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 $\phi^{[1]}, \phi^{[2]} \in V$. Then:

\begin{align} (\alpha \phi^{[1]} + \beta \phi^{[2]})' &= \alpha \phi^{[1]'} + \beta \phi^{[2]'} \\ &= \alpha A(t) \phi^{[1]} + \beta A(t) \phi^{[2]} \\ &= A(t)[\alpha \phi^{[1]} + \beta \phi^{[2]}] \end{align}

- This show that $(\alpha \phi^{[1]} + \beta \phi^{[2]}) \in V$. 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 $\xi^{[1]}, \xi^{[2]}, ..., \xi^{[n]} \in \mathbb{R}^n$ be such that $\{ \xi^{[1]}, \xi^{[2]}, ..., \xi^{[n]} \}$ 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, $\{ \xi^{[1]}, \xi^{[2]}, ..., \xi^{[n]} \}$ 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 $\{ \phi^{[1]}, \phi^{[2]}, ..., \phi^{[n]} \}$ 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:

\begin{align} \quad \alpha_1\phi^{[1]}(t) + \alpha_2\phi^{[2]}(t) + ... + \alpha_n\phi^{[n]}(t) = 0 \end{align}

- Plugging in $t = \tau$ and we get:

\begin{align} \quad \alpha_1 \phi^{[1]}(\tau) + \alpha_2 \phi^{[2]}(\tau) + ... + \alpha_n\phi^{[n]}(\tau) &= 0 \\ \quad \alpha_1 \xi^{[1]} + \alpha_2 \xi^{[2]} + ... + \alpha_n\xi^{[n]} &= 0 \end{align}

- Since $\{ \xi^{[1]}, \xi^{[2]}, ..., \xi^{[n]} \}$ is a linearly independent set of vectors in $\mathbb{R}^n$ the above equation implies that $\alpha_1, \alpha_2, ..., \alpha_n = 0$. Therefore $\{ \phi^{[1]}, \phi^{[2]}, ..., \phi^{[n]} \}$ is a linearly independent set in $V$. Therefore:

\begin{align} \quad \dim V \geq n \end{align}

**Part 3: Showing that $\dim V \leq n$.**

- We use the sets $\{ \xi^{[1]}, \xi^{[2]}, ..., \xi^{[n]} \}$ and $\{ \phi^{[1]}, \phi^{[2]}, ..., \phi^{[n]} \}$ 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 $\{ \xi^{[1]}, \xi^{[2]}, ..., \xi^{[n]} \}$ is a basis of $\mathbb{R}^n$ there exists $\alpha_1, \alpha_2, ..., \alpha_n \in \mathbb{R}^n$ such that:

\begin{align} \quad \xi = \alpha_1\xi^{[1]} + \alpha_2\xi^{[2]} + ... + \alpha_n\xi^{[n]} \end{align}

- Let:

\begin{align} \quad \eta = \alpha_1\phi^{[1]} + \alpha_2\phi^{[2]} + ... + \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:

\begin{align} \quad \eta(\tau) &= \alpha_1\phi^{[1]}(\tau) + \alpha_2\phi^{[2]}(\tau) + ... + \alpha_n\phi^{[n]}(\tau) \\ &=\alpha_1\xi^{[1]} + \alpha_2\xi^{[2]} + ... + \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:

\begin{align} \quad \psi = \alpha_1\phi^{[1]} + \alpha_2\phi^{[2]} + ... + \alpha_n\phi^{[n]} \end{align}

- So every element $\psi \in V$ can be expressed as a linear combination of elements from $\{ \phi^{[1]}, \phi^{[2]}, ..., \phi^{[n]} \}$. So $\{ \phi^{[1]}, \phi^{[2]}, ..., \phi^{[n]} \}$ spans $V$. Hence:

\begin{align} \quad \dim V \leq n \end{align}

**Conclusion:**So $V$ is a vector space with $\dim V = n$. $\blacksquare$