# Matrix Representation of a Linear Transformation of Subspace of Sequences Satisfying Recurrence Relation

## Problem 309

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

\[(a_i)_{i=1}^{\infty}=(a_1, a_2, \dots).\]
Let $U$ be the subspace of $V$ consisting of all real sequences that satisfy the linear recurrence relation $a_{k+2}-5a_{k+1}+3a_{k}=0$ for $k=1, 2, \dots$.

**(a)** Let

\begin{align*}

\mathbf{u}_1&=(1, 0, -3, -15, -66, \dots)\\

\mathbf{u}_2&=(0, 1, 5, 22, 95, \dots)

\end{align*}

be vectors in $U$. Prove that $\{\mathbf{u}_1, \mathbf{u}_2\}$ is a basis of $U$ and conclude that the dimension of $U$ is $2$.

**(b)** Let $T$ be a map from $U$ to $U$ defined by

\[T\big((a_1, a_2, \dots)\big)=(a_2, a_3, \dots). \]
Verify that the map $T$ actually sends a vector $(a_i)_{i=1}^{\infty}\in V$ to a vector $T\big((a_i)_{i=1}^{\infty}\big)$ in $U$, and show that $T$ is a linear transformation from $U$ to $U$.

**(c)**With respect to the basis $\{\mathbf{u}_1, \mathbf{u}_2\}$ obtained in (a), find the matrix representation $A$ of the linear transformation $T:U \to U$ from (b).

Contents

## Proof.

See the post Sequences satisfying linear recurrence relation form a subspace for a proof that $U$ is a subspace of $V$.

### (a) Prove that $\{\mathbf{u}_1, \mathbf{u}_2\}$ is a basis of $U$ and the dimension of $U$ is $2$.

Let

\[(a_i)_{i=1}^{\infty}=(a_1, a_2, a_3, \dots)\]
be an arbitrary vector in $U$.

If we know the first two terms $a_1, a_2$, the remaining terms are determined by the linear recurrence relation $a_{k+2}-5a_{k+1}+3a_{k}=0$. Thus if the first two terms of two sequences of $U$ are equal, then the sequences are equal.

From this observation, we have

\[(a_i)_{i=1}^{\infty}=a_1\mathbf{u}_1+a_2\mathbf{u}_2.\]
Hence $\{\mathbf{u}_1, \mathbf{u}_2\}$ is a spanning set of $U$.

It remains to show that $\{\mathbf{u}_1, \mathbf{u}_2\}$ is a linearly independent set.

Suppose that

\[c_1\mathbf{u}_1+c_2\mathbf{u}_2=\mathbf{0},\]
where $\mathbf{0}$ is the zero sequence.

Then since we have

\begin{align*}

&(0, 0, 0 \dots) \\

&= c_1\mathbf{u}_1+c_2\mathbf{u}_2\\

&=c_1(1, 0, -3, \dots)+c_2(0, 1, 5, \dots)\\

&=(c_1, c_2, -3c_1+5c_2, \dots),

\end{align*}

we must have $c_1=c_2=0$. Hence $\{\mathbf{u}_1, \mathbf{u}_2\}$ is a linearly independent set, and it is a basis. Therefore the dimension of $U$ is $2$.

### (b) $T$ is a linear transformation from $U$ to $U$.

Let us write $T\big((a_i)_{i=1}^{\infty}\big)=(b_i)_{i=1}^{\infty}$. Thus $b_i=a_{i+1}$ by definition of $T$. Then we have

\begin{align*}

b_{k+2}-5b_{k+1}+3b_{k}=a_{k+3}-5a_{k+2}+3a_{k+1}=0

\end{align*}

for $k=1, 2, \dots$ since $(a_i)_{i=1}^{\infty}\in U$, hence satisfies the linear recurrence relation.

Thus the sequence $(b_i)_{i=1}^{\infty}$ also satisfies the recurrence relation, and it is in $U$. Thus $T$ sends a vector in $U$ to a vector in $U$.

Let us prove that $T$ is a linear transformation from $U$ to $U$.

Let $(a_i)_{i=1}^{\infty}, (b_i)_{i=1}^{\infty}$ be vectors in $U$ and let $r\in \R$ be a scalar.

Then we have

\begin{align*}

&T\big( (a_i)_{i=1}^{\infty}+(b_i)_{i=1}^{\infty}\big) \\

&=T\big( (a_i+b_i)_{i=1}^{\infty}\big)\\

&=(a_2+b_2, a_3+b_3, \dots)\\

&=(a_2, a_3, \dots)+(b_2, b_3, \dots)\\

&=T\big( (a_i)_{i=1}^{\infty}\big)+T\big( (b_i)_{i=1}^{\infty}\big)

\end{align*}

Also we have

\begin{align*}

&T\big(c(a_i)_{i=1}^{\infty} \big)\\

&=T\big((ca_i)_{i=1}^{\infty} \big)\\

&=(ca_2, ca_3, \dots)\\

&=c(a_2, a_3, \dots)\\

&=cT\big((a_i)_{i=1}^{\infty} \big)

\end{align*}

Therefore $T$ is a linear transformation.

### (c) The matrix representation $A$ of the linear transformation $T$

We compute the values of the linear transformation of $T$ on basis vectors $\mathbf{u}_1, \mathbf{u}_2$ and express them as a linear combination of $\mathbf{u}_1, \mathbf{u}_2$.

We have

\begin{align*}

T(\mathbf{u}_1)&=T\big((1, 0, -3, -15, -66, \dots) \big)\\

&=(0, -3, -15, -66, \dots)\\

&=-3(0, 1, 5, 22, \dots)\\

&=0\mathbf{u}_1-3 \mathbf{u}_2.

\end{align*}

and also we have

\begin{align*}

T(\mathbf{u}_2)&=T\big((0, 1, 5, 22, 95, \dots) \big)\\

&=(1, 5, 22, 95, \dots)\\

&=(1, 0, -3, -15, \dots)+(0, 5, 25, 110, \dots)\\

&=\mathbf{u}_1+5\mathbf{u}_2.

\end{align*}

Therefore the matrix representation of $T$ with respect to the basis $B=\{\mathbf{u}_1, \mathbf{u}_2\}$ is given by

\[A=[[T(\mathbf{u}_1)]_B, [T(\mathbf{u}_2)]_B]=\begin{bmatrix}

0 & 1\\

-3& 5

\end{bmatrix}.\]
Here $[T(\mathbf{u}_1)]_B=\begin{bmatrix}

0 \\

-3

\end{bmatrix}$ is the coordinate vector of $T(\mathbf{u}_1)$ with respect to the basis $B$. Similarly for $[T(\mathbf{u}_2)]_B]=\begin{bmatrix}

1 \\

5

\end{bmatrix}$.

Remark. Using the matrix $A$ for the linear transformation $T$, we can write formally

\[\begin{bmatrix}

T(\mathbf{u}_1) & T(\mathbf{u}_2)

\end{bmatrix}=\begin{bmatrix}

\mathbf{u}_1 & \mathbf{u}_2

\end{bmatrix}

\begin{bmatrix}

0 & 1\\

-3& 5

\end{bmatrix}.\]

## Related Question.

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

- [Problem 1] The basics about the subspace of sequences satisfying a linear recurrence relations. Sequences satisfying linear recurrence relation form a subspace
- [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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