# 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 $V$. Add to solve later

## 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*}

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 $\R^3$.

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*} 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. Try the next more general question. Problem. Let c_1, c_2,\dots, c_n be mutually distinct real numbers. Show that exponential functions \[e^{c_1x}, e^{c_2x}, \dots, e^{c_nx} are linearly independent over $\R$.

The solution is given in the post ↴
Exponential Functions are Linearly Independent Add to solve later

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1. 10/20/2017

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