Method of Matrix Exponential
Solved Problems
Example 1.
Find the general solution of the system, using the matrix exponential:
\[\frac{{dx}}{{dt}} = 2x + 3y,\; \frac{{dy}}{{dt}} = 3x + 2y.\]
Solution.
We solve this system by following the algorithm described above. Calculate the eigenvalues of the matrix \(A:\)
\[\det \left( {A - \lambda I} \right) = \left| {\begin{array}{*{20}{c}}
{2 - \lambda }&3\\
3&{2 - \lambda }
\end{array}} \right| = 0,\;\; \Rightarrow
{\left( {2 - \lambda } \right)^2} - 9 = 0,\;\; \Rightarrow
4 - 4\lambda + {\lambda ^2} - 9 = 0,\;\; \Rightarrow
{\lambda ^2} - 4\lambda - 5 = 0,\;\; \Rightarrow
{\lambda _1} = 5,\;{\lambda _2} = - 1.\]
We find the corresponding eigenvectors for each of the eigenvalues. For the number \({\lambda _1} = 5\) we have:
\[\left[ {\begin{array}{*{20}{c}}
{2 - 5}&3\\
3&{2 - 5}
\end{array}} \right[\left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
{ - 3}&3\\
3&{ - 3}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
3{V_{11}} - 3{V_{21}} = 0,\;\; \Rightarrow
{V_{11}} - {V_{21}} = 0.\]
By setting \({V_{21}} = t,\) we find the eigenvector \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T}:\)
\[{V_{21}} = t,\;\; \Rightarrow {V_{11}} = {V_{21}} = t,\;\; \Rightarrow
{\mathbf{V}_1} = \left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
t\\
t
\end{array}} \right] = t\left[ {\begin{array}{*{20}{c}}
1\\
1
\end{array}} \right]
\sim \left[ {\begin{array}{*{20}{c}}
1\\
1
\end{array}} \right].\]
Similarly, we find the eigenvector \({\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}}} \right)^T}\) associated with the eigenvalue \({\lambda _2} = -1:\)
\[\left[ {\begin{array}{*{20}{c}}
{2 - \left( { - 1} \right)}&3\\
3&{2 - \left( { - 1} \right)}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{12}}}\\
{{V_{22}}}
\end{array}} \right] = 0,\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
3&3\\
3&3
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{12}}}\\
{{V_{22}}}
\end{array}} \right] = 0,\;\; \Rightarrow
3{V_{12}} + 3{V_{22}} = 0,\;\; \Rightarrow
{V_{12}} + {V_{22}} = 0.\]
Let \({V_{22}} = t.\) Then \({V_{12}} = -{V_{22}}= -t.\) Hence,
\[{\mathbf{V}_2} = \left[ {\begin{array}{*{20}{c}}
{{V_{12}}}\\
{{V_{22}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{ - t}\\
t
\end{array}} \right]
= t\left[ {\begin{array}{*{20}{c}}
{ - 1}\\
1
\end{array}} \right]
\sim \left[ {\begin{array}{*{20}{c}}
{ - 1}\\
1
\end{array}} \right].\]
We compose the matrix \(H\) of the found eigenvectors \({\mathbf{V}_1}\) and \({\mathbf{V}_2}:\)
\[H = \left[ {\begin{array}{*{20}{c}}
1&{ - 1}\\
1&1
\end{array}} \right].\]
Next, we compute the inverse matrix \({H^{ - 1}}:\)
\[\Delta \left( H \right) = \left| {\begin{array}{*{20}{c}}
1&{ - 1}\\
1&1
\end{array}} \right| = 1 + 1 = 2,\]
\[{H^{ - 1}} = \frac{1}{{\Delta \left( H \right)}}{\left[ {\begin{array}{*{20}{c}}
{{H_{11}}}&{{H_{12}}}\\
{{H_{21}}}&{{H_{22}}}
\end{array}} \right]^T}
= \frac{1}{2}{\left[ {\begin{array}{*{20}{c}}
1&{ - 1}\\
1&1
\end{array}} \right]^T}
= \frac{1}{2}\left[ {\begin{array}{*{20}{c}}
1&1\\
{ - 1}&1
\end{array}} \right].\]
As in this example the eigenvalues are simple roots of the characteristic equation, we can immediately write down the Jordan form, which will have a simple diagonal form:
\[J = \left[ {\begin{array}{*{20}{c}}
{{\lambda _1}}&0\\
0&{{\lambda _2}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
\color{blue}5&0\\
0&\color{red}{ - 1}
\end{array}} \right].\]
We verify this by using the formula of the transition from the original matrix \(A\) to the Jordan normal form \(J:\)
\[J = {H^{ - 1}}AH
= \frac{1}{2}\left[ {\begin{array}{*{20}{c}}
1&1\\
{ - 1}&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
2&3\\
3&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&{ - 1}\\
1&1
\end{array}} \right]
= \frac{1}{2}\left[ {\begin{array}{*{20}{c}}
5&5\\
1&{ - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&{ - 1}\\
1&1
\end{array}} \right]
= \frac{1}{2}\left[ {\begin{array}{*{20}{c}}
{5 + 5}&{ - 5 + 5}\\
{1 - 1}&{ - 1 - 1}
\end{array}} \right]
= \frac{1}{2}\left[ {\begin{array}{*{20}{c}}
{10}&0\\
0&{ - 2}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
5&0\\
0&{ - 1}
\end{array}} \right] = J.\]
Now we form the matrix \({e^{tJ}}\) (it can also be called the matrix exponential):
\[{e^{tJ}} = \left[ {\begin{array}{*{20}{c}}
{{e^{5t}}}&0\\
0&{{e^{ - t}}}
\end{array}} \right].\]
Compute the matrix exponential \({e^{tA}}:\)
\[{e^{tA}} = H{e^{tJ}}{H^{ - 1}}
= \left[ {\begin{array}{*{20}{c}}
1&{ - 1}\\
1&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{e^{5t}}}&0\\
0&{{e^{ - t}}}
\end{array}} \right] \cdot \frac{1}{2}\left[ {\begin{array}{*{20}{c}}
1&1\\
{ - 1}&1
\end{array}} \right]
= \frac{1}{2}\left[ {\begin{array}{*{20}{c}}
{{e^{5t}}}&{ - {e^{ - t}}}\\
{{e^{5t}}}&{{e^{ - t}}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&1\\
{ - 1}&1
\end{array}} \right]
= \frac{1}{2}\left[ {\begin{array}{*{20}{c}}
{{e^{5t}} + {e^{ - t}}}&{{e^{5t}} - {e^{ - t}}}\\
{{e^{5t}} - {e^{ - t}}}&{{e^{5t}} + {e^{ - t}}}
\end{array}} \right].\]
The general solution of the system can be written as
\[\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
x\\
y
\end{array}} \right]
= {e^{tA}}\left[ {\begin{array}{*{20}{c}}
{{C_1}}\\
{{C_2}}
\end{array}} \right]
= \frac{1}{2}\left[ {\begin{array}{*{20}{c}}
{{e^{5t}} + {e^{ - t}}}&{{e^{5t}} - {e^{ - t}}}\\
{{e^{5t}} - {e^{ - t}}}&{{e^{5t}} + {e^{ - t}}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{C_1}}\\
{{C_2}}
\end{array}} \right],\]
where \({C_1},{C_2}\) are arbitrary numbers.
This answer can also be expressed in another form:
\[\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
x\\
y
\end{array}} \right] = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}
{{C_1}{e^{5t}} + {C_1}{e^{ - t}} + {C_2}{e^{5t}} - {C_2}{e^{ - t}}}\\
{{C_1}{e^{5t}} - {C_1}{e^{ - t}} + {C_2}{e^{5t}} + {C_2}{e^{ - t}}}
\end{array}} \right] = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}
{{e^{5t}}\left( {{C_1} + {C_2}} \right) + {e^{ - t}}\left( {{C_1} - {C_2}} \right)}\\
{{e^{5t}}\left( {{C_1} + {C_2}} \right) - {e^{ - t}}\left( {{C_1} - {C_2}} \right)}
\end{array}} \right] = \frac{1}{2}\left( {{C_1} + {C_2}} \right){e^{5t}}\left[ {\begin{array}{*{20}{c}}
1\\
1
\end{array}} \right] + \frac{1}{2}\left( {{C_1} - {C_2}} \right){e^{ - t}}\left[ {\begin{array}{*{20}{c}}
1\\
{ - 1}
\end{array}} \right] = {B_1}{e^{5t}}\left[ {\begin{array}{*{20}{c}}
1\\
1
\end{array}} \right] + {B_2}{e^{ - t}}\left[ {\begin{array}{*{20}{c}}
1\\
{ - 1}
\end{array}} \right],\]
where \({B_1}\) and \({B_2}\) denote arbitrary constants associated with \({C_1},{C_2}.\)
Example 2.
Solve the system of equations by the method of matrix exponential:
\[\frac{{dx}}{{dt}} = 4x,\; \frac{{dy}}{{dt}} = x + 4y.\]
Solution.
We solve the auxiliary equation and find the eigenvalues:
\[\det \left( {A - \lambda I} \right) = \left| {\begin{array}{*{20}{c}}
{4 - \lambda }&0\\
1&{4 - \lambda }
\end{array}} \right| = 0,\;\; \Rightarrow
{\left( {4 - \lambda } \right)^2} = 0,\;\; \Rightarrow
{\lambda _1} = 4.\]
So, we have one eigenvalue \({\lambda _1} = 4\) of multiplicity \(2.\) Determine the eigenvector \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T}:\)
\[\left[ {\begin{array}{*{20}{c}}
{4 - 4}&0\\
1&{4 - 4}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
0&0\\
1&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
1 \cdot {V_{11}} + 0 \cdot {V_{21}} = 0.\]
It follows that the coordinate \({V_{11}} = 0,\) and the coordinate \({V_{21}}\) can be any number. We choose for simplicity \({V_{21}} = 1.\) Hence, the eigenvector \({\mathbf{V}_1}\) is equal: \({\mathbf{V}_1} = {\left( {0,1} \right)^T}.\)
The second linearly independent vector is defined as the generalized eigenvector \({\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}}} \right)^T},\) connected to \({\mathbf{V}_1}.\) It can be found from the equation
\[\left( {A - {\lambda _1}I} \right){\mathbf{V}_2} = {\mathbf{V}_1},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
0&0\\
1&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{12}}}\\
{{V_{22}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0\\
1
\end{array}} \right],\;\; \Rightarrow
\left\{ {\begin{array}{*{20}{l}}
{0 \cdot {V_{12}} + 0 \cdot {V_{22}} = 0}\\
{1 \cdot {V_{12}} + 0 \cdot {V_{22}} = 1}
\end{array}} \right..\]
Here the coordinate \({{V_{22}}}\) can be any number. We choose \({{V_{22}}} = 0.\) Then we obtain \({{V_{11}}} = 1.\) Thus, the generalized eigenvector is \({\mathbf{V}_2} = {\left( {1,0} \right)^T}.\)
Now, using the basis vectors, we form the matrix \(H\) - the transition matrix from \(A\) to the Jordan canonical form \(J:\)
\[H = \left[ {\begin{array}{*{20}{c}}
0&1\\
1&0
\end{array}} \right].\]
Calculate the inverse matrix \({H^{ - 1}}:\)
\[\Delta \left( H \right) = \left| {\begin{array}{*{20}{c}}
0&1\\
1&0
\end{array}} \right| = 0 - 1 = - 1,\;\;
{H^{ - 1}} = \frac{1}{{\Delta \left( H \right)}}{\left[ {\begin{array}{*{20}{c}}
{{H_{11}}}&{{H_{12}}}\\
{{H_{21}}}&{{H_{22}}}
\end{array}} \right]^T}.\]
Here \({H_{ij}}\) denote the cofactors of the elements of the matrix \(H.\) In the result of the calculations we find:
\[{H^{ - 1}} = \frac{1}{{\left( { - 1} \right)}}{\left[ {\begin{array}{*{20}{c}}
0&{ - 1}\\
{ - 1}&0
\end{array}} \right]^T}
= \left( { - 1} \right)\left[ {\begin{array}{*{20}{c}}
0&{ - 1}\\
{ - 1}&0
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
0&1\\
1&0
\end{array}} \right].\]
Interestingly, in this case the inverse matrix \({H^{ - 1}}\) coincides with the initial matrix \(H.\) Such an effect is possible, if the square of the original matrix is the identity matrix:
\[{H^2} = {\left[ {\begin{array}{*{20}{c}}
0&1\\
1&0
\end{array}} \right]^2}
= \left[ {\begin{array}{*{20}{c}}
0&1\\
1&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0&1\\
1&0
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{0 + 1}&{0 + 0}\\
{0 + 0}&{1 + 0}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right] = I.\]
The Jordan form \(J\) of the matrix \(A\) is
\[J = {H^{ - 1}}AH
= \left[ {\begin{array}{*{20}{c}}
0&1\\
1&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
4&0\\
1&4
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0&1\\
1&0
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{0 + 1}&{0 + 4}\\
{4 + 0}&{0 + 0}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0&1\\
1&0
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
1&4\\
4&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0&1\\
1&0
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{0 + 4}&{1 + 0}\\
{0 + 0}&{4 + 0}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
\color{blue}4&\color{blue}1\\
\color{blue}0&\color{blue}4
\end{array}} \right].\]
We see that the Jordan form \(J\) consists of a single block of size \(2.\)
Compose the matrix \({e^{tJ}}:\)
\[{e^{tJ}} = \left[ {\begin{array}{*{20}{c}}
{{e^{4t}}}&{t{e^{4t}}}\\
0&{{e^{4t}}}
\end{array}} \right] = {e^{4t}}\left[ {\begin{array}{*{20}{c}}
1&t\\
0&1
\end{array}} \right].\]
Calculate the matrix exponential \({e^{tA}}:\)
\[{e^{tA}} = H{e^{tJ}}{H^{ - 1}}
= {e^{4t}}\left[ {\begin{array}{*{20}{c}}
0&1\\
1&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&t\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0&1\\
1&0
\end{array}} \right]
= {e^{4t}}\left[ {\begin{array}{*{20}{c}}
{0 + 0}&{0 + 1}\\
{1 + 0}&{t + 0}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0&1\\
1&0
\end{array}} \right]
= {e^{4t}}\left[ {\begin{array}{*{20}{c}}
0&1\\
1&t
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0&1\\
1&0
\end{array}} \right]
= {e^{4t}}\left[ {\begin{array}{*{20}{c}}
{0 + 1}&{0 + 0}\\
{0 + t}&{1 + 0}
\end{array}} \right]
= {e^{4t}}\left[ {\begin{array}{*{20}{c}}
1&0\\
t&1
\end{array}} \right].\]
The general solution is written as
\[\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
x\\
y
\end{array}} \right] = {e^{tA}}\mathbf{C}
= {e^{4t}}\left[ {\begin{array}{*{20}{c}}
1&0\\
t&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{C_1}}\\
{{C_2}}
\end{array}} \right],\]
where \({C_1}, {C_2}\) are arbitrary constants.
Example 3.
Solve the system of equations using the matrix exponential:
\[\frac{{dx}}{{dt}} = x + y,\;\ \frac{{dy}}{{dt}} = - x + y.\]
Solution.
In this case, the coefficient matrix \(A\) is
\[A = \left[ {\begin{array}{*{20}{c}}
1&1\\
{ - 1}&1
\end{array}} \right].\]
Calculate its eigenvalues:
\[\det \left( {A - \lambda I} \right) = \left| {\begin{array}{*{20}{c}}
{1 - \lambda }&1\\
{ - 1}&{1 - \lambda }
\end{array}} \right| = 0,\;\; \Rightarrow
{\left( {1 - \lambda } \right)^2} + 1 = 0,\;\; \Rightarrow
{\left( {\lambda - 1} \right)^2} = - 1,\;\; \Rightarrow
\lambda - 1 = \pm i,\;\; \Rightarrow
{\lambda _{1,2}} = 1 \pm i.\]
Hence, the matrix \(A\) has a pair of complex conjugate eigenvalues. For each eigenvalue, we find the corresponding eigenvector (it can have complex coordinates).
Let \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T}\) be an eigenvector, associated with the eigenvalue \({\lambda _1} = 1 + i.\) The coordinates of this vector satisfy the following matrix-vector equation:
\[\left[ {\begin{array}{*{20}{c}}
{1 - \left( {1 + i} \right)}&1\\
{ - 1}&{1 - \left( {1 + i} \right)}
\end{array}} \right] \left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \mathbf{0},\;\;\Rightarrow
\left[ {\begin{array}{*{20}{c}}
{ - i}&1\\
{ - 1}&{ - i}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
\left\{ {\begin{array}{*{20}{l}}
{ - i{V_{11}} + {V_{21}} = 0}\\
{ - {V_{11}} - i{V_{21}} = 0}
\end{array}} \right.,\;\; \Rightarrow
\left. {\left\{ {\begin{array}{*{20}{l}}
{{V_{11}} + i{V_{21}} = 0}\\
{i{V_{11}} - {V_{21}} = 0}
\end{array}} \right.\;} \right|\begin{array}{*{20}{l}}
{}\\
\small{{R_2} - i{R_1}}\normalsize
\end{array},\;\; \Rightarrow
\left\{ {\begin{array}{*{20}{l}}
{{V_{11}} + i{V_{21}} = 0}\\
{0 = 0}
\end{array}} \right.,\;\; \Rightarrow
{V_{11}} + i{V_{21}} = 0.\]
We set \({V_{21}} = t.\) Then \({V_{11}} = -it.\) Hence, the eigenvector \({\mathbf{V}_1}\) is given by
\[{\mathbf{V}_1} = \left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{ - it}\\
t
\end{array}} \right]
= t\left[ {\begin{array}{*{20}{c}}
{ - i}\\
1
\end{array}} \right]
\sim \left[ {\begin{array}{*{20}{c}}
{ - i}\\
1
\end{array}} \right].\]
Similarly, we find the eigenvector \({\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}}} \right)^T},\) associated with the number \({\lambda _2} = 1 - i:\)
\[\left[ {\begin{array}{*{20}{c}}
{1 - \left( {1 - i} \right)}&1\\
{ - 1}&{1 - \left( {1 - i} \right)}
\end{array}} \right] \left[ {\begin{array}{*{20}{c}}
{{V_{12}}}\\
{{V_{22}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
{ i}&1\\
{ - 1}&{i}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{12}}}\\
{{V_{22}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
\left\{ {\begin{array}{*{20}{l}}
{ i{V_{12}} + {V_{22}} = 0}\\
{ -{V_{12}} + i{V_{22}} = 0}
\end{array}} \right.,\;\; \Rightarrow
\left. {\left\{ {\begin{array}{*{20}{l}}
{{V_{12}} - i{V_{22}} = 0}\\
{i{V_{12}} + {V_{22}} = 0}
\end{array}} \right.\;} \right|\begin{array}{*{20}{l}}
{}\\
\small{{R_2} - i{R_1}}\normalsize
\end{array},\;\; \Rightarrow
\left\{ {\begin{array}{*{20}{l}}
{{V_{12}} - i{V_{22}} = 0}\\
{0 = 0}
\end{array}} \right.,\;\; \Rightarrow
{V_{12}} - i{V_{22}} = 0.\]
Here we set \({V_{22}} = t.\) Therefore, \({V_{12}} = it.\) The vector \({\mathbf{V}_2}\) is equal to
\[{\mathbf{V}_2} = \left[ {\begin{array}{*{20}{c}}
{{V_{12}}}\\
{{V_{22}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{it}\\
t
\end{array}} \right]
= t\left[ {\begin{array}{*{20}{c}}
{i}\\
1
\end{array}} \right]
\sim \left[ {\begin{array}{*{20}{c}}
{i}\\
1
\end{array}} \right].\]
We form the matrix \(H\) of the found eigenvectors \({\mathbf{V}_1}\) and \({\mathbf{V}_2}:\)
\[H = \left[ {\begin{array}{*{20}{c}}
{ - i}&i\\
1&1
\end{array}} \right].\]
Calculate the inverse matrix \({H^{ - 1}}\) by the formula
\[{H^{ - 1}} = \frac{1}{{\Delta \left( H \right)}}{\left[ {\begin{array}{*{20}{c}}
{{H_{11}}}&{{H_{12}}}\\
{{H_{21}}}&{{H_{22}}}
\end{array}} \right]^T},\]
where \({\Delta \left( H \right)}\) is the determinant of the matrix \(H,\) and \({H_{ij}}\) are cofactors of the elements of the matrix \(H.\) As a result, we obtain:
\[\Delta \left( H \right) = \left| {\begin{array}{*{20}{c}}
{ - i}&i\\
1&1
\end{array}} \right| = - i - i = - 2i,\]
\[{H^{ - 1}} = \frac{1}{{\left( { - 2i} \right)}}{\left[ {\begin{array}{*{20}{c}}
1&{ - 1}\\
{ - i}&{ - i}
\end{array}} \right]^T}
= \frac{1}{{\left( { - 2i} \right)}}\left[ {\begin{array}{*{20}{c}}
1&{ - i}\\
{ - 1}&{ - i}
\end{array}} \right]
= \frac{1}{{2i}}\left[ {\begin{array}{*{20}{c}}
{ - 1}&i\\
1&i
\end{array}} \right].\]
Now we find the Jordan form \(J\) by the formula
\[J = {H^{ - 1}}AH.\]
By performing calculations, we find:
\[J = \frac{1}{{2i}}\left[ {\begin{array}{*{20}{c}}
{ - 1}&i\\
1&i
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&1\\
{ - 1}&1
\end{array}} \right] \left[ {\begin{array}{*{20}{c}}
{ - i}&i\\
1&1
\end{array}} \right]
= \frac{1}{{2i}}\left[ {\begin{array}{*{20}{c}}
{ - 1 - i}&{ - 1 + i}\\
{1 - i}&{1 + i}
\end{array}} \right] \left[ {\begin{array}{*{20}{c}}
{ - i}&i\\
1&1
\end{array}} \right]
= \frac{1}{{2i}}\left[ {\begin{array}{*{20}{c}}
{2i - 2}&0\\
0&{2i + 2}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{\frac{{i - 1}}{i}}&0\\
0&{\frac{{i + 1}}{i}}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
\color{blue}{1 + i}&0\\
0&\color{red}{1 - i}
\end{array}} \right].\]
Generally speaking, we can immediately write the Jordan form \(J,\) which in this case is diagonal (as the eigenvalues \({\lambda _1},\) \({\lambda _2}\) have multiplicity \(1\)). We will assume that the computation is done to test the Jordan form \(J,\) and the matrices \(H\) and \({H^{ - 1}}\) are needed to define the matrix exponential.
Now we form the matrix \({e^{tJ}}:\)
\[{e^{tJ}} = \left[ {\begin{array}{*{20}{c}}
{{e^{\left( {1 + i} \right)t}}}&0\\
0&{{e^{\left( {1 - i} \right)t}}}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{{e^t}{e^{it}}}&0\\
0&{{e^t}{e^{ - it}}}
\end{array}} \right]
= {e^t}\left[ {\begin{array}{*{20}{c}}
{{e^{it}}}&0\\
0&{{e^{ - it}}}
\end{array}} \right].\]
Compute the matrix exponential \({e^{tA}}:\)
\[{e^{tA}} = H{e^{tJ}}{H^{ - 1}}
= \frac{{{e^t}}}{{2i}}\left[ {\begin{array}{*{20}{c}}
{ - i}&i\\
1&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{e^{it}}}&0\\
0&{{e^{ - it}}}
\end{array}} \right] \left[ {\begin{array}{*{20}{c}}
{ - 1}&i\\
1&i
\end{array}} \right].\]
The exponential functions \({{e^{it}}},\) \({{e^{-it}}},\) can be expanded by Euler's formula:
\[{e^{it}} = \cos t + i\sin t,\;\;{e^{ - it}} = \cos t - i\sin t.\]
We get the following result:
\[{e^{tA}} = \frac{{{e^t}}}{{2i}}\left[ {\begin{array}{*{20}{c}}
{ - i}&i\\
1&1
\end{array}} \right] \left[ {\begin{array}{*{20}{c}}
{\cos t + i\sin t}&0\\
0&{\cos t - i\sin t}
\end{array}} \right] \left[ {\begin{array}{*{20}{c}}
{ - 1}&i\\
1&i
\end{array}} \right]
= \frac{{{e^t}}}{{2i}} \left[ {\begin{array}{*{20}{c}}
{ - i\cos t + \sin t}&{i\cos t + \sin t}\\
{\cos t + i\sin t}&{\cos t - i\sin t}
\end{array}} \right] \left[ {\begin{array}{*{20}{c}}
{ - 1}&i\\
1&i
\end{array}} \right]
= \frac{{{e^t}}}{{2i}}\left[ {\begin{array}{*{20}{c}}
{2i\cos t}&{2i\sin t}\\
{ - 2i\sin t}&{2i\cos t}
\end{array}} \right]
= {e^t}\left[ {\begin{array}{*{20}{c}}
{\cos t}&{\sin t}\\
{ - \sin t}&{\cos t}
\end{array}} \right].\]
The general solution of the system of differential equations is given by
\[\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
x\\
y
\end{array}} \right] = {e^{tA}}\mathbf{C}
= {e^t}\left[ {\begin{array}{*{20}{c}}
{\cos t}&{\sin t}\\
{ - \sin t}&{\cos t}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{C_1}}\\
{{C_2}}
\end{array}} \right],\]
where \(C = {\left( {{C_1},{C_2}} \right)^T}\) is an arbitrary vector.
