Opening the book…
We now consider the system \({\bf y}'=A{\bf y}\), where \(A\) has a complex eigenvalue \(\lambda=\alpha+i\beta\) with \(\beta\ne0\). We continue to assume that \(A\) has real entries, so the characteristic polynomial of \(A\) has real coefficients. This implies that \(\overline\lambda=\alpha-i\beta\) is also an eigenvalue of \(A\).
An eigenvector \({\bf x}\) of \(A\) associated with \(\lambda=\alpha+i\beta\) will have complex entries, so we’ll write
\[{\bf x}={\bf u}+i{\bf v} \nonumber\]
where \({\bf u}\) and \({\bf v}\) have real entries; that is, \({\bf u}\) and \({\bf v}\) are the real and imaginary parts of \({\bf x}\). Since \(A{\bf x}=\lambda {\bf x}\),
\[\label{eq:10.6.1} A({\bf u}+i{\bf v})=(\alpha+i\beta)({\bf u}+i{\bf v}).\]
Taking complex conjugates here and recalling that \(A\) has real entries yields
\[A({\bf u}-i{\bf v})=(\alpha-i\beta)({\bf u}-i{\bf v}), \nonumber\]
which shows that \({\bf x}={\bf u}-i{\bf v}\) is an eigenvector associated with \(\overline\lambda=\alpha-i\beta\). The complex conjugate eigenvalues \(\lambda\) and \(\overline\lambda\) can be separately associated with linearly independent solutions \({\bf y}'=A{\bf y}\); however, we will not pursue this approach, since solutions obtained in this way turn out to be complex–valued. Instead, we’ll obtain solutions of \({\bf y}'=A{\bf y}\) in the form
\[\label{eq:10.6.2} {\bf y}=f_1{\bf u}+f_2{\bf v}\]
where \(f_1\) and \(f_2\) are real–valued scalar functions. The next theorem shows how to do this.
Let \(A\) be an \(n\times n\) matrix with real entries\(.\) Let \(\lambda=\alpha+i\beta\) (\(\beta\ne0\)) be a complex eigenvalue of \(A\) and let \({\bf x}={\bf u}+i{\bf v}\) be an associated eigenvector\(,\) where \({\bf u}\) and \({\bf v}\) have real components\(.\) Then \({\bf u}\) and \({\bf v}\) are both nonzero and
\[{\bf y}_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t) \quad \text{and} \quad {\bf y}_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t), \nonumber\]
which are the real and imaginary parts of
\[\label{eq:10.6.3} e^{\alpha t}(\cos\beta t+i\sin\beta t)({\bf u}+i{\bf v}),\]
are linearly independent solutions of \({\bf y}'=A{\bf y}\).
- Proof
-
A function of the form Equation \ref{eq:10.6.2} is a solution of \({\bf y}'=A{\bf y}\) if and only if
\[\label{eq:10.6.4} f_1'{\bf u}+f_2'{\bf v}=f_1A{\bf u}+f_2A{\bf v}.\]
Carrying out the multiplication indicated on the right side of Equation \ref{eq:10.6.1} and collecting the real and imaginary parts of the result yields
\[A({\bf u}+i{\bf v})=(\alpha{\bf u}-\beta{\bf v})+i(\alpha{\bf v}+\beta{\bf u}). \nonumber\]
Equating real and imaginary parts on the two sides of this equation yields
\[\begin{array}{rcl} A{\bf u}=\alpha{\bf u}-\beta{\bf v}\\ A{\bf v}=\alpha{\bf v}+\beta{\bf u}. \end{array}\nonumber\]
We leave it to you to show from this that \({\bf u}\) and \({\bf v}\) are both nonzero. Substituting from these equations into Equation \ref{eq:10.6.4} yields
\[\begin{aligned} f_1'{\bf u}+f_2'{\bf v} &=f_1(\alpha{\bf u}-\beta{\bf v})+f_2(\alpha{\bf v}+\beta{\bf u})\\ &=(\alpha f_1+\beta f_2){\bf u}+(-\beta f_1+\alpha f_2){\bf v}.\end{aligned}\nonumber\]
This is true if
\[\begin{array}{rcr} f_1'=\alpha f_1+\beta f_2\phantom{,}\\ f_2'=-\beta f_1+\alpha f_2, \end{array} \quad \text{or equivalently} \quad \begin{array}{rcr} f_1'-\alpha f_1=\phantom{-}\beta f_2\phantom{.}\\ f_2'-\alpha f_2=-\beta f_1. \end{array}\nonumber\]
If we let \(f_1=g_1e^{\alpha t}\) and \(f_2=g_2e^{\alpha t}\), where \(g_1\) and \(g_2\) are to be determined, then the last two equations become
\[\begin{array}{rcr} g_1'=\beta g_2\phantom{.}\\ g_2'=-\beta g_1, \end{array}\nonumber\]
which implies that
\[g_1''=\beta g_2'=-\beta^2 g_1, \nonumber\]
so
\[g_1''+\beta^2 g_1=0. \nonumber\]
The general solution of this equation is
\[g_1=c_1\cos\beta t+c_2\sin\beta t. \nonumber\]
Moreover, since \(g_2=g_1'/\beta\),
\[g_2=-c_1\sin\beta t+c_2\cos\beta t. \nonumber\]
Multiplying \(g_1\) and \(g_2\) by \(e^{\alpha t}\) shows that
\[\begin{aligned} f_1&=e^{\alpha t}(\phantom{-}c_1\cos\beta t+c_2\sin\beta t ),\\ f_2&=e^{\alpha t}(-c_1\sin\beta t+c_2\cos\beta t).\end{aligned}\nonumber\]
Substituting these into Equation \ref{eq:10.6.2} shows that
\[\label{eq:10.6.5} \begin{array}{rcl} {\bf y}&=&e^{\alpha t}\left[(c_1\cos\beta t+c_2\sin\beta t){\bf u} +(-c_1\sin\beta t+c_2\cos\beta t){\bf v}\right]\\ &=&c_1e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t) +c_2e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t) \end{array}\]
is a solution of \({\bf y}'=A{\bf y}\) for any choice of the constants \(c_1\) and \(c_2\). In particular, by first taking \(c_1=1\) and \(c_2=0\) and then taking \(c_1=0\) and \(c_2=1\), we see that \({\bf y}_1\) and \({\bf y}_2\) are solutions of \({\bf y}'=A{\bf y}\).
Find the general solution of
\[\label{eq:10.6.6} {\bf y}'=\left[\begin{array}{cc}{4}&{-5}\\{5}&{-2}\end{array}\right]{\bf y}.\]
Solution
The characteristic polynomial of the coefficient matrix \(A\) in Equation \ref{eq:10.6.6} is
\[\left|\begin{array}{cc} 4-\lambda&-5\\ 5&-2-\lambda \end{array}\right|=(\lambda-1)^2+16. \nonumber\]
Hence, \(\lambda=1+4i\) is an eigenvalue of \(A\). The associated eigenvectors satisfy \(\left(A-\left(1+4i\right)I\right){\bf x}={\bf 0}\). The augmented matrix of this system is
\[\left[\begin{array}{cccr} 3-4i&-5&\vdots&0\\ 5&-3-4i&\vdots&0 \end{array}\right], \nonumber\]
which is row equivalent to
\[\left[\begin{array}{cccr} 1&-{3+4i\over5}&\vdots&0\\ 0&0&\vdots&0 \end{array}\right]. \nonumber\]
Therefore \(x_1=(3+4i)x_2/5\). Taking \(x_2=5\) yields \(x_1=3+4i\), so
\[{\bf x}=\left[\begin{array}{c}3+4i\\5\end{array}\right] \nonumber\]
is an eigenvector. The real and imaginary parts of
\[e^t(\cos4t+i\sin4t)\left[\begin{array}{c}3+4i\\5\end{array}\right] \nonumber\]
are
\[{\bf y}_1=e^t\left[\begin{array}{c}3\cos4t-4\sin 4t\\5\cos4t\end{array}\right]\quad\text{ and }\quad {\bf y}_2=e^t\left[\begin{array}{c}3\sin4t+4\cos4t\\5\sin 4t\end{array}\right], \nonumber\]
which are linearly independent solutions of Equation \ref{eq:10.6.6}. The general solution of Equation \ref{eq:10.6.6} is
\[{\bf y}= c_1e^t\left[\begin{array}{c}3\cos4t-4\sin 4t\\5\cos4t\end{array}\right]+ c_2e^t\left[\begin{array}{c}3\sin4t+4\cos4t\\5\sin 4t\end{array}\right]. \nonumber\]
Find the general solution of
\[\label{eq:10.6.7} {\bf y}'=\left[\begin{array}{cc}{-14}&{39}\\{-6}&{16}\end{array}\right]{\bf y}.\]
Solution
The characteristic polynomial of the coefficient matrix \(A\) in Equation \ref{eq:10.6.7} is
\[\left|\begin{array}{cc}-14-\lambda&39\\-6&16-\lambda \end{array}\right|=(\lambda-1)^2+9. \nonumber\]
Hence, \(\lambda=1+3i\) is an eigenvalue of \(A\). The associated eigenvectors satisfy \(\left(A-(1+3i)I\right){\bf x}={\bf 0}\). The augmented augmented matrix of this system is
\[\left[\begin{array}{cccr}-15-3i&39&\vdots&0\\ -6&15-3i&\vdots&0 \end{array}\right], \nonumber\]
which is row equivalent to
\[\left[\begin{array}{cccr} 1&{-5+i\over2}&\vdots&0\\ 0&0&\vdots&0 \end{array}\right]. \nonumber\]
Therefore \(x_1=(5-i)/2\). Taking \(x_2=2\) yields \(x_1=5-i\), so
\[{\bf x}=\left[\begin{array}{c}5-i\\2\end{array}\right] \nonumber\]
is an eigenvector. The real and imaginary parts of
\[e^t(\cos3t+i\sin3t)\left[\begin{array}{c}5-i\\2\end{array}\right] \nonumber\]
are
\[{\bf y}_1=e^t\left[\begin{array}{c}\sin3t+5\cos3t\\2\cos 3t\end{array}\right]\quad\text{ and }\quad {\bf y}_2=e^t\left[\begin{array}{c}-\cos3t+5\sin3t\\2\sin 3t\end{array}\right], \nonumber\]
which are linearly independent solutions of Equation \ref{eq:10.6.7}. The general solution of Equation \ref{eq:10.6.7} is
\[{\bf y}=c_1e^t\left[\begin{array}{c}\sin3t+5\cos3t\\2\cos 3t\end{array}\right]+ c_2e^t\left[\begin{array}{c}-\cos3t+5\sin3t\\2\sin 3t\end{array}\right]. \nonumber\]
Find the general solution of
\[\label{eq:10.6.8} {\bf y}'=\left[\begin{array}{ccc}{-5}&{5}&{4}\\{-8}&{7}&{6}\\{1}&{0}&{0}\end{array}\right]{\bf y}.\]
Solution
The characteristic polynomial of the coefficient matrix \(A\) in Equation \ref{eq:10.6.8} is
\[\left|\begin{array}{ccc}-5-\lambda&5&4\\-8&7-\lambda& 6\\ \phantom{-}1 &0&-\lambda\end{array}\right|=-(\lambda-2)(\lambda^2+1). \nonumber\]
Hence, the eigenvalues of \(A\) are \(\lambda_1=2\), \(\lambda_2=i\), and \(\lambda_3=-i\). The augmented matrix of \((A-2I){\bf x=0}\) is
\[\left[\begin{array}{rrrcr}-7&5&4&\vdots&0\\-8& 5&6&\vdots&0\\ 1&0&-2&\vdots&0 \end{array}\right], \nonumber\]
which is row equivalent to
\[\left[\begin{array}{rrrcr} 1&0&-2&\vdots&0\\ 0&1&-2& \vdots&0\\ 0&0&0&\vdots&0\end{array}\right]. \nonumber\]
Therefore \(x_1=x_2=2x_3\). Taking \(x_3=1\) yields
\[{\bf x}_1=\threecol221, \nonumber\]
so
\[{\bf y}_1=\threecol221e^{2t} \nonumber\]
is a solution of Equation \ref{eq:10.6.8}.
The augmented matrix of \((A-iI){\bf x=0}\) is
\[\left[\begin{array}{ccrccc}-5-i&5&4&\vdots&0\\-8& 7-i&6&\vdots&0\\ \phantom{-}1&0&-i&\vdots&0 \end{array}\right], \nonumber\]
which is row equivalent to
\[\left[\begin{array}{ccccc} 1&0&-i&\vdots&0\\ 0&1&1-i& \vdots&0\\ 0&0&0&\vdots&0\end{array}\right]. \nonumber\]
Therefore \(x_1=ix_3\) and \(x_2=-(1-i)x_3\). Taking \(x_3=1\) yields the eigenvector
\[{\bf x}_2=\left[\begin{array}{c} i\\-1+i\\ 1\end{array} \right]. \nonumber\]
The real and imaginary parts of
\[(\cos t+i\sin t)\left[\begin{array}{c}i\\-1+i\\1\end{array}\right] \nonumber\]
are
\[{\bf y}_2= \left[\begin{array}{c}-\sin t\\-\cos t-\sin t\\\cos t\end{array}\right] \quad\text{ and }\quad {\bf y}_3=\left[\begin{array}{c}\cos t\\\cos t-\sin t\\\sin t\end{array}\right], \nonumber\]
which are solutions of Equation \ref{eq:10.6.8}. Since the Wronskian of \(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) at \(t=0\) is
\[\left|\begin{array}{rrr} 2&0&1\\2&-1&1\\1&1&0\end{array}\right|=1, \nonumber\]
\(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) is a fundamental set of solutions of Equation \ref{eq:10.6.8}. The general solution of Equation \ref{eq:10.6.8} is
\[{\bf y}=c_1 \threecol221e^{2t} +c_2\left[\begin{array}{c}-\sin t\\-\cos t-\sin t\\\cos t\end{array}\right] +c_3\left[\begin{array}{c}\cos t\\\cos t-\sin t\\\sin t\end{array}\right]. \nonumber\]
Find the general solution of
\[\label{eq:10.6.9} {\bf y}'=\left[\begin{array}{ccc}{1}&{-1}&{-2}\\{1}&{3}&{2}\\{1}&{-1}&{2}\end{array}\right]{\bf y}.\]
Solution
The characteristic polynomial of the coefficient matrix \(A\) in Equation \ref{eq:10.6.9} is
\[\left|\begin{array}{ccc} 1-\lambda&-1&-2\\ 1&3-\lambda& \phantom{-}2\\ 1 &-1&2-\lambda\end{array}\right|= -(\lambda-2)\left((\lambda-2)^2+4\right). \nonumber\]
Hence, the eigenvalues of \(A\) are \(\lambda_1=2\), \(\lambda_2=2+2i\), and \(\lambda_3=2-2i\). The augmented matrix of \((A-2I){\bf x=0}\) is
\[\left[\begin{array}{rrrcr}-1&-1&-2&\vdots&0\\1& 1&2&\vdots&0\\ 1&-1&0&\vdots&0 \end{array}\right], \nonumber\]
which is row equivalent to
\[\left[\begin{array}{rrrcr} 1&0&1&\vdots&0\\ 0&1&1& \vdots&0\\ 0&0&0&\vdots&0\end{array}\right]. \nonumber\]
Therefore \(x_1=x_2=-x_3\). Taking \(x_3=1\) yields
\[{\bf x}_1=\threecol{-1}{-1}1, \nonumber\]
so
\[{\bf y}_1=\threecol{-1}{-1}1e^{2t} \nonumber\]
is a solution of Equation \ref{eq:10.6.9}. The augmented matrix of \(\left(A-(2+2i)I\right){\bf x=0}\) is
\[\left[\begin{array}{ccrcc}-1-2i&-1&-2&\vdots&0\\ 1& 1-2i&\phantom{-}2&\vdots&0\\ 1&-1&-2i&\vdots&0 \end{array}\right], \nonumber\]
which is row equivalent to
\[\left[\begin{array}{rrrcr} 1&0&-i&\vdots&0\\ 0&1&i& \vdots&0\\ 0&0&0&\vdots&0\end{array}\right]. \nonumber\]
Therefore \(x_1=ix_3\) and \(x_2=-ix_3\). Taking \(x_3=1\) yields the eigenvector
\[{\bf x}_2=\threecol i{-i}1 \nonumber\]
The real and imaginary parts of
\[e^{2t}(\cos2t+i\sin2t)\threecol i{-i}1 \nonumber\]
are
\[{\bf y}_2=e^{2t}\left[\begin{array}{r}-\sin2t\\\sin2t\\\cos 2t\end{array}\right]\quad\text{ and }\quad {\bf y}_2=e^{2t}\left[\begin{array}{r}\cos2t\\-\cos2t\\\sin 2t\end{array}\right], \nonumber\]
which are solutions of Equation \ref{eq:10.6.9}. Since the Wronskian of \(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) at \(t=0\) is
\[\left|\begin{array}{rrr} -1&0&1\\-1&0&-1\\1&1&0\end{array}\right|=-2, \nonumber\]
\(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) is a fundamental set of solutions of Equation \ref{eq:10.6.9}. The general solution of Equation \ref{eq:10.6.9} is
\[{\bf y}=c_1\threecol{-1}{-1}1e^{2t}+ c_2e^{2t}\left[\begin{array}{r}-\sin2t\\\sin2t\\\cos 2t\end{array}\right]+ c_3e^{2t}\left[\begin{array}{r}\cos2t\\-\cos2t\\\sin 2t\end{array}\right]. \nonumber\]
This page titled 10.5: Constant Coefficient Homogeneous Systems III was last modified on Sat, 19 Nov 2022 22:30:40 GMT and is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform.
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