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 $k=1, 2, \dots$.
Let $T$ be the linear transformation from $U$ to $U$ defined by
\[T\big((a_1, a_2, \dots)\big)=(a_2, a_3, \dots). \]

Let $B=\{\mathbf{u}_1, \mathbf{u}_2\}$ be a basis of $U$, where
\begin{align*}
\mathbf{u}_1&=(1, 0, -3, -15, -66, \dots)\\
\mathbf{u}_2&=(0, 1, 5, 22, 95, \dots).
\end{align*}
Let $A$ be the matrix representation of the linear transformation $T: U \to U$ with respect to the basis $B$.

(a) Find the eigenvalues and eigenvectors of $T$.

(b) Use the result of (a), find a sequence $(a_i)_{i=1}^{\infty}$ satisfying the linear recurrence relation $a_{k+2}-5a_{k+1}+3a_{k}=0$ and the initial condition $a_1=1, a_2=1$.

(c) Find the formula for the sequences $(a_i)_{i=1}^{\infty}$ satisfying the linear recurrence relation $a_{k+2}-5a_{k+1}+3a_{k}=0$ and express it using $a_1, a_2$.

Related Problems.

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

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

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

Proof.

(a) Find the eigenvalues and eigenvectors of $T$.

By problem 2, the matrix representation of the linear transformation $T: U\to U$ with respect to the basis $B=\{\mathbf{u}_1, \mathbf{u}_2\}$ of $U$ is given by
\[A=\begin{bmatrix}
0 & 1\\
-3& 5
\end{bmatrix}.\] The eigenvalues of $T$ is the same as the eigenvalues of the matrix $A$.
The characteristic polynomial for $A$ is
\begin{align*}
p(t)=\det(A-tI)=\begin{vmatrix}
-t & 1\\
-3& 5-t
\end{vmatrix}=t^2-5t+3.
\end{align*}
Solving $p(t)=0$, the eigenvalues of $T$ are
\[t=\frac{5\pm \sqrt{13}}{2}.\]

Let us find an eigenvector corresponding to $t=\frac{5+ \sqrt{13}}{2}$.
If $(a_i)_{i=1}^{\infty}$ is an eigenvalue, then we have
\[T\big( (a_i)_{i=1}^{\infty}\big)=\frac{5+ \sqrt{13}}{2} (a_i)_{i=1}^{\infty}.\] That is,
\begin{align*}
(a_2, a_3, a_4, \dots)=\frac{5+ \sqrt{13}}{2}(a_1, a_2, a_3, \dots).
\end{align*}
Thus, the eigenvector satisfies
\[a_{i+1}=\frac{5+ \sqrt{13}}{2}a_i,\] hence it is a geometric progression with initial value $a_1$ and common ratio $\frac{5+ \sqrt{13}}{2}$. Thus, we have
\[(a_i)_{i=1}^{\infty}=\left( \left(\frac{5+ \sqrt{13}}{2} \right )^{i-1} a_1 \right)_{i=1}^{\infty}\] is an eigenvector for any $a_1\in \R$.

Similarly, we see that the eigenvectors for $t=\frac{5- \sqrt{13}}{2}$ is
\[(a_i)_{i=1}^{\infty}=\left( \left(\frac{5- \sqrt{13}}{2} \right )^{i-1} a_1 \right)_{i=1}^{\infty}\] for any $a_1\in \R$.
Since $U$ is a $2$-dimensional vector space, these are all the eigenvectors.

(b) Solve linear recurrence relation $a_{k+2}-5a_{k+1}+3a_{k}=0$ with $a_1=1, a_2=1$.

By the result of (a), we know that
\[\mathbf{v}_1=\left( \left(\frac{5+ \sqrt{13}}{2} \right )^{i-1} \right)_{i=1}^{\infty}\] and
\[\mathbf{v}_2=\left( \left(\frac{5- \sqrt{13}}{2} \right )^{i-1} \right)_{i=1}^{\infty}\] are eigenvectors of $t=\frac{5\pm \sqrt{13}}{2}$, respectively. (We took the initial value $a_1=1$.)
Thus, the set $B’=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis of $U$.

Then any sequence $(a_i)_{i=1}^{\infty} \in U$ can be written as a linear combination of $\mathbf{v}_1, \mathbf{v}_2$. We have
\[(a_i)_{i=1}^{\infty}=c_1\mathbf{v}_1+c_2 \mathbf{v}_2,\] for some scalars $c_1, c_2$. Comparing the first two terms, we have
\begin{align*}
1&=a_1=c_1+c_2\\
1&=a_2=c_1 \left(\frac{5+ \sqrt{13}}{2} \right )+c_2 \left(\frac{5- \sqrt{13}}{2} \right ).
\end{align*}
Solving these equations, we obtain
\[c_1=\frac{13-3\sqrt{13}}{26}, c_2=\frac{13+3\sqrt{13}}{26}.\] Therefore, the sequence $(a_i)_{i=1}^{\infty} \in U$ with initial condition $a_1=1, a_2=1$ is
\begin{align*}
\frac{13-3\sqrt{13}}{26}\left( \left(\frac{5+ \sqrt{13}}{2} \right )^{i-1} \right)_{i=1}^{\infty}+\frac{13+3\sqrt{13}}{26}\left( \left(\frac{5- \sqrt{13}}{2} \right )^{i-1} \right)_{i=1}^{\infty}
\end{align*}

(c) Find the formula for the sequences $(a_i)_{i=1}^{\infty} \in U$.

Let $(a_i)_{i=1}^{\infty}\in U$. Using the basis vectors in $B=\{\mathbf{u}_1, \mathbf{u}_2\}$, we have
\[(a_i)_{i=1}^{\infty} = a_1 \mathbf{u}_1+ a_2\mathbf{u}_2.\] Since $B’=\{\mathbf{v}_1, \mathbf{v}_2\}$ is also a basis for $U$, we can also write
\[(a_i)_{i=1}^{\infty} = b_1 \mathbf{v}_1+ b_2\mathbf{v}_2\] for some scalars $b_1, b_2$.

We want to express $b_1, b_2$ in terms of $a_1, a_2$.
To do this, we first express $\mathbf{u}_1$ and $\mathbf{u}_2$ as linear combinations of $\mathbf{v}_1, \mathbf{v}_2$.

Let $\mathbf{u}_1=c_1\mathbf{v}_1+c_2\mathbf{v}_2$ for some $c_1, c_2$. We determine $c_1, c_2$.
Comparing the first two terms, we have
\begin{align*}
1&=c_1+c_2\\
0&=c_1 \left(\frac{5+ \sqrt{13}}{2} \right )+c_2 \left(\frac{5- \sqrt{13}}{2} \right ).
\end{align*}
Solving these equations, we have
\[c_1=\frac{13-5\sqrt{13}}{26}, c_2=\frac{13+5\sqrt{13}}{26}.\] Thus we have obtained the linear combination
\[\mathbf{u}_1=\frac{13-5\sqrt{13}}{26} \mathbf{v}_1+\frac{13+5\sqrt{13}}{26}\mathbf{v}_2. \tag{*}\]

Similarly, if $\mathbf{u}_2=c_1\mathbf{v}_1+c_2\mathbf{v}_2$ for some $c_1, c_2$, then comparing the first two terms we have
\begin{align*}
0&=c_1+c_2\\
1&=c_1 \left(\frac{5+ \sqrt{13}}{2} \right )+c_2 \left(\frac{5- \sqrt{13}}{2} \right ).
\end{align*}
and this gives
\[c_1=\frac{\sqrt{13}}{13}, c_2=-\frac{\sqrt{13}}{13}.\] Thus we have
\[\mathbf{u}_2=\frac{\sqrt{13}}{13} \mathbf{v}_1-\frac{\sqrt{13}}{13}\mathbf{v}_2. \tag{**}\]

Using (*), (**), we have
\begin{align*}
&(a_i)_{i=1}^{\infty} = a_1 \mathbf{u}_1+ a_2\mathbf{u}_2\\
&=a_1\left(\frac{13-5\sqrt{13}}{26} \mathbf{v}_1+\frac{13+5\sqrt{13}}{26}\mathbf{v}_2 \right)+a_2\left(\frac{\sqrt{13}}{13} \mathbf{v}_1-\frac{\sqrt{13}}{13}\mathbf{v}_2 \right)\\
&=\left(\frac{13-5\sqrt{13}}{26}a_1+\frac{\sqrt{13}}{13}a_2\right) \mathbf{v}_1
+\left( \frac{13+5\sqrt{13}}{26}a_1-\frac{\sqrt{13}}{13}a_2 \right) \mathbf{v}_2.
\end{align*}
The coefficients in the last linear combination are the $b_1, b_2$ that we wanted to find.

Therefore, the general formula for $(a_i)_{i=1}^{\infty} \in U$ using $a_1, a_2$ is given by
\begin{align*}
&(a_i)_{i=1}^{\infty} = \\
&=\left(\frac{13-5\sqrt{13}}{26}a_1+\frac{\sqrt{13}}{13}a_2\right) \left( \left(\frac{5+ \sqrt{13}}{2} \right )^{i-1} \right)_{i=1}^{\infty}\\
&+\left( \frac{13+5\sqrt{13}}{26}a_1-\frac{\sqrt{13}}{13}a_2 \right) \left( \left(\frac{5- \sqrt{13}}{2} \right )^{i-1} \right)_{i=1}^{\infty}.
\end{align*}

Remark 1.

For example, if the initial condition is $a_1=1, a_2=1$, then this formula gives the same formula we obtained in part (b).

Remark 2.

Note that we could have solved as follows.
We had
\[(a_i)_{i=1}^{\infty} = a_1 \mathbf{u}_1+ a_2\mathbf{u}_2 = b_1 \mathbf{v}_1+ b_2\mathbf{v}_2\] and wanted to express $b_1, b_2$ in terms of $a_1, a_2$.
If $P$ is the change of coordinates matrix from $B’$ to $B$, we have
\[\begin{bmatrix}
a_1 \\
a_2
\end{bmatrix}_B=P\begin{bmatrix}
b_1 \\
b_2
\end{bmatrix}_{B’},\] and
\[P=\begin{bmatrix}
1 & 1\\[6pt] \frac{5+ \sqrt{13}}{2} & \frac{5- \sqrt{13}}{2}
\end{bmatrix}.\] The inverse matix is
\[P^{-1}=\begin{bmatrix}
\frac{13-5 \sqrt{13}}{26} & \frac{\sqrt{13}}{13}\\[6pt] \frac{13+5\sqrt{13}}{26} & -\frac{\sqrt{13}}{13}
\end{bmatrix}\] and thus
\[\begin{bmatrix}
b_1 \\
b_2
\end{bmatrix}=P^{-1}\begin{bmatrix}
a_1 \\
a_2
\end{bmatrix},\] and we obtain the same $b_1, b_2$ as before.

The post Solve Linear Recurrence Relation Using Linear Algebra (Eigenvalues and Eigenvectors) first appeared on Problems in Mathematics.

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).

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)

The post Matrix Representation of a Linear Transformation of Subspace of Sequences Satisfying Recurrence Relation first appeared on Problems in Mathematics.

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 $V$.

Proof.

We prove the subspace criteria.

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.

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.

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)

The post Sequences Satisfying Linear Recurrence Relation Form a Subspace first appeared on Problems in Mathematics.