## Section3.4Eigenvalue method

Note: 2 lectures, §5.2 in [EP], part of §7.3, §7.5, and §7.6 in [BD]

In this section we will learn how to solve linear homogeneous constant coefficient systems of ODEs by the eigenvalue method. Suppose we have such a system

\begin{equation*} {\vec{x}}' = P\vec{x} , \end{equation*}

where $P$ is a constant square matrix. We wish to adapt the method for the single constant coefficient equation by trying the function $e^{\lambda t}\text{.}$ However, $\vec{x}$ is a vector. So we try $\vec{x} = \vec{v} e^{\lambda t}\text{,}$ where $\vec{v}$ is an arbitrary constant vector. We plug this $\vec{x}$ into the equation to get

\begin{equation*} \underbrace{\lambda \vec{v} e^{\lambda t}}_{{\vec{x}}'} = \underbrace{P\vec{v} e^{\lambda t}}_{P\vec{x}} . \end{equation*}

We divide by $e^{\lambda t}$ and notice that we are looking for a scalar $\lambda$ and a vector $\vec{v}$ that satisfy the equation

\begin{equation*} \lambda \vec{v} = P\vec{v} . \end{equation*}

To solve this equation we need a little bit more linear algebra, which we now review.

### Subsection3.4.1Eigenvalues and eigenvectors of a matrix

Let $A$ be a constant square matrix. Suppose there is a scalar $\lambda$ and a nonzero vector $\vec{v}$ such that

\begin{equation*} A \vec{v} = \lambda \vec{v}. \end{equation*}

We call $\lambda$ an eigenvalue of $A$ and we call $\vec{v}$ a corresponding eigenvector.

###### Example3.4.1.

The matrix $\left[ \begin{smallmatrix} 2 & 1 \\ 0 & 1 \end{smallmatrix} \right]$ has an eigenvalue $\lambda = 2$ with a corresponding eigenvector $\left[ \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \right]$ as

\begin{equation*} \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix} = 2 \begin{bmatrix} 1 \\ 0 \end{bmatrix} . \end{equation*}

Let us see how to compute eigenvalues for any matrix. Rewrite the equation for an eigenvalue as

\begin{equation*} (A - \lambda I)\vec{v} = \vec{0} . \end{equation*}

This equation has a nonzero solution $\vec{v}$ only if $A - \lambda I$ is not invertible. Were it invertible, we could write ${(A - \lambda I)}^{-1}(A - \lambda I)\vec{v} = {(A-\lambda I)}^{-1}\vec{0}\text{,}$ which implies $\vec{v} = \vec{0}\text{.}$ Therefore, $A$ has the eigenvalue $\lambda$ if and only if $\lambda$ solves the equation

\begin{equation*} \det (A-\lambda I) = 0 . \end{equation*}

Consequently, we will be able to find an eigenvalue of $A$ without finding a corresponding eigenvector. An eigenvector will have to be found later, once $\lambda$ is known.

###### Example3.4.2.

Find all eigenvalues of $\left[ \begin{smallmatrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{smallmatrix} \right]\text{.}$

We write

\begin{multline*} \det \left( \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \right) = \det \left( \begin{bmatrix} 2-\lambda & 1 & 1 \\ 1 & 2-\lambda & 0 \\ 0 & 0 & 2-\lambda \end{bmatrix} \right) = \\ = (2-\lambda) \bigl({(2-\lambda)}^2 - 1\bigr) = -(\lambda -1)(\lambda -2)(\lambda-3) . \end{multline*}

So the eigenvalues are $\lambda = 1\text{,}$ $\lambda = 2\text{,}$ and $\lambda = 3\text{.}$

For an $n \times n$ matrix, the polynomial we get by computing $\det(A - \lambda I)$ is of degree $n\text{,}$ and hence in general, we have $n$ eigenvalues. Some may be repeated, some may be complex.

To find an eigenvector corresponding to an eigenvalue $\lambda\text{,}$ we write

\begin{equation*} (A-\lambda I) \vec{v} = \vec{0} , \end{equation*}

and solve for a nontrivial (nonzero) vector $\vec{v}\text{.}$ If $\lambda$ is an eigenvalue, there will be at least one free variable, and so for each distinct eigenvalue $\lambda\text{,}$ we can always find an eigenvector.

###### Example3.4.3.

Find an eigenvector of $\left[ \begin{smallmatrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{smallmatrix} \right]$ corresponding to the eigenvalue $\lambda = 3\text{.}$

We write

\begin{equation*} (A-\lambda I) \vec{v} = \left( \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} - 3 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \right) \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \vec{0} . \end{equation*}

It is easy to solve this system of linear equations. We write down the augmented matrix

\begin{equation*} \left[ \begin{array}{ccc|c} -1 & 1 & 1 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \end{array} \right] , \end{equation*}

and perform row operations (exercise: which ones?) until we get:

\begin{equation*} \left[ \begin{array}{ccc|c} 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] . \end{equation*}

The entries of $\vec{v}$ have to satisfy the equations $v_1 - v_2 = 0\text{,}$ $v_3 = 0\text{,}$ and $v_2$ is a free variable. We can pick $v_2$ to be arbitrary (but nonzero), let $v_1 = v_2\text{,}$ and of course $v_3 = 0\text{.}$ For example, if we pick $v_2 = 1\text{,}$ then $\vec{v} = \left[ \begin{smallmatrix} 1 \\ 1 \\ 0 \end{smallmatrix} \right]\text{.}$ Let us verify that $\vec{v}$ really is an eigenvector corresponding to $\lambda = 3\text{:}$

\begin{equation*} \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 0 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} . \end{equation*}

Yay! It worked.

###### Exercise3.4.1.

(easy)   Are eigenvectors unique? Can you find a different eigenvector for $\lambda = 3$ in the example above? How are the two eigenvectors related?

###### Exercise3.4.2.

When the matrix is $2 \times 2$ you do not need to do row operations when computing an eigenvector, you can read it off from $A-\lambda I$ (if you have computed the eigenvalues correctly). Can you see why? Explain. Try it for the matrix $\left[ \begin{smallmatrix} 2 & 1 \\ 1 & 2 \end{smallmatrix} \right]\text{.}$

### Subsection3.4.2The eigenvalue method with distinct real eigenvalues

OK. We have the system of equations

\begin{equation*} {\vec{x}}' = P\vec{x} . \end{equation*}

We find the eigenvalues $\lambda_1\text{,}$ $\lambda_2\text{,}$ ..., $\lambda_n$ of the matrix $P\text{,}$ and corresponding eigenvectors $\vec{v}_1\text{,}$ $\vec{v}_2\text{,}$ ..., $\vec{v}_n\text{.}$ Now we notice that the functions $\vec{v}_1 e^{\lambda_1 t}\text{,}$ $\vec{v}_2 e^{\lambda_2 t}\text{,}$ ..., $\vec{v}_n e^{\lambda_n t}$ are solutions of the system of equations and hence $\vec{x} = c_1 \vec{v}_1 e^{\lambda_1 t} + c_2 \vec{v}_2 e^{\lambda_2 t} + \cdots + c_n \vec{v}_n e^{\lambda_n t}$ is a solution.

The corresponding fundamental matrix solution is

\begin{equation*} X(t) = \bigl[\, \vec{v}_1 e^{\lambda_1 t} \quad \vec{v}_2 e^{\lambda_2 t} \quad \cdots \quad \vec{v}_n e^{\lambda_n t} \,\bigr]. \end{equation*}

That is, $X(t)$ is the matrix whose $j^{\text{th}}$ column is $\vec{v}_j e^{\lambda_j t}\text{.}$

###### Example3.4.4.

Consider the system

\begin{equation*} {\vec{x}}' = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \vec{x} . \end{equation*}

Find the general solution.

Earlier, we found the eigenvalues are $1,2,3\text{.}$ We found the eigenvector $\left[ \begin{smallmatrix} 1 \\ 1 \\ 0 \end{smallmatrix} \right]$ for the eigenvalue 3. Similarly we find the eigenvector $\left[ \begin{smallmatrix} 1 \\ -1 \\ 0 \end{smallmatrix} \right]$ for the eigenvalue 1, and $\left[ \begin{smallmatrix} 0 \\ 1 \\ -1 \end{smallmatrix} \right]$ for the eigenvalue 2 (exercise: check). Hence our general solution is

\begin{equation*} \vec{x} = c_1 \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} e^t + c_2 \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} e^{2t} + c_3 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} e^{3t} = \begin{bmatrix} c_1 e^t+c_3 e^{3t} \\ -c_1 e^t + c_2 e^{2t} + c_3 e^{3t} \\ - c_2 e^{2t} \end{bmatrix} . \end{equation*}

In terms of a fundamental matrix solution,

\begin{equation*} \vec{x} = X(t)\, \vec{c} = \begin{bmatrix} e^t & 0 & e^{3t} \\ -e^t & e^{2t} & e^{3t} \\ 0 & -e^{2t} & 0 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} . \end{equation*}
###### Exercise3.4.3.

Check that this $\vec{x}$ really solves the system.

Note: If we write a single homogeneous linear constant coefficient $n^{\text{th}}$ order equation as a first order system (as we did in Section 3.1), then the eigenvalue equation

\begin{equation*} \det(P - \lambda I) = 0 \end{equation*}

is essentially the same as the characteristic equation we got in Section 2.2 and Section 2.3.

### Subsection3.4.3Complex eigenvalues

A matrix may very well have complex eigenvalues even if all the entries are real. Take, for example,

\begin{equation*} {\vec{x}}' = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \vec{x} . \end{equation*}

Let us compute the eigenvalues of the matrix $P = \left[ \begin{smallmatrix} 1 & 1 \\ -1 & 1 \end{smallmatrix} \right]\text{.}$

\begin{equation*} \det(P - \lambda I) = \det\left( \begin{bmatrix} 1-\lambda & 1 \\ -1 & 1-\lambda \end{bmatrix} \right) = {(1-\lambda)}^2 + 1 = \lambda^2 - 2 \lambda + 2 = 0 . \end{equation*}

Thus $\lambda = 1 \pm i\text{.}$ Corresponding eigenvectors are also complex. Start with $\lambda = 1-i\text{.}$

\begin{equation*} \begin{aligned} \bigl(P-(1-i) I\bigr) \vec{v} & = \vec{0} , \\ \begin{bmatrix} i & 1 \\ -1 & i \end{bmatrix} \vec{v} = \vec{0}. \end{aligned} \end{equation*}

The equations $i v_1 + v_2 = 0$ and $-v_1 + iv_2 = 0$ are multiples of each other. So we only need to consider one of them. After picking $v_2 = 1\text{,}$ for example, we have an eigenvector $\vec{v} = \left[ \begin{smallmatrix} i \\ 1 \end{smallmatrix} \right]\text{.}$ In similar fashion we find that $\left[ \begin{smallmatrix} -i \\ 1 \end{smallmatrix} \right]$ is an eigenvector corresponding to the eigenvalue $1+i\text{.}$

We could write the solution as

\begin{equation*} \vec{x} = c_1 \begin{bmatrix} i \\ 1 \end{bmatrix} e^{(1-i)t} + c_2 \begin{bmatrix} -i \\ 1 \end{bmatrix} e^{(1+i)t} = \begin{bmatrix} c_1 i e^{(1-i)t} - c_2 i e^{(1+i)t} \\ c_1 e^{(1-i)t} + c_2 e^{(1+i)t} \end{bmatrix} . \end{equation*}

We would then need to look for complex values $c_1$ and $c_2$ to solve any initial conditions. It is perhaps not completely clear that we get a real solution. After solving for $c_1$ and $c_2\text{,}$ we could use Euler's formula and do the whole song and dance we did before, but we will not. We will apply the formula in a smarter way first to find independent real solutions.

We claim that we did not have to look for a second eigenvector (nor for the second eigenvalue). All complex eigenvalues come in pairs (because the matrix $P$ is real).

First a small detour. The real part of a complex number $z$ can be computed as $\frac{z + \bar{z}}{2}\text{,}$ where the bar above $z$ means $\overline{a+ib} = a -ib\text{.}$ This operation is called the complex conjugate. If $a$ is a real number, then $\bar{a} = a\text{.}$ Similarly we bar whole vectors or matrices by taking the complex conjugate of every entry. Suppose a matrix $P$ is real. Then $\overline{P} = P\text{,}$ and so $\overline{P\vec{x}} = \overline{P} \, \overline{\vec{x}} = P \overline{\vec{x}}\text{.}$ Also the complex conjugate of 0 is still 0, therefore,

\begin{equation*} \vec{0} = \overline{\vec{0}} = \overline{(P-\lambda I)\vec{v}} = (P-\bar{\lambda} I)\overline{\vec{v}} . \end{equation*}

In other words, if $\lambda = a+ib$ is an eigenvalue, then so is $\bar{\lambda} = a-ib\text{.}$ And if $\vec{v}$ is an eigenvector corresponding to the eigenvalue $\lambda\text{,}$ then $\overline{\vec{v}}$ is an eigenvector corresponding to the eigenvalue $\bar{\lambda}\text{.}$

Suppose $a + ib$ is a complex eigenvalue of $P\text{,}$ and $\vec{v}$ is a corresponding eigenvector. Then

\begin{equation*} \vec{x}_1 = \vec{v} e^{(a+ib)t} \end{equation*}

is a solution (complex-valued) of ${\vec{x}}' = P \vec{x}\text{.}$ Euler's formula shows that $\overline{e^{a+ib}} = e^{a-ib}\text{,}$ and so

\begin{equation*} \vec{x}_2 = \overline{\vec{x}_1} = \overline{\vec{v}} e^{(a-ib)t} \end{equation*}

is also a solution. As $\vec{x}_1$ and $\vec{x}_2$ are solutions, the function

\begin{equation*} \vec{x}_3 = \operatorname{Re} \vec{x}_1 = \operatorname{Re} \vec{v} e^{(a+ib)t} = \frac{\vec{x}_1 + \overline{\vec{x}_1}}{2} = \frac{\vec{x}_1 + \vec{x}_2}{2} = \frac{1}{2} \vec{x}_1 + \frac{1}{2}\vec{x}_2 \end{equation*}

is also a solution. And $\vec{x}_3$ is real-valued! Similarly as $\operatorname{Im} z = \frac{z-\bar{z}}{2i}$ is the imaginary part, we find that

\begin{equation*} \vec{x}_4 = \operatorname{Im} \vec{x}_1 = \frac{\vec{x}_1 - \overline{\vec{x}_1}}{2i} = \frac{\vec{x}_1 - \vec{x}_2}{2i} . \end{equation*}

is also a real-valued solution. It turns out that $\vec{x}_3$ and $\vec{x}_4$ are linearly independent. We will use Euler's formula to separate out the real and imaginary part.

Returning to our problem,

\begin{equation*} \vec{x}_1 = \begin{bmatrix} i \\ 1 \end{bmatrix} e^{(1-i)t} = \begin{bmatrix} i \\ 1 \end{bmatrix} \left( e^t \cos t - i e^t \sin t \right) = \begin{bmatrix} i e^t \cos t + e^t \sin t \\ e^t \cos t - i e^t \sin t \end{bmatrix} = \begin{bmatrix} e^t \sin t \\ e^t \cos t \end{bmatrix} + i \begin{bmatrix} e^t \cos t \\ - e^t \sin t \end{bmatrix} . \end{equation*}

Then

\begin{equation*} \operatorname{Re} \vec{x}_1 = \begin{bmatrix} e^t \sin t \\ e^t \cos t \end{bmatrix} , \qquad \text{and} \qquad \operatorname{Im} \vec{x}_1 = \begin{bmatrix} e^t \cos t \\ - e^t \sin t \end{bmatrix} , \end{equation*}

are the two real-valued linearly independent solutions we seek.

###### Exercise3.4.4.

Check that these really are solutions.

The general solution is

\begin{equation*} \vec{x} = c_1 \begin{bmatrix} e^t \sin t \\ e^t \cos t \end{bmatrix} + c_2 \begin{bmatrix} e^t \cos t \\ -e^t \sin t \end{bmatrix} = \begin{bmatrix} c_1 e^t \sin t + c_2 e^t \cos t \\ c_1 e^t \cos t - c_2 e^t \sin t \end{bmatrix} . \end{equation*}

This solution is real-valued for real $c_1$ and $c_2\text{.}$ We now solve for any initial conditions we may have.

Let us summarize as a theorem.

For each pair of complex eigenvalues $a+ib$ and $a-ib\text{,}$ we get two real-valued linearly independent solutions. We then go on to the next eigenvalue, which is either a real eigenvalue or another complex eigenvalue pair. If we have $n$ distinct eigenvalues (real or complex), then we end up with $n$ linearly independent solutions. If we had only two equations ($n=2$) as in the example above, then once we found two solutions we are finished, and our general solution is

\begin{equation*} \vec{x} = c_1 \vec{x}_1 + c_2 \vec{x}_2 = c_1 \bigl( \operatorname{Re} \vec{v} e^{(a+ib)t} \bigr) + c_2 \bigl( \operatorname{Im} \vec{v} e^{(a+ib)t} \bigr) . \end{equation*}

We can now find a real-valued general solution to any homogeneous system where the matrix has distinct eigenvalues. When we have repeated eigenvalues, matters get a bit more complicated and we will look at that situation in Section 3.7.

### Subsection3.4.4Exercises

###### Exercise3.4.5.

(easy)   Let $A$ be a $3 \times 3$ matrix with an eigenvalue of 3 and a corresponding eigenvector $\vec{v} = \left[ \begin{smallmatrix} 1 \\ -1 \\ 3 \end{smallmatrix} \right]\text{.}$ Find $A \vec{v}\text{.}$

###### Exercise3.4.6.

1. Find the general solution of $x_1' = 2 x_1\text{,}$ $x_2' = 3 x_2$ using the eigenvalue method (first write the system in the form ${\vec{x}}' = A \vec{x}$).

2. Solve the system by solving each equation separately and verify you get the same general solution.

###### Exercise3.4.7.

Find the general solution of $x_1' = 3 x_1 + x_2\text{,}$ $x_2' = 2 x_1 + 4 x_2$ using the eigenvalue method.

###### Exercise3.4.8.

Find the general solution of $x_1' = x_1 -2 x_2\text{,}$ $x_2' = 2 x_1 + x_2$ using the eigenvalue method. Do not use complex exponentials in your solution.

###### Exercise3.4.9.

1. Compute eigenvalues and eigenvectors of $A = \left[ \begin{smallmatrix} 9 & -2 & -6 \\ -8 & 3 & 6 \\ 10 & -2 & -6 \end{smallmatrix} \right]\text{.}$

2. Find the general solution of ${\vec{x}}' = A \vec{x}\text{.}$

###### Exercise3.4.10.

Compute eigenvalues and eigenvectors of $\left[ \begin{smallmatrix} -2 & -1 & -1 \\ 3 & 2 & 1 \\ -3 & -1 & 0 \\ \end{smallmatrix} \right]\text{.}$

###### Exercise3.4.11.

Let $a,b,c,d,e,f$ be numbers. Find the eigenvalues of $\left[ \begin{smallmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f \\ \end{smallmatrix} \right]\text{.}$

###### Exercise3.4.101.

1. Compute eigenvalues and eigenvectors of $A= \left[ \begin{smallmatrix} 1 & 0 & 3 \\ -1 & 0 & 1 \\ 2 & 0 & 2 \end{smallmatrix}\right]\text{.}$

2. Solve the system $\vec{x}\,' = A \vec{x}\text{.}$

a) Eigenvalues: $4,0,-1$     Eigenvectors: $\left[ \begin{smallmatrix} 1 \\ 0 \\ 1 \end{smallmatrix}\right]\text{,}$ $\left[ \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix}\right]\text{,}$ $\left[ \begin{smallmatrix} 3 \\ 5 \\ -2 \end{smallmatrix}\right]$
b) $\vec{x} = C_1 \left[ \begin{smallmatrix} 1 \\ 0 \\ 1 \end{smallmatrix}\right] e^{4t} + C_2 \left[ \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix}\right] + C_3 \left[ \begin{smallmatrix} 3 \\ 5 \\ -2 \end{smallmatrix}\right] e^{-t}$

###### Exercise3.4.102.

1. Compute eigenvalues and eigenvectors of $A=\left[ \begin{smallmatrix} 1 & 1 \\ -1 & 0 \end{smallmatrix}\right]\text{.}$

2. Solve the system $\vec{x}\,' = A\vec{x}\text{.}$

a) Eigenvalues: $\frac{1+\sqrt{3}i}{2}, \frac{1-\sqrt{3}i}{2}\text{,}$     Eigenvectors: $\left[ \begin{smallmatrix} -2 \\ 1-\sqrt{3}i \end{smallmatrix}\right]\text{,}$ $\left[ \begin{smallmatrix} -2 \\ 1+\sqrt{3}i \end{smallmatrix}\right]$
b) $\vec{x} = C_1 e^{t/2} \left[ \begin{smallmatrix} -2\cos\bigl(\frac{\sqrt{3}t}{2}\bigr) \\ \cos\bigl(\frac{\sqrt{3}t}{2}\bigr) + \sqrt{3}\sin\bigl(\frac{\sqrt{3}t}{2}\bigr) \end{smallmatrix}\right] + C_2 e^{t/2} \left[ \begin{smallmatrix} - 2\sin\bigl(\frac{\sqrt{3}t}{2}\bigr) \\ \sin\bigl(\frac{\sqrt{3}t}{2}\bigr) -\sqrt{3}\cos\bigl(\frac{\sqrt{3}t}{2}\bigr) \end{smallmatrix}\right]$

###### Exercise3.4.103.

Solve $x_1' = x_2\text{,}$ $x_2' = x_1$ using the eigenvalue method.

$\vec{x} = C_1 \left[ \begin{smallmatrix} 1 \\ 1 \end{smallmatrix}\right] e^{t} + C_2 \left[ \begin{smallmatrix} 1 \\ -1 \end{smallmatrix}\right] e^{-t}$
Solve $x_1' = x_2\text{,}$ $x_2' = -x_1$ using the eigenvalue method.
$\vec{x} = C_1 \left[ \begin{smallmatrix} \cos(t) \\ -\sin(t) \end{smallmatrix}\right] + C_2 \left[ \begin{smallmatrix} \sin(t) \\ \cos(t) \end{smallmatrix}\right]$