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Question

Construct the general solution of x'=Ax involving complex eigenfunctions and then obtain the general real solution. Describe the shapes of typical trajectories.

A=[306423112396154]A=\left[ \begin{array}{rrr}{30} & {64} & {23} \\ {-11} & {-23} & {-9} \\ {6} & {15} & {4}\end{array}\right]

Solution

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Answered 9 months ago
Answered 9 months ago
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Let

A=[306423112396154]A=\left[\begin{array}{rrr}{30} & {64} & {23} \\ {-11} & {-23} & {-9} \\ {6} & {15} & {4}\end{array}\right]

Then

(AxI3)=[x30642311x+239615x4](A-xI_3)=\begin{bmatrix} \phantom{-}x-30 & {-64} & {-23} \\ {\phantom{-}11} & {\phantom{-}x+23} & {\phantom{-}9} \\ {-6} & {-15} & {\phantom{-}x-4} \end{bmatrix}

Let characteristic polynomial of AA be p(x)p(x). Then

p(x)= det(xI3A)=x311x2+39x29\begin{align*} p(x) &=\text{ det}(xI_3-A)\\ &=x^3-11x^2+39x-29 \end{align*}

Thus the characteristic polynomial is x311x2+39x29x^3-11x^2+39x-29.

Now x311x2+39x29=(x1)[x(52i)][x(5+2i)]x^3-11x^2+39x-29=(x-1)[x-(5-2i)][x-(5+2i)]. This implies that the roots of p(x)p(x) are 1,52i,5+2i1,5-2i,5+2i. Hence the eigenvalues of AA are 1,52i,5+2i1,5-2i,5+2i.

Now let

X1=[x1x2x3]X_1=\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}

be an eigenvector corresponding to the eigenvalue 5+2i5+2i. Therefore AX1=(5+2i)X1AX_1=(5+2i)X_1 which means [(5+2i)I3A]X1=0[(5+2i)I_3-A]X_1=0. From this equation we get that

[25+2i64231128+2i961512i][x1x2x3]=[000]\begin{bmatrix} -25+2i & {-64} & {-23} \\ {\phantom{-}11} & {\phantom{-}28+2i} & {\phantom{-}9} \\ {-6} & {-15} & {-1-2i} \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}

By Gaussian Elimination, we have x1=2334i3x3x_1=\frac{23-34i}{3}x_3 and x2=9+14i3x3x_2=\frac{-9+14i}{3}x_3. Therefore

[(2334i)r(9+14i)r3r]\begin{bmatrix} (23-34i)r\\ (-9+14i)r\\ 3r \end{bmatrix}

is an eigenvector corresponding to the eigenvalue 5+2i5+2i for any non-zero real number rr. By putting r=1r=1, we get that

v1=[2334i9+14i3]\mathbf{v_1}=\begin{bmatrix} 23-34i\\ -9+14i\\ 3 \end{bmatrix}

is an eigenvector corresponding to the eigenvalue 5+2i5+2i.

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