# Sequences Satisfying Linear Recurrence Relation Form a Subspace

## Problem 308

Let $V$ be a real vector space of all real sequences

\[(a_i)_{i=1}^{\infty}=(a_1, a_2, \cdots).\]
Let $U$ be the subset of $V$ defined by

\[U=\{ (a_i)_{i=1}^{\infty} \in V \mid a_{k+2}-5a_{k+1}+3a_{k}=0, k=1, 2, \dots \}.\]

Prove that $U$ is a subspace of $V$.

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## Proof.

We prove the subspace criteria.

**Subspace criteria**

- The zero vector in $V$ is in $U$.
- For any two elements $(a_i)_{i=1}^{\infty}, (b_i)_{i=1}^{\infty}\in U$, we have $(a_i)_{i=1}^{\infty}+(b_i)_{i=1}^{\infty} \in U$.
- For any scalar $c$ and any element $(a_i)_{i=1}^{\infty} \in U$, we have $c(a_i)_{i=1}^{\infty} \in U$.

**Condition 1**.

The zero vector in $V$ is the zero sequence $(0)=(0,0,0,\dots)$. Clearly, this sequence satisfies the recurrence relation $a_{k+2}-5a_{k+1}+3a_{k}=0$. Thus the zero vector $(0)\in U$, hence condition 1 is met.

**Condition 2**.

To check condition 2, take $(a_i)_{i=1}^{\infty}, (b_i)_{i=1}^{\infty}\in U$.

We want to show that the sum

\[(a_i)_{i=1}^{\infty}+(b_i)_{i=1}^{\infty}=(a_i+b_i)_{i=1}^{\infty}\]
is also in $U$.

Since $(a_i)_{i=1}^{\infty}, (b_i)_{i=1}^{\infty}\in U$, these sequences satisfy the recurrence relation

\[a_{k+2}-5a_{k+1}+3a_{k}=0 \tag{*}\]
and

\[b_{k+2}-5b_{k+1}+3b_{k}=0\]
for $k=1, 2, \dots$.

Using these two relations, we have

\begin{align*}

&(a_{k+2}+b_{k+2})-5(a_{k+1}+b_{k+1})+3(a_{k}+b_k)\\

&=(a_{k+2}-5a_{k+1}+3a_{k})+(b_{k+2}-5b_{k+1}+3b_{k})\\

&=0+0=0.

\end{align*}

Thus the sum $(a_i)_{i=1}^{\infty}+(b_i)_{i=1}^{\infty}=(a_i+b_i)_{i=1}^{\infty}$ satisfies the recurrence relation and it is also in $U$. Hence condition 2 is met.

**Condition 3**.

To check condition 3, take $(a_i)_{i=1}^{\infty} \in U$ and let $c\in \R$ be a scalar.

Then the sequence $(a_i)_{i=1}^{\infty}$ satisfies (*). Multiplying (*) by $c$, we have

\[(ca_{k+2})-5(ca_{k+1})+3(ca_{k})=0.\]
This implies that the scalar product $c(a_i)_{i=1}^{\infty}=(ca_i)_{i=1}^{\infty}$ satisfies the recurrence relation, and hence it is in $U$. Thus condition 3 is satisfied.

We have checked all subspace criteria, and thus $U$ is a subspace of the vector space $V$.

## Related Question.

This is the first problem of three problems about a linear recurrence relation and linear algebra.

- [Problem 2] The problems/solutions of finding a basis and dimension of $U$ and finding a matrix representation of some linear transformation of $U$ to itself are given in the post

Matrix representation of a linear transformation of subspace of sequences satisfying recurrence relation. - [Problem 3] Want to know how to find a general formula for a sequence satisfying a linear recurrence relation using linear algebra? Check out the post Solve linear recurrence relation using linear algebra (eigenvalues and eigenvectors)

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