Banguizb

Answered

2022-07-07

Solve the following system of equations

$\{\begin{array}{l}{x}_{1}^{{}^{\prime}}(t)={x}_{1}(t)+3{x}_{2}(t)\\ {x}_{2}^{{}^{\prime}}(t)=3{x}_{1}(t)-2{x}_{2}(t)-{x}_{3}(t)\\ {x}_{3}^{{}^{\prime}}=-{x}_{2}(t)+{x}_{3}(t)\end{array}$

First, I create the column vectors $X$ and ${X}^{{}^{\prime}}$. Then the matrix

$A=\left[\begin{array}{ccc}1& 3& 0\\ 3& -2& -1\\ 0& -1& 1\end{array}\right]$

Now, I find the eigenvalues, $-4,3,1$ and their corresponding eigenvectors $(-3,5,1{)}^{T}(-3,-2,1{)}^{T}(1,0,3{)}^{T}$.

Then, what?

$\{\begin{array}{l}{x}_{1}^{{}^{\prime}}(t)={x}_{1}(t)+3{x}_{2}(t)\\ {x}_{2}^{{}^{\prime}}(t)=3{x}_{1}(t)-2{x}_{2}(t)-{x}_{3}(t)\\ {x}_{3}^{{}^{\prime}}=-{x}_{2}(t)+{x}_{3}(t)\end{array}$

First, I create the column vectors $X$ and ${X}^{{}^{\prime}}$. Then the matrix

$A=\left[\begin{array}{ccc}1& 3& 0\\ 3& -2& -1\\ 0& -1& 1\end{array}\right]$

Now, I find the eigenvalues, $-4,3,1$ and their corresponding eigenvectors $(-3,5,1{)}^{T}(-3,-2,1{)}^{T}(1,0,3{)}^{T}$.

Then, what?

Answer & Explanation

Oliver Shepherd

Expert

2022-07-08Added 24 answers

The matrix $A$ that you have is symmetric. So it has an orthonormal basis of eigenvectors. The eigenvectors you have found are mutually orthogonal (which they must be because they correspond to different eigenvalues.) So, if you normalize your eigenvectors and make those normalized vectors the columms of a matrix $U$, then ${U}^{T}U=I$ is automatic, and ${U}^{T}AU=D$ is diagonal. Explicitly,

$U=\left[\begin{array}{ccc}-\frac{3}{\sqrt{35}}& -\frac{3}{\sqrt{14}}& \frac{1}{\sqrt{10}}\\ \frac{5}{\sqrt{35}}& -\frac{2}{\sqrt{14}}& 0\\ \frac{1}{\sqrt{35}}& \frac{1}{\sqrt{14}}& \frac{3}{\sqrt{10}}\end{array}\right]$

The inverse of $U$ is the transpose ${U}^{T}$ of $U$. And,

${U}^{T}AU=\left[\begin{array}{ccc}-4& 0& 0\\ 0& 3& 0\\ 0& 0& 1\end{array}\right]=D.$

Equivalently,

$A=U\left[\begin{array}{ccc}-4& 0& 0\\ 0& 3& 0\\ 0& 0& 1\end{array}\right]{U}^{T}=UD{U}^{T}.$

The general solution is expressed in terms of $a={x}_{1}(0)$, $b={x}_{2}(0)$, $c={x}_{3}(0)$ as

$\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]={e}^{tA}\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=U{e}^{tD}{U}^{T}\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=U\left[\begin{array}{ccc}{e}^{-4t}& 0& 0\\ 0& {e}^{3t}& 0\\ 0& 0& {e}^{t}\end{array}\right]{U}^{T}\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$

$U=\left[\begin{array}{ccc}-\frac{3}{\sqrt{35}}& -\frac{3}{\sqrt{14}}& \frac{1}{\sqrt{10}}\\ \frac{5}{\sqrt{35}}& -\frac{2}{\sqrt{14}}& 0\\ \frac{1}{\sqrt{35}}& \frac{1}{\sqrt{14}}& \frac{3}{\sqrt{10}}\end{array}\right]$

The inverse of $U$ is the transpose ${U}^{T}$ of $U$. And,

${U}^{T}AU=\left[\begin{array}{ccc}-4& 0& 0\\ 0& 3& 0\\ 0& 0& 1\end{array}\right]=D.$

Equivalently,

$A=U\left[\begin{array}{ccc}-4& 0& 0\\ 0& 3& 0\\ 0& 0& 1\end{array}\right]{U}^{T}=UD{U}^{T}.$

The general solution is expressed in terms of $a={x}_{1}(0)$, $b={x}_{2}(0)$, $c={x}_{3}(0)$ as

$\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]={e}^{tA}\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=U{e}^{tD}{U}^{T}\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=U\left[\begin{array}{ccc}{e}^{-4t}& 0& 0\\ 0& {e}^{3t}& 0\\ 0& 0& {e}^{t}\end{array}\right]{U}^{T}\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$

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