Matrix Exponentials - Differential Equations

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Question

Use the definition of matrix exponential,

$e^{At}=I+At+A^2\frac{t^2}{2!}+...+A^k\frac{t^k}{k!}+...=\sum_{k=0}^\infty A^k\frac{t^k}{k!}$

to compute $e^{At}$ of the following matrix.

$A=\begin{pmatrix} 0& 1\ 2&1 \end{pmatrix}$

Answer

Given the matrix,

$A=\begin{pmatrix} 0& 1\ 2&1 \end{pmatrix}$

and using the definition of matrix exponential,

$e^{At}=I+At+A^2\frac{t^2}{2!}+...+A^k\frac{t^k}{k!}+...=\sum_{k=0}^\infty A^k\frac{t^k}{k!}$

calculate

$A^2=AA=\begin{pmatrix} 0&1 \ 2&1 \end{pmatrix} \begin{pmatrix} 0&1 \ 2&1 \end{pmatrix}=\begin{pmatrix} 0& 1\ 4& 1 \end{pmatrix}$

$A^3=A^2A=\begin{pmatrix} 0&1 \ 4&1 \end{pmatrix} \begin{pmatrix} 0&1 \ 2&1 \end{pmatrix}=\begin{pmatrix} 0&1 \ 8&1 \end{pmatrix}$

Therefore

$A^k=\begin{pmatrix} 0&1 \ 2^k&1 \end{pmatrix}; k=1,2,3,...$

$e^{At}=I+\frac{A}{1!}t+\frac{A^2}{2!}t^2+...$

$\begin{pmatrix} 0& 1\ 2&1 \end{pmatrix}+\frac{1}{1!}\begin{pmatrix} 0& 1\ 4&1 \end{pmatrix}+\frac{1}{3!}\begin{pmatrix} 0& 1\ 8&1 \end{pmatrix}$

$e^{At}=\begin{pmatrix} 0&e^t \ e^{2t}&e^t \end{pmatrix}$

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Question

Given the matrix $A =\begin{bmatrix} -1 & -1& -3 \ -6& 1& 0 \ 0& 1 &2 \end{bmatrix}$ , calculate the matrix exponential, $e^{tA}$ . You may leave your answer diagonalized: i.e. it may contain matrices multiplied together and inverted.

Answer

First we find our eigenvalues by finding the characteristic equation, which is the determinant of $(\lambda I - A)$ (or $(A - \lambda I)$ ). $| \lambda I -A| =\begin{vmatrix} \lambda +1 & 1& 3 \ 6& \lambda-1& 0 \ 0& -1 & \lambda -2 \end{vmatrix}$ Expansion down column one yields

$(\lambda + 1)\begin{vmatrix} \lambda - 1& 0 \ -1& \lambda - 2 \end{vmatrix} - 6\begin{vmatrix} 1&3 \ -1& \lambda -2 \end{vmatrix} =(\lambda+1)(\lambda-2)(\lambda-1) - 6(\lambda-2 + 3)$

Simplifying and factoring out a $(\lambda+1)$ , we have

$(\lambda+1)((\lambda -1)(\lambda-2) - 6) = (\lambda+1)(\lambda^2 - 3\lambda - 4) = (\lambda+1)(\lambda - 4)(\lambda + 1)$

So our eigenvalues are $\lambda = -1, 4$

To find the eigenvectors, we find the basis for the null space of $(\lambda I - A)$ for each lambda.

lambda = -1

$\begin{bmatrix} 0 & 1& 3 & \vline & 0 \ 6& -2& 0 & \vline & 0 \ 0& -1 & -3 & \vline & 0 \end{bmatrix}$

Adding row 1 to row 3 and placing into row 3, dividing row two by 6, and swapping rows two and 1 gives us our reduced row echelon form. For our purposes, it suffices just to do the first step and look at the resulting system.

$\begin{bmatrix} 0 & 1& 3 & \vline & 0 \ 6& -2& 0 & \vline & 0 \ 0& 0 & 0 & \vline & 0 \end{bmatrix}$

So that $x_2 + 3x_3 = 0\6x_1 - 2x_2 = 0$

Which has solutions $\begin{bmatrix} \frac{1}{3}\ 1 \ -\frac{1}{3} \end{bmatrix}$ . Thus, a clean eigenvector here would be $\begin{bmatrix} 1\ 3 \ -1 \end{bmatrix}$

For lambda = 4, we have

$\begin{bmatrix} 5 & 1& 3 & \vline & 0 \ 6& 3& 0 & \vline & 0 \ 0& -1 & 2 & \vline & 0 \end{bmatrix}\rightarrow \begin{bmatrix} 5 & 0& 5 & \vline & 0 \ 6& 3& 0 & \vline & 0 \ 0& -1 & 2 & \vline & 0 \end{bmatrix}\rightarrow \begin{bmatrix} 5 & 0& 5 & \vline & 0 \ 6& 0& 6 & \vline & 0 \ 0& -1 & 2 & \vline & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 0& 1 & \vline & 0 \ 0& 1& -2 & \vline & 0 \ 0& 0 & 0 & \vline & 0 \end{bmatrix}$

Step 1: Add row 3 to row 1.

Step 2: Add 3 row 3 to row 2

Step 3: Add -6/5 row 1 in to row 2. Swap and divide as necessary to get proper pivots.

This gives us

$\begin{bmatrix} 1 & 0& 1 & \vline & 0 \ 0& 1& -2 & \vline & 0 \ 0& 0 & 0 & \vline & 0 \end{bmatrix}$

So that $x_1 + x_3 = 0\x_2 - 2x_3 = 0$

Which has solutions $\begin{bmatrix} -1\ 2 \ 1 \end{bmatrix}$ . Thus, a clean eigenvector here would be $\begin{bmatrix} -1\ 2 \ 1 \end{bmatrix}$ .

As we only ended up with two eigenvectors, we'll need to grab a generalized eigenvector as well. To do this, we will solve

$(-I - A)w = \begin{bmatrix} -1\-3 \ 1 \end{bmatrix}$

(for lambda = 1, and we set it equal to the negation of our eigenvector for 1.)

This gives us

$\begin{bmatrix} 0 & 1& 3 & \vline & -1 \ 6& -2& 0 & \vline & -3 \ 0& -1 & -3 & \vline & 1 \end{bmatrix}$

And the steps to solve this are identical to the steps to solving for the eigenvector for -1. Following them once more, and further reducing, we get.

$\begin{bmatrix} 1 & 0& 1 & \vline & -\frac{5}{6} \ 0& 1& 3 & \vline & -1 \ 0& 0 & 0 & \vline & 0 \end{bmatrix}$

Solving the system, our generalized eigenvector is given by $\begin{bmatrix} \frac{5}{6}\ 1 \0 \end{bmatrix}$ . Decomposing into the Jordan matrix gives us

$A =\begin{bmatrix} -1 & -1& \frac{5}{6} \ 2& -3& 1 \ 1& 1 &0 \end{bmatrix} \begin{bmatrix} 4 & 0& 0 \ 0& -1& 1 \ 0& 0 &-1 \end{bmatrix} \begin{bmatrix} -1 & -1& \frac{5}{6} \ 2& -3& 1 \ 1& 1 &0 \end{bmatrix}^{-1}$ ,

When we exponentiate this in the above form, we only need to find the matrix exponential of the Jordan matrix. This is done by exponentiating the entries on the main diagonal, and making the entries on the super diagonals of each Jordan block powers of t over the proper factorials. Thus, the matrix exponential is given by

$e^{tA} =\begin{bmatrix} -1 & -1& \frac{5}{6} \ 2& -3& 1 \ 1& 1 &0 \end{bmatrix} \begin{bmatrix} e^{4t} & 0& 0 \ 0& e^{-t}& \frac{t^1}{1!} \ 0& 0 &e^{-t} \end{bmatrix} \begin{bmatrix} -1 & -1& \frac{5}{6} \ 2& -3& 1 \ 1& 1 &0 \end{bmatrix}^{-1}$

From here, it would just be a matter of inverting and multiplying together -- daunting algebraically, but conceptually quite easy.

Note: In the final form above, anything with the same entries, but the columns switched is okay. I.e., it's okay to have the first eigenvector in the last column of the last two matrices, and $e^{4t}$ be in the lower right hand corner of the second matrix.

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Question

Given the matrix $A =\begin{bmatrix} 3 & 1 \ 1& 3 \end{bmatrix}$ , calculate the matrix exponential, $e^{tA}$ .

Answer

First we find our eigenvalues by finding the characteristic equation, which is the determinant of $(\lambda I - A)$ (or $(A - \lambda I)$ ).

$|\lambda I - A| = \begin{vmatrix} \lambda - 3& -1 \ -1& \lambda - 3 \end{vmatrix} = (\lambda - 3)^2 -1 = \lambda^2 - 6\lambda + 8 = (\lambda - 2)(\lambda - 4)$

Thus, we have eigenvalues of 4 and 2. Solving for the eigenvectors by finding the bases of the eigenspaces, we have

lambda = 4

$\begin{bmatrix} 1&-1 &\vline &0 \ -1& 1&\vline &0 \end{bmatrix}$

Adding Row1 into Row 2, we're left with

$\begin{bmatrix} 1&-1 &\vline &0 \ 0& 0&\vline &0 \end{bmatrix}$

So that

And have an eigenvector of $\begin{bmatrix} 1\1 \end{bmatrix}$ .

For lambda = 2, we have

$\begin{bmatrix} -1&-1 &\vline &0 \ -1& -1&\vline &0 \end{bmatrix}$

Adding -1 Row 1 into Row 2, we have

$\begin{bmatrix} -1 & -1 &\vline &0 \ 0& 0&\vline &0 \end{bmatrix}$

So that

and $\begin{bmatrix} 1\-1 \end{bmatrix}$ is an eigenvector.

Constructing our diagonalized matrix, we have

$P = \begin{bmatrix} 1&1 \ 1&-1 \end{bmatrix}$ $D = \begin{bmatrix} 4&0 \ 0&2 \end{bmatrix}$

Using the formula for calculating the inverses of 2x2 matrices, we have

$P^{-1} =-\frac{1}{2}\begin{bmatrix} -1&-1 \ -1&1 \end{bmatrix}$

To calculate the matrix exponential, we can just find the matrix exponential of and multiply and $P^{-1}$ back in. So $e^{tA} =Pe^{tD}P^{-1}$ .

$e^{tD}$ is just found by taking the entries on the diagonal and exponentiating. Thus, $e^{tA} = \begin{bmatrix} 1&1 \ 1&-1 \end{bmatrix}\begin{bmatrix} e^{4t}&0 \ 0&e^{2t} \end{bmatrix}\begin{bmatrix} \frac{1}{2}&\frac{1}{2} \ \frac{1}{2}&-\frac{1}{2} \end{bmatrix}$

Multiplying together, we get

$e^{tA} =\begin{bmatrix} 1&1 \ 1&-1 \end{bmatrix}\begin{bmatrix} e^{4t}&0 \ 0&e^{2t} \end{bmatrix}\begin{bmatrix} \frac{1}{2}&\frac{1}{2} \ \frac{1}{2}&-\frac{1}{2} \end{bmatrix}= \frac{1}{2} \begin{bmatrix} e^{4t} + e^{2t}&e^{4t} - e^{2t} \ e^{4t} - e^{2t}&e^{4t} + e^{2t} \end{bmatrix}$

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Question

Use the definition of matrix exponential,

$e^{At}=I+At+A^2\frac{t^2}{2!}+...+A^k\frac{t^k}{k!}+...=\sum_{k=0}^\infty A^k\frac{t^k}{k!}$

to compute $e^{At}$ of the following matrix.

$A=\begin{pmatrix} 0& 1\ 2&1 \end{pmatrix}$

Answer

Given the matrix,

$A=\begin{pmatrix} 0& 1\ 2&1 \end{pmatrix}$

and using the definition of matrix exponential,

$e^{At}=I+At+A^2\frac{t^2}{2!}+...+A^k\frac{t^k}{k!}+...=\sum_{k=0}^\infty A^k\frac{t^k}{k!}$

calculate

$A^2=AA=\begin{pmatrix} 0&1 \ 2&1 \end{pmatrix} \begin{pmatrix} 0&1 \ 2&1 \end{pmatrix}=\begin{pmatrix} 0& 1\ 4& 1 \end{pmatrix}$

$A^3=A^2A=\begin{pmatrix} 0&1 \ 4&1 \end{pmatrix} \begin{pmatrix} 0&1 \ 2&1 \end{pmatrix}=\begin{pmatrix} 0&1 \ 8&1 \end{pmatrix}$

Therefore

$A^k=\begin{pmatrix} 0&1 \ 2^k&1 \end{pmatrix}; k=1,2,3,...$

$e^{At}=I+\frac{A}{1!}t+\frac{A^2}{2!}t^2+...$

$\begin{pmatrix} 0& 1\ 2&1 \end{pmatrix}+\frac{1}{1!}\begin{pmatrix} 0& 1\ 4&1 \end{pmatrix}+\frac{1}{3!}\begin{pmatrix} 0& 1\ 8&1 \end{pmatrix}$

$e^{At}=\begin{pmatrix} 0&e^t \ e^{2t}&e^t \end{pmatrix}$

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Question

Calculate the matrix exponential, $e^{tA}$ , for the following matrix: $A = \begin{bmatrix} -7& 3\-1 &-3 \end{bmatrix}$ .

Answer

To get the matrix exponential, we will have to diagonalize the matrix, which requires us to find the eigenvalues and eigenvectors. Thus, we have

$\begin{vmatrix} \lambda+7 &-3 \ 1& \lambda +3 \end{vmatrix} = (\lambda+7)(\lambda+3) +3 = \lambda^2 + 10\lambda +24 = (\lambda + 6)(\lambda + 4) = 0$

Using $\lambda = -4, -6$ , we then find the eigenvectors by solving for the eigenspace.

$\lambda = -4$

$\begin{bmatrix} 3 & -3 &\vline &0 \ 1& -1 &\vline & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & -1 &\vline &0 \ 0& 0 &\vline & 0 \end{bmatrix}$

This has solutions , or $x_1\begin{bmatrix} 1\ 1 \end{bmatrix}$ . So a suitable eigenvector is simply $\begin{bmatrix} 1\1 \end{bmatrix}$ .

Repeating for $\lambda = -6$ ,

$\begin{bmatrix} 1 & -3 &\vline &0 \ 1& -3 &\vline & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & -3 &\vline &0 \ 0& 0 &\vline & 0 \end{bmatrix}$

This has solutions , and thus a suitable eigenvector is $\begin{bmatrix} 3\1 \end{bmatrix}$ .

Thus, we have $P = \begin{bmatrix} 1 & 3\ 1 & 1 \end{bmatrix}$ , $D = \begin{bmatrix} -4 &0 \ 0& -6 \end{bmatrix}$ , and using the inverse formula for 2x2 matrices, $P^{-1} = -\frac{1}{2}\begin{bmatrix} 1 &-3 \ -1 &1 \end{bmatrix}$ . Now we just take the matrix exponential of and multiply the three matrices back together. Thus,

$e^{tA} = Pe^{tD}P^{-1} = -\frac{1}{2}\begin{bmatrix} 1 &3 \ 1& 1 \end{bmatrix}\begin{bmatrix} e^{-4t} &0 \ 0& e^{-6t} \end{bmatrix}\begin{bmatrix} 1 &-3 \ -1&1 \end{bmatrix}$

Multiplying these out yields $e^{tA} = \frac{1}{2}\begin{bmatrix} 3e^{-6t}-e^{-4t} & -3e^{-6t} + 3 e^{-4t} \ e^{-6t}-e^{-4t}&-e^{-6t} + 3e^{-4t} \end{bmatrix}$

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Question

Find the general solution to

$y^{'''} + y^{''} + 3y' - 5y = 0$

Answer

The auxiliary equation is

The roots are

$r = 1, -1 \pm 2i$

Our solution is

$y(x) = C_1e^x + C_2e^{-x}\cos{2x} + C_3e^{-x}\sin{2x}$

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