Vector space of sequences

Question

Let $V$ be vector space for all real sequences $\{a_i\}_{i=0}^\infty$ and let $M \subset V$ that satisfies for all $n \ge 3$ recurrence relation $ a_n = Aa_{n-1} + Ba_{n-2} + Ca_{n-3}$, where $A,B,C$ are real constants. Prove that $M$ is vector subspace of $V$ and its dimension is $3$. I do not have any idea how to work with sequences. Thank you for help and sorry for my English.

Answer 1

HINT Observe that for a sequence in $M$, all entries after the third are determined by the first three (i.e. the $0,1,$ and $2$ entries). Given this, try and show that $M$ has as a basis $\{e_1,e_2,e_3\}$, where \begin{align} e_1&=(1,0,0,C,\ldots), \\ e_2&=(0,1,0,B,\ldots), \\ e_3&=(0,0,1,A,\ldots). \end{align}

Answer 2

Always use the definitions.

First, define the sum of the sequences $(a_i)_i$ and $(b_i)_i$ as the sequence $(a_i+b_i)_i$. And for real $r$ define $r(a_i)_i=(ra_i)_i.$ (This is like $\mathbb R^n$ except that we have an infinite string of co-ordinates instead of $n$ co-ordinates.) Then verify that $V$ satisfies all the requirements of the definition for being a vector space over $\mathbb R.$ The $0$-vector of $V$ has $a_i=0$ for all $i.$

Second, verify that $M$ meets all the requirements of the definition for being a vector space over $\mathbb R.$ So $M$ is a vector subspace of $V.$

Third, observe that any $(a_i)_i\in M$ is completely and uniquely determined by the values $a_0,a_1,a_2.$

Fourth, for $j\in \{0,1,2\}$ let $v(j)=(v_{j,i})_i\in M$ where $$1=v_{0,0}=v_{1,1}=v_{2,2}$$ and $$0=v_{0,1}=v_{0,2}=v_{1,0}=v_{1,2}=v_{2,0}=v_{2,1}.$$ Any $a=(a_i)_i\in M$ is a linear combination of $v(0),v(1), v(2)$. This is because the sequence $(b_i)_i=(a_0v_{0,i}+a_1v_{1,i}+a_2v_{2,i})_i\;$.... which is equal to the vector $a_0v(0) +a_1v(1)+a_2v(2)\;$.... also belongs to $M,$ and we have $b_i=a_i$ for $i\in \{0,1,2\},$ so we have $(b_i)_i=(a_i)_i.$

Therefore the dimension of $M$ is at most $3.$

Now $v(0),v(1),v(2)$ are linearly independent vectors. For if $d_0,d_1,d_2$ are real numbers and if $(x_i)_i=d_0v(0)+d_1v(1)+d_2v(2)$ is equal to the $0$-vector, then for $i\in \{0,1,2\}$ we have $0=x_i=d_i.$

Therefore the dimension of $M$ is at least $3.$