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.
