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Flashcards: Matrix Exponentials

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.

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

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

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

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

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Flashcards: Matrix Exponentials

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Explanation:

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 x_2 $\begin{bmatrix} \frac{1}{3}\\ 1 \\ -\frac{1}{3} \end{bmatrix}$ . 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 x_3 $\begin{bmatrix} -1\\ 2 \\ 1 \end{bmatrix}$ . 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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Flashcards: Matrix Exponentials

Flashcards: Matrix Exponentials

Flashcards: Matrix Exponentials

Flashcards: Matrix Exponentials

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Flashcards: Matrix Exponentials

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