by using the Vandermonde determinant formula as follows.
Observe that the coefficient matrix of the system is a Vandermonde matrix:
\[A:=\begin{bmatrix}
1 & 1 & 1 \\
1 &2 &3 \\
1^2 & 2^2 & 3^2
\end{bmatrix}.\] The Vandermonde determinant formula yields that
\[\det(A)=(3-1)(3-2)(2-1)=2\neq 0.\] Hence the coefficient matrix $A$ is nonsingular.
Thus we obtain the solution $c_1=c_2=c_3=0$.

(c) Prove that $B’=\{e^x-2e^{3x}, e^x+e^{2x}+2e^{3x}, 3e^{2x}+e^{3x}\}$ is a basis for $V$.

We consider the coordinate vectors of vectors in $B’$ with respect to the basis $B$.
The coordinate vectors with respect to basis $B$ are
\[[e^x-2e^{3x}]_B=\begin{bmatrix}
1 \\
0 \\
-2
\end{bmatrix}, [e^x+e^{2x}+2e^{3x}]_B=\begin{bmatrix}
1 \\
1 \\
2
\end{bmatrix}, [3e^{2x}+e^{3x}]_B=\begin{bmatrix}
0 \\
3 \\
1
\end{bmatrix}.\] Let $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ be these vectors and let $T=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$.
Then we know that $B’$ is a basis for $V$ if and only if $T$ is a basis for $\R^3$.

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We claim that $T$ is linearly independent.
Consider the matrix whose column vectors are $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$:
\begin{align*}
\begin{bmatrix}
1 & 1 & 0 \\
0 &1 &3 \\
-2 & 2 & 1
\end{bmatrix}
\xrightarrow{R_3+2R_1}
\begin{bmatrix}
1 & 1 & 0 \\
0 &1 &3 \\
0 & 4 & 1
\end{bmatrix}
\xrightarrow[R_3-4R_1]{R_1-R_2}\\[6pt] \begin{bmatrix}
1 & 0 & -3 \\
0 &1 &3 \\
0 & 0 & -11
\end{bmatrix}
\xrightarrow{-\frac{1}{11}R_3}
\begin{bmatrix}
1 & 0 & -3 \\
0 &1 &3 \\
0 & 0 & 1
\end{bmatrix}
\xrightarrow[R_2-3R_3]{R_1+3R_3}
\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix}.
\end{align*}

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Thus, the matrix is nonsingular and hence the column vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly independent.
As $T$ consists of three linearly independent vectors in the three-dimensional vector space $\R^3$, we conclude that $T$ is a basis for $\R^3$.
Therefore, by the correspondence of the coordinates, we see that $B’$ is a basis for $V$.

Related Question.

If you know the Wronskian, then you may use the Wronskian to prove that the exponential functions $e^x, e^{2x}, e^{3x}$ are linearly independent.

See the post
Using the Wronskian for Exponential Functions, Determine Whether the Set is Linearly Independent for the details.

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Try the next more general question.

The solution is given in the post ↴
Exponential Functions are Linearly Independent

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