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

Linear algebra problems and solutions

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).

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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.


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3 Responses

  1. 02/18/2017

    […] 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 rela…. […]

  2. 02/19/2017

    […] Matrix representation of a linear transformation of subspace of sequences satisfying recurrence rela… […]

  3. 03/02/2017

    […] Matrix representation of a linear transformation of subspace of sequences satisfying recurrence rela… […]

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