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[University Differential Equations] Calculating the exponential of a Matrix

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Maybe you meant to say that (i±1,±1) is an eigenvector for the eigenvalue ±i; the + case works, but the − case for your solution does not work, because

* i−1−2=i−3≠−i\*(i−1) and i−1−1=i−2≠−i\*1, while
* i−1+2=i+1=−i\*(i−1) and i−1+1=i=−i\*(−1).

With that said, Symbolab's solution looks like yours, so it looks as if your work following the explicit eigenvectors went as if you had written them out correctly.

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An alternative way to find the matrix exponential, the one that I learned, is to first diagonalize the matrix (which you can do in this case because the matrix has a full eigenbasis, because its eigenvalues are distinct), then exponentiate the diagonals, then multiply out; this works because the power series for exponentiation is also valid for matrices, although *why* it's valid is a bit beyond the scope of your course, and then for the diagonalizable case, the P^(−1)P factors cancel out in each term, leaving terms of the form PD^(n)P^(−1) for each whole number n, and then the P and P^(−1) terms can be factored out, each power of a diagonal matrix is the matrix with its diagonal entries raised to that power, and so it reduces to ordinary exponentiation.

Okay, in case you haven't diagonalized a matrix before, the P here is a matrix with each eigenvector as a column, and then the D is a diagonal matrix with the corresponding eigenvalue as the corresponding diagonal entry:

    P=⌈i+1 i−1⌉ D=⌈i  0⌉
      ⌊ 1  −1 ⌋   ⌊0 −i⌋

The idea is that, say, (i+1,1) will be transformed to (1,0), then to (i,0), then to i\*(i+1,1).

Then det(P)=−2i, so

    P^−1=⌈−½i ½i+½⌉
         ⌊−½i ½i−½⌋

and P^(−1) does work as expected.

Then the diagonal elements of e^(Dt) are e^(it) and e^(−it), and

    Pe^(Dt)=⌈(i+1)e^(it) (i−1)e^(−it)⌉
            ⌊   e^(it)     −e^(−it)  ⌋

and finally,

    e^(At)=Pe^(Dt)P^(−1)=⌈−½(i−1)e^(it)+½(i+1)e^(−it)      ie^(it)−ie^(−it)     ⌉
                         ⌊    −½ie^(it)+½ie^(−it)     ½(i+1)e^(it)−½(i−1)e^(−it)⌋

This also agrees with Symbolab.

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