The V.S. of the Sol. Set to a Lin. Homo. System of First Order ODEs

The Vector Space of the Solution Set to a Linear Homogeneous System of First Order ODEs

Recall from the Linear Homogeneous and Linear Nonhomogeneous Systems of First Order ODEs page that if $J = (a, b)$ and $A : J \to \mathbb{R}^{n \times n}$ is a continuous function then a linear homogeneous system of $n$ first order ODEs is of the form:

(1)
\begin{align} \quad \mathbf{x}' = A(t) \mathbf{x} \end{align}

We now prove a very important result which says that all of the solutions to the linear homogeneous system above forms an $n$-dimensional vector space.

 Theorem 1: The set of all solutions to the linear homogeneous system of first order ODEs $\mathbf{x}' = A(t) \mathbf{x}$ forms an $n$-dimensional vector space with the operations of function addition $+$ and scalar multiplication $\cdot$ over the field $\mathbb{R}$.
• Proof: We break this proof up into three parts. In the first part we show that the set of solutions to $\mathbf{x}' = A(t)\mathbf{x}$ under the operations of function addition $+$ and scalar multiplication $\cdot$ is a vector space over the field $\mathbb{R}$.
• In the second part we show that this vector space has dimension of at least $n$.
• In the third part we show that this vector space has dimension of at most $n$.
• Let $V$ denote the solution set to the linear homogeneous system of first order ODEs $\mathbf{x}' = A(t)\mathbf{x}$.
• Part 1: Showing that $V$ is a vector space over $\mathbb{R}$. We first note that $V$ is a collection of continuous functions from $J$ to $\mathbb{R}^n$ and $C(J, \mathbb{R}^n)$ is itself a vector space over $\mathbb{R}$. Thus we show that $V$ is a vector subspace over $\mathbb{R}$.
• Clearly $0 \in V$ since $0$ is the trivial solution to $\mathbf{x}' = A(t)\mathbf{x}$.
• Now let $\alpha, \beta \in \mathbb{R}$ and let $\phi^{}, \phi^{} \in V$. Then:
(2)
\begin{align} (\alpha \phi^{} + \beta \phi^{})' &= \alpha \phi^{'} + \beta \phi^{'} \\ &= \alpha A(t) \phi^{} + \beta A(t) \phi^{} \\ &= A(t)[\alpha \phi^{} + \beta \phi^{}] \end{align}
• This show that $(\alpha \phi^{} + \beta \phi^{}) \in V$. So $V$ is closed under function addition $+$ and scalar multiplication $\cdot$. Therefore $V$ is a subspace of $C(J, \mathbb{R})$ and is itself a vector space.
• Part 2: Showing that $\dim V \geq n$.
• Let $\xi^{}, \xi^{}, ..., \xi^{[n]} \in \mathbb{R}^n$ be such that $\{ \xi^{}, \xi^{}, ..., \xi^{[n]} \}$ is a linearly independent set in $\mathbb{R}^n$. (It is possible to choose such vectors in $\mathbb{R}^n$ since $\dim \mathbb{R}^n = n$) Furthermore, $\{ \xi^{}, \xi^{}, ..., \xi^{[n]} \}$ must be a basis of $\mathbb{R}^n$.
• Let $\tau \in J$. For each $i \in \{ 1, 2, ..., n \}$, the initial value problem $\mathbf{x}' = A(t)\mathbf{x}$ with $x(\tau) = \xi^{[i]}$ has a solution, call it $\phi^{[i]}$. So $\{ \phi^{}, \phi^{}, ..., \phi^{[n]} \}$ is a collection of elements from $V$. We will show this is a linearly independent set of elements from $V$.
• Let $\alpha_1, \alpha_2, ..., \alpha_n \in \mathbb{R}$ and suppose that for all $t \in J$ we have that:
(3)
\begin{align} \quad \alpha_1\phi^{}(t) + \alpha_2\phi^{}(t) + ... + \alpha_n\phi^{[n]}(t) = 0 \end{align}
• Plugging in $t = \tau$ and we get:
(4)
\begin{align} \quad \alpha_1 \phi^{}(\tau) + \alpha_2 \phi^{}(\tau) + ... + \alpha_n\phi^{[n]}(\tau) &= 0 \\ \quad \alpha_1 \xi^{} + \alpha_2 \xi^{} + ... + \alpha_n\xi^{[n]} &= 0 \end{align}
• Since $\{ \xi^{}, \xi^{}, ..., \xi^{[n]} \}$ is a linearly independent set of vectors in $\mathbb{R}^n$ the above equation implies that $\alpha_1, \alpha_2, ..., \alpha_n = 0$. Therefore $\{ \phi^{}, \phi^{}, ..., \phi^{[n]} \}$ is a linearly independent set in $V$. Therefore:
(5)
\begin{align} \quad \dim V \geq n \end{align}
• Part 3: Showing that $\dim V \leq n$.
• We use the sets $\{ \xi^{}, \xi^{}, ..., \xi^{[n]} \}$ and $\{ \phi^{}, \phi^{}, ..., \phi^{[n]} \}$ as before.
• Let $\psi \in V$. Then $\psi$ solves the linear homogeneous system $\mathbf{x}' = A(t) \mathbf{x}$. Furthermore, for the element $\tau \in J$ there is an element $\xi \in \mathbb{R}^n$ such that $\psi(\tau) = \xi$.
• Now since $\{ \xi^{}, \xi^{}, ..., \xi^{[n]} \}$ is a basis of $\mathbb{R}^n$ there exists $\alpha_1, \alpha_2, ..., \alpha_n \in \mathbb{R}^n$ such that:
(6)
\begin{align} \quad \xi = \alpha_1\xi^{} + \alpha_2\xi^{} + ... + \alpha_n\xi^{[n]} \end{align}
• Let:
(7)
\begin{align} \quad \eta = \alpha_1\phi^{} + \alpha_2\phi^{} + ... + \alpha_n\phi^{[n]} \end{align}
• Note that $\eta \in V$ as it is a linear combination of elements in $V$. So $\eta$ solves the linear homogeneous system $\mathbf{x} = A(t) \mathbf{x}$ and furthermore:
(8)
\begin{align} \quad \eta(\tau) &= \alpha_1\phi^{}(\tau) + \alpha_2\phi^{}(\tau) + ... + \alpha_n\phi^{[n]}(\tau) \\ &=\alpha_1\xi^{} + \alpha_2\xi^{} + ... + \alpha_n\xi^{[n]} \\ &= \xi \end{align}
• Since $\psi$ and $\eta$ are solutions to the IVP $\mathbf{x}' = A(t)\mathbf{x}$ with $x(\tau) = \xi$, we have by the uniqueness theorem that $\psi = \eta$. Therefore:
(9)
\begin{align} \quad \psi = \alpha_1\phi^{} + \alpha_2\phi^{} + ... + \alpha_n\phi^{[n]} \end{align}
• So every element $\psi \in V$ can be expressed as a linear combination of elements from $\{ \phi^{}, \phi^{}, ..., \phi^{[n]} \}$. So $\{ \phi^{}, \phi^{}, ..., \phi^{[n]} \}$ spans $V$. Hence:
(10)
\begin{align} \quad \dim V \leq n \end{align}
• Conclusion: So $V$ is a vector space with $\dim V = n$. $\blacksquare$