Title: Default for 22.2 Applications of Eigenvalues and Eigenvectors
Course: HELM Workbooks 2023-2024

Diagonalization of a matrix with distinct eigenvalues

Diagonalization means transforming a non-diagonal matrix into an equivalent matrix which is diagonal and hence is simpler to deal with.

A matrix \(A\) with distinct eigenvalues has eigenvectors which are linearly independent. If we form a matrix \(P\) whose columns are these eigenvectors, it can be shown that \( \det (P)\neq 0\) so that \(P^{-1}\) exists.

The product \(P^{-1}AP\) is then a diagonal matrix \(D\) whose diagonal elements are the eigenvalues of \(A\). Thus if \(\lambda_1,\lambda_2,\ldots \lambda_n\) are the distinct eigenvalues of \(A\) with associated eigenvectors \(\underline{x}_1,\underline{x}_2,\ldots , \underline{x}_n\) respectively, then

\[P = [\underline{x}_1 : \underline{x}_2 : \ \cdots\ : \underline{x}_n]\]

will produce a product

\[P^{-1}AP = D = \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & \cdots & \lambda_n \end{bmatrix}\]

We see that the order of the eigenvalues in \(D\) matches the order in which \(P\) is formed from the eigenvectors.

Note that:

1. The matrix \(P\) is called the modal matrix of \(A\).
2. Since \(D\) is a diagonal matrix with eigenvalues \(\lambda_1,\lambda_2,\ldots,\lambda_n\) which are the same as those of \(A\), then the matrices \(D\) and \(A\) are said to be similar.
3. The transformation of \(A\) into \(D\) using \(P^{-1}AP=D\) is said to be a similarity transformation.

Example

Let \(A=\left[\begin{array}{cc}2&3\\ 3&2\end{array}\right]\). Obtain the modal matrix \(P\) and calculate the product \(P^{-1}AP\).

Solution

The matrix \(A\) has two distinct eigenvalues \(\lambda_1=-1\), \(\lambda_2=5\) with corresponding eigenvectors \(\underline{x_1}=\left[\begin{array}{r}x\\ -x\end{array}\right]\) and \(\underline{x_2}=\left[\begin{array}{c}x\\ x\end{array}\right]\). We can therefore form the modal matrix from the simplest eigenvectors of these forms:

\[ P=\left[\begin{array}{rr}1&1\\ -1&1\end{array}\right]\]

(Other eigenvectors would be acceptable e.g. we could use \(P=\left[\begin{array}{rr}2&3\\ -2&3\end{array}\right]\) but there is no reason to over complicate the calculation.)

It is easy to obtain the inverse of this \(2\times 2\) matrix \(P\) and the reader should confirm that:

\[P^{-1}=\dfrac{1}{\text{det}(P)}\ \text{adj} (P)=\dfrac{1}{2}\left[\begin{array}{rr}1&1\\ -1&1\end{array}\right]^T=\dfrac{1}{2}\left[\begin{array}{rr}1&-1\\ 1&1\end{array}\right]\]

We can now construct the product \(P^{-1}AP\):

\[\begin{aligned} P^{-1}AP&=\dfrac{1}{2}\left[\begin{array}{rr}1&-1\\ 1&1\end{array}\right]\left[\begin{array}{cc}2&3\\ 3&2\end{array}\right]\left[\begin{array}{rr}1&1\\ -1&1\end{array}\right]\\ &=\dfrac{1}{2}\left[\begin{array}{rr}1&-1\\ 1&1\end{array}\right]\left[\begin{array}{rr}-1&5\\ 1&5\end{array}\right]\\ \\ &= \dfrac{1}{2}\left[\begin{array}{rr}-2&0\\ 0&10\end{array}\right]\\ &= \left[\begin{array}{rr}-1&0\\ 0&5\end{array}\right]\end{aligned}\]

which is a diagonal matrix with entries that are the eigenvalues, as expected. Show (by repeating the method outlined above) that had we defined \(P=\left[\begin{array}{rr}1&1\\ 1&-1\end{array}\right]\) (i.e. interchanged the order in which the eigenvectors were taken) we would find \(P^{-1}AP=\left[\begin{array}{rr}5&0\\ 0&-1\end{array}\right]\) (i.e. the resulting diagonal elements would also be interchanged.)

Matrix powers

If \(P^{-1}AP=D\) then we can obtain \(A\) (i.e. make \(A\) the subject of this matrix equation) as follows:

Multiplying on the left by \(P\) and on the right by \(P^{-1}\) we obtain

\[PP^{-1}APP^{-1}=PDP^{-1}\]

Now using the fact that  \(PP^{-1}=P^{-1}P=I\) we obtain

\[IAI=PDP^{-1}\]

and so

\[A=PDP^{-1}\]

We can use this result to obtain the powers of a square matrix, a process which is sometimes useful in control theory. Note that

\(A^2 = A \cdot A\),   \(A^3=A \cdot A \cdot A\)   etc.

Clearly, obtaining high powers of \(A\) directly would in general involve many multiplications. The process is quite straightforward, however, for a diagonal matrix \(D\), as this next Exercise shows.

We now use the relation \(\ \ A=PDP^{-1}\)  to obtain a formula for powers of \(A\) in terms of the easily calculated powers of the diagonal matrix \(D\):

\begin{align*} A^ = A \cdot A & = (PDP^{-1})(PDP^{-1}) \\ & = PD(P^{-1}P)DP^{-1} \\ & = PDIDP^{-1} \\ & = PD^2P^{-1} \end{align*}

Similarly,

\begin{align*} A^3 = A^2 \cdot A & (PD^2P^{-1})(PDP^{-1}) \\ & = PD^2(P^{-1}P)DP^{-1} \\ & = PD^3P^{-1} \end{align*}

The general result is given in the following Key Point:

Key Point

For a matrix \(A\) with distinct eigenvalues \(\lambda_1,\lambda_2,\dots ,\lambda_n\) and associated eigenvectors \(\underline{x}_1, \underline{x}_2,\dots ,\underline{x}_n\), if

\[P=[\underline{x}_1 : \underline{x}_2 : \ \cdots\ : \underline{x}_n]\]

then \(D=P^{-1}AP\) is a diagonal matrix such that \[ D=\left[\begin{matrix}\lambda_1&&&&\\ &\lambda_2&&&\\ &&\ddots &&\\ &&&\lambda_n\end{matrix}\right]\qquad\text{and}\qquad A^k=PD^kP^{-1}\]

Example

If \(A=\left[\begin{array}{cc}2&3\\ 3&2\end{array}\right]\), find \(A^{23}\). (Use the results of the previous Example.)

Solution

We know from the previous Example that if \(P=\left[\begin{array}{rr}1&1\\-1&1\end{array}\right]\) then \( P^{-1}AP=\left[\begin{array}{rr}-1&0\\ 0&5\end{array}\right]= D\) where \(P^{-1}=\dfrac{1}{2}\left[\begin{array}{rr}\hfill 1&\hfill -1\\ \hfill 1&\hfill 1\end{array}\right]\).

Therefore,

\(A=PDP^{-1}\)   and   \(A^{23}=PD^{23}P^{-1}\)   using the general result in the Key Point, i.e.,

\[A^{23}= \dfrac{1}{2} \left[\begin{array}{rr}\hfill 1&\hfill 1\\ \hfill -1&\hfill 1\end{array}\right] \left[\begin{array}{rr}\hfill -1&\hfill 0\\ \hfill 0&\hfill 5^{23}\end{array}\right] \left[\begin{array}{rr}\hfill 1&\hfill -1\\ \hfill 1&\hfill 1\end{array}\right]\]

which is easily evaluated, especially compared to calculating the product \(A^{23}\) directly!

Exercise

The matrix \([value]\) has eigenvalues \([value]\) and \([value]\) with respective eigenvectors  \([value]\) and \([value]\).
If \( P_1=[value], \ P_2=[value]\) and \( P_3=[value],\) write down the following products (You may not need to do detailed calculations).

1. \(P_1^{-1}AP_1=\)

2. \(P_2^{-1}AP_2=\)

3. \(P_3^{-1}AP_3=\)
Feedback: Since \(P_1,P_2\) and \(P_3\) are all modal matrices of \(A\), \(\qquad P_1^{-1}AP_1=[value]= [value] = D_1\) \(\qquad P_2^{-1}AP_2=[value]= [value] = D_2\) \(\qquad P_3^{-1}AP_3=[value]= [value]= D_3\) Note that \(D_1=D_3,\) demonstrating that any eigenvectors of \(A\) can be used to form \(P\). Note also that since the columns of \(P_1\) have been interchanged in forming \(P_2\) then so have the eigenvalues in \(D_2\) as compared with \(D_1\).

Exercise

Let \(D=[value]\) and write down the following powers of \(D\):

1. \(D^2=\)

2. \(D^3=\)

3. \(D^{[value]}=\)
Feedback: \(\qquad D^2=[value][value]=[value]=[value]\). \(\qquad D^3=[value][value]=[value]=[value]\). Continuing in this way: \(\qquad D^[value]=[value]=[value]\).

Exercise

For each of the following matrices, find a modal matrix \(P\):

1. \(A=[value]\)

\( P=\)

2. \(A=[value]\)

\( P=\)
Feedback: 1. The eigenvalues of \(A=[value]\) are \([value]\) and \([value]\), with corresponding eigenvectors \([value]\) and \([value]\), respectively. Hence a modal matrix to \([value]\) is \(P=[value]\). We can confirm that \(\qquad P^{-1}AP=[value][value][value]=[value],\) hence \([value]\) is a modal matrix of \(A\). 2. For \(A=[value]\) we find eigenvalues \([value],[value]\), and \([value]\) with corresponding eigenvectors \([value],[value]\), and \([value]\). Hence a modal matrix is \(P=[value]\). We can confirm that \(\qquad P^{-1}AP=[value][value][value]=[value],\) hence \([value]\) is a modal matrix of \(A\).

Systems of first order differential equations

Systems of first order ordinary differential equations arise in many areas of mathematics and engineering, for example in control theory and in the analysis of electrical circuits. In each case the basic unknowns are each a function of the time variable \(t\). A number of techniques have been developed to solve such systems of equations; for example the Laplace transform. Here we shall use eigenvalues and eigenvectors to obtain the solution. Our first step will be to recast the system of ordinary differential equations in the matrix form \(\dot{\underline{x}}=A \underline{x}\) where \(A\) is an \(n\times n\) coefficient matrix of constants, \(\underline{x}\) is the \(n\times 1\) column vector of unknown functions and \(\dot {\underline{x}}\) is the \(n\times 1\) column vector containing the derivatives of the unknowns. The main step will be to use the modal matrix of \(A\) to diagonalise the system of differential equations. This process will transform \(\dot {\underline{x}}=A \underline{x}\) into the form \(\dot{\underline{y}}=D \underline{x}\) where \(D\) is a diagonal matrix. We shall find that this new diagonal system of differential equations can be easily solved. This special solution will allow us to obtain the solution of the original system.

Exercise

Obtain the solutions of the pair of first order differential equations

\(\qquad \left.\begin{array}{l}\dot x=[value]x\\ \dot y=[value]y\end{array}\right\}\ \ \)

given the initial conditions

\(\qquad \begin{array}{lll}x(0)=[value]&&\text{ i.e.}\ \ x=[value]\ \ \text{ at}\ \ t=0\\ y(0)=[value]&&\text{ i.e.}\ \ y=[value]\ \ \text{ at}\ \ t=0\end{array}\)

(The notation is that \( \dot x\equiv \dfrac{dx}{dt},\ \ \dot y\equiv \dfrac{dy}{dt}\) )
[Hint: Recall, from your study of differential equations, that the general solution of the differential equation \( \dfrac{dy}{dt}=Ky\) is \(y=y_0 e^{Kt}\).]

\(x=\)
\(y=\)
Feedback: Using the hint: \(x=x_0e^{[value]t}\qquad y=y_0e^{[value]t}\) where \(x_0=x(0)\) and \(y_0=y(0)\). From the given initial condition \(x_0=[value]\qquad y_0=[value]\quad\), so finally \(x=[value]\qquad y=[value]\).

In the previous Exercise although we had two differential equations to solve they were really quite separate. We needed no knowledge of matrix theory to solve them. However, we should note that the two differential equations can be written in matrix form.

Thus if
\(\qquad \underline{x}=\left[\begin{array}{c}x\\ y\end{array}\right]\qquad \dot{ \underline{x}}=\left[\begin{array}{c}\dot x\\ \dot y\end{array}\right]\quad A=\left[\begin{array}{rr}-2&0\\ 0&-5\end{array}\right]\)

the equations
\(\qquad\left.\begin{array}{l}\dot x=-2x\\ \dot y=-5y\end{array}\right\}\)
can be written as
\(\qquad \left[\begin{array}{c}\dot x\\ \dot y\end{array}\right]=\left[\begin{array}{rr}-2&0\\ 0&-5\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]\)
i.e. \(\dot {\underline{x}}=A \underline{x}\).

Exercise

Write in matrix form the pair of coupled differential equations

\(\qquad \left.\begin{array}{l}\dot x=[value]\\ \dot y=[value]\end{array}\right\}\ \ \)

\(\left[\begin{array}{c}\dot x\\ \dot y\end{array}\right]=\) \(\left[\begin{array}{c}x\\ y\end{array}\right]\)
Feedback: \(\qquad\left[\begin{array}{c}\dot x\\ \dot y\end{array}\right]=[value]\left[\begin{array}{c}x\\ y\end{array}\right]\) \(\qquad \ \ \dot{\underline{x}}\ \ \ \ =\ \ \ A\ \underline{x} \qquad\qquad\ \ \ \!\!\!\!\)

The essential difference between the two pairs of differential equations just considered is that the first pair were really separate equations whereas the second pair were coupled:

1. The first equation involving only the unknown \(x\), the second involving only \(y\). In matrix terms this corresponded to a diagonal matrix \(A\) in the system \(\dot {\underline{x}}=A \underline{x}\).
2. The pair were coupled in that both equations involved both \(x\) and \(y\). This corresponded to the non-diagonal matrix \(A\) in the system \(\dot{\underline{x}}=A \underline{x}.\)

Clearly the second system here is more difficult to deal with than the first and this is where we can use our knowledge of diagonalization.

Consider a system of differential equations written in matrix form: \(\dot{\underline{x}}=A \underline{x}\) where
\(\qquad \underline{x}=\left[\begin{array}{c}x(t)\\ y(t)\end{array}\right]\qquad \text{ and}\qquad \dot{\underline{x}}=\left[\begin{array}{c}\dot x (t)\\ \dot y (t)\end{array}\right]\) We now introduce a new column vector of unknowns \(Y=\left[\begin{array}{c}r(t)\\ s(t)\end{array}\right]\) through the relation \( \underline{x}=P\underline{y}\) where \(P\) is the modal matrix of \(A\). Then, since \(P\) is a matrix of constants:
\(\qquad\dot{\underline{x}}= P\dot{\underline{y}}\qquad \text{ so}\quad \dot{\underline{x}}=A\underline{x}\quad \text{ becomes}\quad P\dot{\underline{y}}=A(P \underline{y})\)
Then multiplying by \( P^{-1}\) on the left,
\(\qquad \dot{\underline{y}}= (P^{-1}AP) \underline{y}\)
But, because of the properties of the modal matrix, we know that \(P^{-1}AP\) is a diagonal matrix. Thus if \(\lambda_1,\lambda_2\) are distinct eigenvalues of \(A\) then:
\(\qquad P^{-1}AP=\left[\begin{array}{cc}\lambda_1&0\\ 0&\lambda_2\end{array}\right]\)
Hence \(\dot{\underline{y}}=(P^{-1}AP) \underline{y}\) becomes
\(\qquad \left[\begin{array}{c}\dot r\\ \dot s\end{array}\right]=\left[\begin{array}{cc}\lambda_1&0\\ 0&\lambda_2\end{array}\right]\left[\begin{array}{c}r\\ s\end{array}\right].\)
That is, when written out we have
\(\qquad \dot r = \lambda_1 r\) \(\qquad \dot s = \lambda_1 s\)
These equations are decoupled. The first equation only involves the unknown function \(r(t)\) and has solution \(r(t)=Ce^{\lambda_1t}\). The second equation only involves the unknown function \(s(t)\) and has solution \(s(t)=Ke^{\lambda_2t}\). \([C,K\) are arbitrary constants.]

Once \(r,s\) are known the original unknowns \(x,y\) can be found from the relation \(\underline{x} =P \underline{y}\). Note that the theory outlined above is more widely applicable as specified in the next Key Point:

Key Point

For any system of differential equations of the form \(\dot{\underline{x}} =A \underline{x},\) where \(A\) is an \(n\times n\) matrix with distinct eigenvalues \(\lambda_1,\lambda_2,\dots ,\lambda_n\), and \(t\) is the independent variable, the solution is \[\underline{x}=P \underline{y},\] where \(P\) is the modal matrix of \(A\) and \[Y=[C_1 e^{\lambda_1t},\ C_2 e^{\lambda_2t},\dots ,C_n e^{\lambda_nt}]^T.\]

Example

Find the solution of the coupled differential equations
\(\qquad\dot x =4x+2y\) \(\qquad \dot y =-x+y\)
with initial conditions \(\qquad x(0)=1\qquad y(0)=0\) Here \(\dot x\equiv \dfrac{dx}{dt}\) and \(\dot y\equiv \dfrac{dy}{dt}\).

Solution

Here \(\quad A=\left[\begin{array}{cc}\hfill 4&\hfill 2\\ \hfill -1&\hfill 1\end{array}\right].\) It is easily checked that \(A\) has distinct eigenvalues \(\lambda_1=3\) and \(\lambda_2=2\) and corresponding eigenvectors \(\underline{x_1}=\left[\begin{array}{c}\hfill -2\\ \hfill 1\end{array}\right],\ \underline{x_2}=\left[\begin{array}{c}\hfill 1\\ \hfill -1\end{array}\right]\). Therefore, taking \(\quad P=\left[\begin{array}{cc}\hfill -2&\hfill 1\\ \hfill 1&\hfill -1\end{array}\right]\qquad\text{ then}\qquad P^{-1}AP=\left[\begin{array}{cc}3&0\\ 0&2\end{array}\right]\) and using the previous Key Point,
\(\quad r(t)=Ce^{3t}\qquad s(t)=Ke^{2t},\) so
\[\begin{aligned} \left[\begin{array}{c}x\\ y\end{array}\right]\equiv \underline{x}=P\underline{y}=\left[\begin{array}{cc}\hfill -2&\hfill 1\\ \hfill 1&\hfill -1\end{array}\right]\left[\begin{array}{c}r\\ s\end{array}\right] &=&\left[\begin{array}{cc}\hfill -2&\hfill 1\\ \hfill 1&\hfill -1\end{array}\right]\left[\begin{array}{c}Ce^{3t}\\ Ke^{2t}\end{array}\right]\\ &=&\left[\begin{array}{cc}-2Ce^{3t}+Ke^{2t}\\ Ce^{3t}-Ke^{2t}\end{array}\right].\end{aligned}\] Therefore \(x=-2Ce^{3t}+Ke^{2t}\) and \(y=Ce^{3t}-Ke^{2t}.\) We can now impose the initial conditions \(x(0)=1\) and \(y(0)=0\) to give \[\begin{aligned} 1&=-2C+K\\ 0&=C-K.\end{aligned}\] Thus \(C=K=-1\) and the solution to the original system of differential equations is \[\begin{aligned} x(t)&=2e^{3t}-e^{2t}\\ y(t)&=-e^{3t}+e^{2t}\end{aligned}\]

The approach we have demonstrated in the previous Example can be extended to

1. Systems of first order differential equations with \(n\) unknowns
2. Systems of second order differential equations (described in the next subsection).

The only restriction, as we have said, is that the matrix \(A\) in the system \(\dot{\underline{x}}=A \underline{x}\) has distinct eigenvalues.

Systems of second order differential equations

The decoupling method discussed above can be readily extended to this situation which could arise, for example, in a mechanical system consisting of coupled springs.

A typical example of such a system with two unknowns has the form

\(\qquad \ddot x=ax+by\)
\(\qquad \ddot y= cx+dy\)

or, in matrix form,

\(\qquad \ddot{\underline{x}}=A \underline{x}\) where
\(\qquad \underline{x}=\left[\begin{array}{c}x\\ y\end{array}\right]\qquad A=\left[\begin{array}{cc}a&b\\ c&d\end{array}\right],\qquad \ddot x=\dfrac{d^2x}{dt^2},\quad \ddot y=\dfrac{d^2y}{dt^2}.\)

Exercise

Make the substitution \(\underline{x}=P\underline{y}\) in the equation \(\ddot{\underline{x}}=A \underline{x}\) where \(\underline{y}=\left[\begin{array}{c}r(t)\\ s(t)\end{array}\right]\) and \(P\) is the modal matrix of \(A\). \(A\) is assumed to have distinct eigenvalues \(\lambda_1\) and \(\lambda_2\).
Let \(D=[value]\).

1. What is the result of the substitution?

Now suppose \(\lambda_1\) and \(\lambda_2\) are both negative. You may let \(\lambda_1=-\omega_1^2\),\(\lambda_2=-\omega_2^2\) for \(\omega_1, \omega_2 >0\). Solve the pair of decoupled equations for \(r(t)\) and \(s(t)\) in terms of \(\omega_1, \omega_2\). You can type \(\omega_1\) and \(\omega_2\) as "omega_1" and "omega_2", respectively.
2. \(r(t)=\)
\(s(t)=\)
Feedback: (a) Putting \(\underline{x}=P\underline{y}\) into the second order system \(\ddot{\underline{x}}=A \underline{x}\) gives \(\ddot{\underline{y}} =P^{-1}AP \underline{y}\), that is \(\ddot{\underline{y}}=D\underline{y}\) where \(\quad D=\left[\begin{array}{cc}\lambda_1&0\\ 0&\lambda_2\end{array}\right]\) and \(\ddot{\underline{y}}= \left[\begin{array}{c}\ddot r\\ \ddot s\end{array}\right]\). (b) Hence \(\qquad\left[\begin{array}{c}\ddot r\\ \ddot s\end{array}\right]=\left[\begin{array}{cc}\lambda_1&0\\ 0&\lambda_2\end{array}\right]\left[\begin{array}{c}r\\ s\end{array}\right]\). That is, \(\ddot r=\lambda_1r=-\omega_1^2 r\) and \(\ddot s=\lambda_2 s=-\omega_2^2s\) (where \(\lambda_1\) and \(\lambda_2\) are both negative.) The two decoupled equations are of the form of the differential equation governing simple harmonic motion, hence the general solution is \(r=K\cos\omega_1t+L\sin \omega_1t\) and \(s=M\cos\omega _2t+N\sin \omega_2 t\). The solutions for \(x\) and \(y\) could then be obtained by use of \(\underline{x}=P\underline{y}.\) Note that in this second order case four initial conditions, two each for both \(x\) and \(y\), are required because four constants \(K,L,M,N\) arise in the solution.

Exercise

Solve by decoupling the following first order systems: \(\dfrac{d \underline{x}}{dt}=A\underline{x}\) where \(A=[value], \underline{x}(0)=[value]\).

\(\underline{x}=\)
Feedback: (c) We find that [value] only has two eigenvalues: [value],[value], where the latter is repeated. However, we find a full set of linearly independent eigenvectors \([value]^T,[value]^T,[value]^T\), the latter two corresponding to the eigenvalue [value]. Hence we proceed as normal, substituting \(\underline{x}=[value]\underline{y}\) to get \(\dot{\underline{y}}=[value]\underline{y}\), which is solved by \(\underline{y}=[value]\). Written in terms of \(\underline{x}\), this becomes \(\underline{x}=P\underline{y}=[value][value]=[value]\). To find the specific solution we solve \(\underline{x}(0)=[value]\), which leads to the final solution \(\underline{x}=[value]\).

Exercise

Solve by decoupling the following first order system:

\(\dfrac{d \underline{x}}{dt}=A\underline{x}\) where \(A=[value], \underline{x}(0)=[value]\).

\(\underline{x}=\)
Feedback: We calculate that [value] has eigenvalues [value] and [value], with corresponding eigenvectors \([value]^T\) and \([value]^T\). Hence we make the substitution \(\underline{x}=P\underline{y}\) with \(P=[value]\) to get \(\dot{\underline{y}}=D\underline{y}\), where \(D=[value]\). This is then a decoupled equation, with general solution \(\underline{y}=[value]\). To get the solution in terms of \(\underline{x}\), we use the equation \(\underline{x}=P\underline{y}=[value][value]=[value]\). We then find the particular solution by letting \(t=0\) and solving \(\underline{x}(0)=[value]=[value]\). This is solved by \([value],[value]\), leading to the specific solution \(\underline{x}=[value]\).

Exercise

Solve by decoupling the following first order system:

\(\dot x_1= [value] \qquad \dot x_2= [value] \qquad \dot x_3= [value] \qquad\) with \(x_1(0)=[value], \qquad x_2(0)=[value], \qquad x_3(0)=[value].\)

\(\underline{x}=\)
Feedback: We first translate the system of equations into a single matrix equation: \(\ddot{\underline{x}} =[value]\underline{x}\). The eigenvalues are \([value],[value]\) and \([value]\), the eigenvectors are \([value]^T,[value]^T,[value]^T\) and hence we substitute \(\underline{x}=[value]\underline{y}\) to get \(\dot{\underline{y}}=[value]\underline{y}\). We solve the decoupled equations to get \(\underline{y}=[value]\), then write that in terms of \(\underline{x}\) as \(\underline{x}=[value][value]=[value]\). To find the specific solution we solve \(\underline{x}(0)=[value]\), which leads to the final solution \(\underline{x}=[value]\).

Exercise

Solve by decoupling the following first order systems: \(\dfrac{d \underline{x}}{dt}=A \underline{x}\) where \(A=[value], \underline{x}(0)=[value]\).

\(\underline{x}=\)
Feedback: The eigenvalues are \([value],[value]\) and \([value]\), the eigenvectors are \([value]^T,[value]^T,[value]^T\) and hence we substitute \(\underline{x}=[value]\underline{y}\) to get \(\dot{\underline{y}}=[value]\underline{y}\). We solve the decoupled equations to get \(\underline{y}=[value]\), then write that in terms of \(\underline{x}\) as \(\underline{x}=[value][value]=[value]\). To find the specific solution we solve \(\underline{x}(0)=[value]\), which leads to the final solution \(\underline{x}=[value]\).

Exercise

Matrix methods can be used to solve systems of second order differential equations such as might arise with coupled electrical or mechanical systems. For example the motion of two masses \(m_1\) and \(m_2\) vibrating on coupled springs, neglecting damping and spring masses, is governed by \[\begin{aligned} m_1{\ddot y}_1&=-k_1y_1+k_2(y_2-y_1)\\ m_2{\ddot y}_2&=-k_2(y_2-y_1)\end{aligned}\] where dots denote derivatives with respect to time. Write this system as a matrix equation \({\ddot{\underline{y}}}=A\underline{y}\) and use the decoupling method to find \(\underline{y}\).
1. \(m_1=m_2=1,\ k_1=[value],\ k_2=[value]\) with initial conditions \(y_1(0)=[value], y_2(0)=[value],{\dot y_1}(0)=[value],{\dot y}_2(0)=[value].\)
\(Y=\)
2. \(m_1=m_2=1,\ k_1=[value],\ k_2=[value]\) with initial conditions \(y_1(0)=[value], y_2(0)=[value],{\dot y_1}(0)=[value],{\dot y}_2(0)=[value].\)

\(\underline{y}=\)
Feedback: (a) First we convert the problem into matrix form: \(\ddot{\underline{y}}=A\underline{y}\) where \(A=[value]=[value]\). We find the eigenvalues of \(A\) to be [value] and [value] with corresponding eigenvectors \([value]^T\) and \([value]^T\). Letting \(P=[value]\) we substitute \(\underline{z}=P\underline{y}\) to get \(\ddot{\underline{z}}=D\underline{z}\) where \(D=[value]\). This can be solved by the method of decoupled equations for \(\underline{z}=[value]\). We then convert back to the original system with \(\underline{y}=P\underline{z}=[value][value]=[value]\). To find the specific solution, we solve for \(A,B,C,D\) in \(\underline{y}(0)=[value]\) and \(\dot{\underline{y}}(0)=[value]\) to get \(A=[value],B=[value],C=[value],D=[value]\) and a specific solution of \(\underline{y}=[value]\). (b) Similarly, we start by converting to matrix form: \(\ddot{\underline{y}}=A\underline{y}\) where \(A=[value]\). We find the eigenvalues of \(A\) to be [value] and [value] with corresponding eigenvectors \([value]^T\) and \([value]^T\). Letting \(P=[value]\) we substitute \(\underline{z}=P\underline{y}\) to get \(\ddot{\underline{z}}=D\underline{z}\) where \(D=[value]\). This can be solved by the method of decoupled equations for \(\underline{z}=[value]\). We then convert back to the original system with \(\underline{y}=P\underline{z}=[value][value]=[value]\). To find the specific solution, we solve for \(A,B,C,D\) in \(\underline{y}(0)=[value]\) and \(\dot{\underline{y}}(0)=[value]\) to get \(A=[value],B=[value],C=[value],D=[value]\) and a specific solution of \(\underline{y}=[value]\).