Part 1: Euler’s Number Raised to a Matrix | Gustafson, G. B. (n.d.). Differential equations and linear algebra

Matrix Exponential — Jordan Form, Part 1

Book Source:

Gustafson, G. B. (n.d.). Differential equations and linear algebra: A course for science and engineering (Vols. I–II). University of Utah.

https://math.utah.edu/~gustafso/debook/deBookGG.pdf

QUESTION 11.4.15

$$ e^{\begin{pmatrix}1&1\\0&2\end{pmatrix}t} = \begin{pmatrix} e^t & e^{2t}-e^t\\ 0 & e^{2t} \end{pmatrix} $$

ANSWER [There do not appear to be any answers in this text.]

Matrix Exponential Identities. Verify from exponential identities.

$\mathcal{F}_1$ [Formulae] First, we find a necessary formula. After reviewing the book, we find the following formula. We will start with this, but there are many more.

$$ e^{At}=\sum_{n=0}^{\infty}A^n\frac{t^n}{n!} $$

$\mathcal{F}_2$ [Formulate] Whenever verifying or showing, you often start with the left side and move through the steps until it equals the right side.

$$ e^{{\color{#12345a}{\begin{pmatrix}1&1\\0&2\end{pmatrix}}}t} = \sum_{n=0}^{\infty} {\color{#12345a}{\begin{pmatrix}1&1\\0&2\end{pmatrix}}}^{n} \frac{t^n}{n!} $$

At this point, what do we do? There could be a simple linear algebra trick, but likely, we need to diagonalize. However, we must reference what we are allowed to use in the book. Let us check it again.

In this text, the “Jordan Form” appears to match our situation.

$$ e^{{\color{#12345a}A}t}=P^{-1}e^{Jt}P \quad\leftrightarrow\quad J=P{\color{#12345a}A}P^{-1} $$

NOTE To the untrained student, you may look at this and think it is diagonalization. However, on page 721, you will find that it is referring to a triangular matrix, which we have. Recognize that it is upper triangular.

Pg. 721: “Because the Jordan form is a triangular matrix, then $T=J$ gives an algorithm for generation of columns in matrix $P$.”

From this goose chase, we see that the previous formula is $J=P{\color{#12345a}A}P^{-1}=T$. Then,

$$ e^{{\color{#12345a}A}t}=P^{-1}e^{Jt}P=P^{-1}e^{Tt}P $$

From Theorem 9.15 (Jordan’s Extension), $T$ is the diagonal elements. Then,

$$ A=\begin{pmatrix}1&1\\0&2\end{pmatrix} \quad\Rightarrow\quad T=\begin{bmatrix}1&0\\0&2\end{bmatrix}. $$

MISTAKE ↓ We have two $P$’s …

Now, if $A=PTP^{-1}$, we do some basic linear algebra arithmetic.

$$ \begin{aligned} A=PTP^{-1} &\Rightarrow \left[A=P(PT^{-1})^{-1}\right]\\ &\Rightarrow \left[A=P(P{\color{#326b9e}T^{-1}})^{-1}\right] (P{\color{#326b9e}T^{-1}})\\[6pt] &\Rightarrow A(P{\color{#326b9e}T^{-1}}) = P(P{\color{#326b9e}T^{-1}})^{-1} (P{\color{#326b9e}T^{-1}})\\ &\Rightarrow A(P{\color{#326b9e}T^{-1}})=PI\\[6pt] &\Rightarrow A(P{\color{#326b9e}T^{-1}})=P \end{aligned} $$

END MISTAKE

Backtracking: Pg. 721 explains that setting $T=J$ gives an algorithm for generating the columns of $P$.

Now, the book appears to be pointing to using a software program—as this is for engineers. Page 721 discusses finding $J$ and $P$ with computer algebra systems and directs us to page 898 for a Maple example.

Before I go forward, I would like to know what Maple is. I first consult the book, then use DuckDuckGo to look it up. Maple is a mathematical software package with symbolic computation capabilities.

This textbook appears to want us to use Maple to find $P$. The following example is transcribed from page 898.

Jordan Decomposition Using Maple

The matrix

$$ A= \begin{pmatrix} 4&-2&5\\ -2&4&-3\\ 0&0&2 \end{pmatrix} $$

has a Jordan decomposition

$$ A=PJP^{-1} = \begin{pmatrix} 1&4&-7\\ -1&4&1\\ 0&0&4 \end{pmatrix} \begin{pmatrix} 6&0&0\\ 0&2&1\\ 0&0&2 \end{pmatrix} \begin{pmatrix} \frac12&-\frac12&1\\ \frac18&\frac18&\frac{3}{16}\\ 0&0&\frac14 \end{pmatrix}. $$

# Maple: Find the Jordan form of matrix A
A := Matrix([[4, -2, 5], [-2, 4, -3], [0, 0, 2]]);
factor(LinearAlgebra[CharacteristicPolynomial](A, lambda));
# Answer: (lambda-6)*(lambda-2)^2
J, P := LinearAlgebra[JordanForm](A, output=['J', 'Q']);
zero := A.P-P.J; # Zero matrix expected

Now, this is a special case where we would use A.I., unless otherwise specified by your professor to use an alternative software. Since I do not have the requested software available, I will ask ChatGPT to use the example with our matrix $A$ and output a transformation matrix $P$ for us.

TIME TRAVELING MATH

From ChatGPT, we got the following Maple-style output:

Using $A=\begin{pmatrix}1&1\\0&2\end{pmatrix}$, a transformation matrix is

Plain text

     [1    1]
P := [      ]
     [0    1]

Its columns are eigenvectors for 1 and 2, respectively, giving $AP=PJ$ with $J=\operatorname{diag}(1,2)$.

Now, we have $P=\begin{bmatrix}1&1\\0&1\end{bmatrix}$. Since this is a $2\times2$ matrix, we can compute the inverse directly.

MISTAKE BEGINS ↓

$$ A=\begin{bmatrix}a&b\\c&d\end{bmatrix} \quad\Rightarrow\quad A^{-1} = \frac{\operatorname{adj}(A)}{\det(A)} = {\color{#12345a}{ \frac{\begin{bmatrix}1&0\\-1&1\end{bmatrix}} {(1)(1)-(1)(0)} = \begin{bmatrix}1&0\\-1&1\end{bmatrix} \quad\text{WRONG.} }} $$

Therefore,

$$ A=\begin{pmatrix}1&1\\0&2\end{pmatrix} \quad\Rightarrow\quad PTP^{-1} = \begin{bmatrix}1&1\\0&1\end{bmatrix} \begin{bmatrix}1&0\\0&2\end{bmatrix} \begin{bmatrix}1&0\\-1&1\end{bmatrix}. $$

Now, we continue:

$$ \begin{aligned} e^{{\color{#12345a}A}t} &=P^{-1}e^{Jt}P=P^{-1}e^{Tt}P\\ &\Rightarrow P^{-1}e^{Tt}P = \begin{bmatrix}1&0\\-1&1\end{bmatrix} e^{\begin{bmatrix}1&0\\0&2\end{bmatrix}t} \begin{bmatrix}1&1\\0&1\end{bmatrix}. \end{aligned} $$

We extract

$$ e^{\begin{bmatrix}1&0\\0&2\end{bmatrix}t} $$

and simplify. We have the provided formulae, so we can work through this.

$$ \begin{aligned} e^{\begin{bmatrix}1&0\\0&2\end{bmatrix}t} &= \sum_{n=0}^{\infty} \begin{bmatrix}1&0\\0&2\end{bmatrix}^{n} \frac{t^n}{n!}\\[6pt] &= \sum_{n=0}^{\infty} \begin{bmatrix}1^n&0\\0&2^n\end{bmatrix} \frac{t^n}{n!}\\[6pt] &= \sum_{n=0}^{\infty} \begin{bmatrix} 1^n\frac{t^n}{n!}&0\\ 0&2^n\frac{t^n}{n!} \end{bmatrix}\\[6pt] &= \begin{bmatrix} \displaystyle\sum_{n=0}^{\infty}1^n\frac{t^n}{n!}&0\\ 0&\displaystyle\sum_{n=0}^{\infty}2^n\frac{t^n}{n!} \end{bmatrix}\\[6pt] &= \begin{bmatrix}e^t&0\\0&e^{2t}\end{bmatrix}. \end{aligned} $$

Next,

$$ \begin{aligned} &\begin{bmatrix}1&0\\-1&1\end{bmatrix} e^{\begin{bmatrix}1&0\\0&2\end{bmatrix}t} \begin{bmatrix}1&1\\0&1\end{bmatrix}\\[6pt] &= \begin{bmatrix}1&0\\-1&1\end{bmatrix} \begin{bmatrix}e^t&0\\0&e^{2t}\end{bmatrix} \begin{bmatrix}1&1\\0&1\end{bmatrix}\\[6pt] &= \begin{bmatrix}e^t&e^{2t}\\-e^t&e^{2t}\end{bmatrix} \begin{bmatrix}1&1\\0&1\end{bmatrix}\\[6pt] &= \begin{bmatrix} e^t&e^t+e^{2t}\\ -e^t&-e^t+e^{2t} \end{bmatrix}. \end{aligned} $$

MISTAKE ↑ or TYPO

$$ \begin{pmatrix}1&0\\-1&1\end{pmatrix} \begin{pmatrix}e^t&0\\0&e^{2t}\end{pmatrix} \begin{pmatrix}1&1\\0&1\end{pmatrix} = \boxed{ \begin{pmatrix} e^t&e^t\\ -e^t&e^{2t}-e^t \end{pmatrix} } $$

I secondly asked ChatGPT WITHOUT ASKING FOR A SOLUTION.

NOTE: I want to point out that this particular Jordan decomposition is also a diagonalization, because its Jordan form is diagonal. I had ChatGPT diagonalize $A$.

$$ A=PJP^{-1} = \begin{bmatrix}1&1\\0&1\end{bmatrix} \begin{bmatrix}1&0\\0&2\end{bmatrix} \begin{bmatrix}1&-1\\0&1\end{bmatrix} $$

***SO FREAKING IMPORTANT***

What I did was not write out the formula for the inverse. I did it in my head.

$$ A=\begin{bmatrix}a&b\\c&d\end{bmatrix} \quad\Rightarrow\quad A^{-1} = \frac{\operatorname{adj}(A)}{\det(A)} = \frac{\begin{bmatrix}d&-b\\-c&a\end{bmatrix}} {ad-bc}, \qquad ad-bc\ne0. $$

The typo I made was flipping the $b$ and $c$. All because I was an arrogant egotistical cocky douchebag who thought he could do basic multiplication in his head.

Now, we have to redo the whole thing from that point. I leave this for you to do.

Going back.

$\mathcal{F}_3$ [Finalize]

Matrix Exponential — Jordan Form, Part 1

Jordan Decomposition Using Maple

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