The Matrix Exponential

Algebra Level 5

We know the infinite series: e a = 1 + a + a 2 2 ! + a 3 3 ! + . . . e^a = 1 + a +\frac{a^2}{2!} + \frac{a^3}{3!} + ...

This series can be generalised for square matrices. Consider a matrix A A . Its exponential can be defined as:

e A = I + A + A 2 2 ! + A 3 3 ! + . . . e^A = I + A +\frac{A^2}{2!} + \frac{A^3}{3!} + ...

Here, I I is the Identity matrix having the same size as A A .

Let us consider a situation where: A = [ 2 0 0 3 ] A = \begin{bmatrix} 2 & 0\\ 0 & 3\\ \end{bmatrix}

The exponential for this matrix turns out to be the matrix B B . This implies: e A = B e^A = B

Now consider a matrix T T being defined as such:

T = [ 1 2 3 4 ] T = \begin{bmatrix} 1 & 2\\ 3 & 4\\ \end{bmatrix}

The goal is to find the matrix exponential of the matrix T A T − 1 TAT^{-1} .

e T A T − 1 = C e^{TAT^{-1}} = C

Compute B + C B + C . The result is of the form:

B + C = [ w x y z ] B + C = \begin{bmatrix} w & x\\ y & z\\ \end{bmatrix}

Enter your answer as ⌊ w ⌋ + ⌊ x ⌋ + ⌊ y ⌋ + ⌊ z ⌋ \lfloor{w}\rfloor+\lfloor{x}\rfloor+\lfloor{y}\rfloor+\lfloor{z}\rfloor , where w w , x x , y y and z z are real numbers.

Note: ⌊ Q ⌋ \lfloor{Q}\rfloor denotes the greatest integer less than or equal to a real number Q Q .

Hint: Use the series expansions to arrive at general results and use those results in your computations. First, arrive at a general result for the matrix exponential of a diagonal matrix, and then attempt to find a general result for the matrix defined as C C .The solver may use a calculator in the last steps of the problem.


The answer is 117.

Karan Chatrath by Karan Chatrath , 27, Netherlands

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1 solution

Karan Chatrath
May 14, 2019

Consider any diagonal matrix A.

A = [ a 1 0 0 a 2 ] A = \begin{bmatrix} a_1 & 0\\ 0 & a_2\\ \end{bmatrix}

Using the definition of the matrix exponential: e A = [ 1 0 0 1 ] + [ a 1 0 0 a 2 ] + [ a 1 2 0 0 a 2 2 ] 2 ! + [ a 1 3 0 0 a 2 3 ] 3 ! + . . . e^A = \begin{bmatrix}1 & 0\\0 & 1\\\end{bmatrix} + \begin{bmatrix}a_1 & 0\\0 & a_2\\\end{bmatrix} +\frac{\begin{bmatrix}a_1^2 & 0\\0 & a_2^2\\\end{bmatrix}}{2!} + \frac{\begin{bmatrix}a_1^3 & 0\\0 & a_2^3\\\end{bmatrix}}{3!} + ...

After some trivial simplification:

e A = [ ( 1 + a 1 + a 1 2 2 ! + a 1 3 3 ! + … ) 0 0 ( 1 + a 2 + a 2 2 2 ! + a 2 3 3 ! + … ) ] e^A = \begin{bmatrix}(1+a_1+\frac{a_1^2}{2!}+\frac{a_1^3}{3!}+\dots) & 0\\0 & (1+a_2+\frac{a_2^2}{2!}+\frac{a_2^3}{3!}+\dots)\\\end{bmatrix}

Which turns out to be:

e A = [ e a 1 0 0 e a 2 ] e^A = \begin{bmatrix}e^{a_1} & 0\\0 & e^{a_2}\\\end{bmatrix}

So, in the case where a 1 = 2 a_1=2 and a 2 = 3 a_2=3 , we have

B = e A = [ e 2 0 0 e 3 ] B = e^A = \begin{bmatrix}e^{2} & 0\\0 & e^{3}\\\end{bmatrix}

Now, we consider an arbitrary matrix T T . The need is to compute the matrix exponential of T A T − 1 TAT^{-1} .

To do this, we again use the series expansion:

e T A T − 1 = I + T A T − 1 + ( T A T − 1 ) 2 2 ! + ( T A T − 1 ) 3 3 ! + … e^{TAT^{-1}} = I + TAT^{-1} + \frac{(TAT^{-1})^2}{2!} + \frac{(TAT^{-1})^3}{3!} + \dots

Remember that

T T − 1 = I TT^{-1} = I

T − 1 T = I T^{-1}T = I

The above matrix exponential can be rewritten as:

e T A T − 1 = T I T − 1 + T A T − 1 + T A T − 1 T A T − 1 2 ! + T A T − 1 T A T − 1 T A T − 1 3 ! + … e^{TAT^{-1}} = TIT^{-1} + TAT^{-1} + \frac{TAT^{-1}T AT^{-1}}{2!} + \frac{TAT^{-1}TAT^{-1}TAT^{-1}}{3!} + \dots

All the terms having T − 1 T T^{-1}T combined cancel to be the identity matrix. The expression simplifies to:

e T A T − 1 = T I T − 1 + T A T − 1 + T A 2 T − 1 2 ! + T A 3 T − 1 3 ! + … e^{TAT^{-1}} = TIT^{-1} + TAT^{-1} + \frac{TA^2T^{-1}}{2!} + \frac{TA^3T^{-1}}{3!} + \dots

or:

e T A T − 1 = T ( I + A + A 2 2 ! + A 3 3 ! + … ) T − 1 e^{TAT^{-1}} = T(I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \dots)T^{-1}

This gives us the result:

e T A T − 1 = T e A T − 1 e^{TAT^{-1}} = Te^AT^{-1}

Using this result, the matrix C can be computed as such:

C = e T A T − 1 = [ 1 2 3 4 ] [ e 2 0 0 e 3 ] [ 1 2 3 4 ] − 1 C = e^{TAT^{-1}} = \begin{bmatrix}1 & 2\\3 & 4\\\end{bmatrix}\begin{bmatrix}e^{2} & 0\\0 & e^{3}\\\end{bmatrix}\begin{bmatrix}1 & 2\\3 & 4\\\end{bmatrix}^{-1}

Having found B B and C C , the final result can be computed. The answer is 117 .

8 Helpful 2 Interesting 3 Brilliant 0 Confused

@Karan Chatrath
I want to share something with you. Recent problem of Steven sir is very easy.
Posting its solution feels me time waste.
Steven sir itself knows the solution still he wants to post it and the problem is also a very straight forward .
There is nothing logical in the problem.
I request you to put solutions of those problem in which the author is facing difficulty.
The problems which I am posting from last 5 days , I also don't know how to solve them .
The physics book which I have is very hard.
No solutions are available on internet .
Therefore I upload them.
As I am JEE and Physics Olympiad Aspirant.
I want to solve tough questions, but don't know how to solve them.
My Coaching FIITJEE physics teacher is basically a Chutiya.
He doesn't able to solve tough questions.
He always try to escape from me , I have reported this problem to FIITJEE Manager but, this all people are working under the hand of teacher, so there is no hope for good.
I hope you understand my situation.


Talulah Riley - 9 months, 1 week ago

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I agree that the problems you have posted are hard. Some of them, I cannot solve. I don't consider myself to be very good at solving problems. But you are looking at this in the wrong way. I see Brilliant as a 'knowledge sharing' platform rather than it being a 'taking shortcut' platform. I would not encourage you to post problems that you cannot solve and expect solutions on a silver platter. I recommend that you post them as discussions where multiple members can share their thoughts, when possible. Also, always share your attempt . I think this could be far more conceptually enriching for you and for all of us, who enjoy physics.

Do not hurry when it comes to solving problems. Take your time to think it through. Apply yourself.

As for recent problems posted on this website, note that most of the authors post original problems, which is not easy to do. And I also don't agree that they are 'very easy'. Maybe you have improved at problem-solving significantly to make such a claim, but the problems are always conceptually enriching. Instead of just solving a problem, try to think through it more carefully. Think of ways to solve it differently. Think of variations of a problem. Don't just blindly solve. That should not be the objective.

Karan Chatrath - 9 months, 1 week ago

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