Let $P_1$ be the vector space of all real polynomials of degree $1$ or less. Consider the linear transformation $T: P_1 \to P_1$ defined by
\[T(ax+b)=(3a+b)x+a+3,\] for any $ax+b\in P_1$.

(a) With respect to the basis $B=\{1, x\}$, find the matrix of the linear transformation $T$.

(b) Find a basis $B’$ of the vector space $P_1$ such that the matrix of $T$ with respect to $B’$ is a diagonal matrix.

(c) Express $f(x)=5x+3$ as a linear combination of basis vectors of $B’$.

(a) Find the matrix of the linear transformation $T$ with respect to the basis $B=\{1, x\}$.

The matrix $A$ for $T$ with respect to the basis $B=\{1, x\}$ is given by
\[A=[\, [T(1)]_B, [T(x)]_B\,].\] Since we have
\[T(1)=x+3 \text{ and } T(x)=3x+4,\] the coordinate vectors of these vector with respect to $B$ are
\begin{align*}
[T(1)]_B=\begin{bmatrix}
3 \\
1
\end{bmatrix} \text{ and } [T(x)]_B=\begin{bmatrix}
4 \\
3
\end{bmatrix}.
\end{align*}
Therefore, the matrix $A$ for $T$ with respect to $B$ is
\[A=\begin{bmatrix}
3 & 4\\
1& 3
\end{bmatrix}.\]

(b) Find a basis $B’$ of the vector space $P_1$ such that the matrix of $T$ with respect to $B’$ is a diagonal matrix.

We first diagonalize the matrix $A$ obtained in (a).
Solving the characteristic polynomial
\begin{align*}
p_A(t)=\det(A-tI)=\begin{bmatrix}
3-t & 4\\
1& 3-t
\end{bmatrix}\\
=t^2-6t+5=(t-1)(t-5),
\end{align*}
the eigenvalues are $1$ and $5$. Since $A$ has two distinct eigenvalues, it is diagonalizable.

We find the eigenvectors corresponding to the eigenvalue $1$.
The eigenvectors are nonzero solutions of $(A-I)\mathbf{x}=\mathbf{0}$.
From this, we find that $\begin{bmatrix}
2 \\
-1
\end{bmatrix}$ is an eigenvector.
Similarly, we find that $\begin{bmatrix}
2 \\
1
\end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $5$.

Thus, if we set
\[P=\begin{bmatrix}
2 & 2\\
-1& 1
\end{bmatrix},\] then we have the diagonalization of $A$:
\[\begin{bmatrix}
1 & 0\\
0 & 5
\end{bmatrix}=P^{-1}AP.\] It follows that $\begin{bmatrix}
1 & 0\\
0 & 5
\end{bmatrix}$ is the matrix representation for $T$ with respect to the basis
\[B’=\{\mathbf{v}_1, \mathbf{v}_2\},\] where
\[[\mathbf{v}_1]_B=\begin{bmatrix}
2 \\
-1
\end{bmatrix} \text{ and } [\mathbf{v}_2]_B=\begin{bmatrix}
2 \\
1
\end{bmatrix},\] and $P$ is the transition matrix (change of basis) from $B’$ to $B$.

(Note that
\[[\mathbf{v}_1]_B=P[\mathbf{v}_1]_{B’}=P\begin{bmatrix}
1 \\
0
\end{bmatrix}=\begin{bmatrix}
2 \\
-1
\end{bmatrix}.\] Similarly for $[\mathbf{v}_2]_B$.)

Hence, $\mathbf{v}_1=-x+2$ and $\mathbf{v}_2=x+2$ form the basis $B’=\{\mathbf{v}_1, \mathbf{v}_2\}$ with respect to which the matrix for $T$ is the diagonal matrix $\begin{bmatrix}
1 & 0\\
0 & 5
\end{bmatrix}$.

(c) Express $f(x)=5x+3$ as a linear combination of basis vectors of $B’$.

Note that since $P$ is the transition matrix from $B’$ to $B$, we have the following relation between coordinate vectors. For any $\mathbf{v}\in P_1$, we have
\[P[\mathbf{v}]_{B’}=[\mathbf{v}]_B.\] Using this, we have
\begin{align*}
[f(x)]_{B’}&=P^{-1}[f(x)]_B\\
&=\frac{1}{4}\begin{bmatrix}
1 & -2\\
1 & 2
\end{bmatrix}\begin{bmatrix}
3 \\
5
\end{bmatrix}=\begin{bmatrix}
-1 \\
4
\end{bmatrix}.
\end{align*}
Hence we obtain the linear combination of $f(x)$ with $\mathbf{v}_1, \mathbf{v}_2$:
\begin{align*}
f(x)&=\frac{-7}{4}\mathbf{v}_1 + \frac{13}{4} \mathbf{v}_2\\
\end{align*}

(You may confirm that $5x+3= \frac{-7}{4} (-x+2) + \frac{13}{4} (x+2)$.)

The post Basis with Respect to Which the Matrix for Linear Transformation is Diagonal 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 $k=1, 2, \dots$.
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.