Exam (elaborations)
Che 505 Final Exam Part 2 (Oral) With Complete Solutions Latest Update
- Course
- CHE 505
- Institution
- Arizona State University
Che 505 Final Exam Part 2 (Oral) With Complete Solutions Latest Update
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Solution 2024/2025Pepper
Che 505 Final Exam Part 2 (Oral) With
Complete Solutions Latest Update
Eigenproblem ANS✔✔ Ax=lambda*x
x = eigenvector (operator doesnt change orientation of eigenvector)
lambda = eigenvalue
how to find eigenvalue ANS✔✔ characteristic equation:
det(A-lambda*I) = 0
for nxn matrix, characteristic equation will be nth order polynomial to find
lambda
determinant doesn't scale very well
how to find eigenvectors ANS✔✔ (A-lambda*I)x = 0
basis vectors ANS✔✔ vectors x
n linearly independent n-dimensional vectors
can represent any n dimensional vector as a linear combination of basis
vectors
any vector y is a linear combination of xi
, Solution 2024/2025
Pepper
Reciprocal vectors ANS✔✔ vectors z
<zi,xj> = zi+*xj = delta(ij)
= 0 when i is not equal to j
=1 when i=j
X is a matrix where the basis vectors are columns
zi+ must be the rows of X^-1
orthonormal basis ANS✔✔ special cause where xi (the basis vectors) are
orthonormal
xi+*xj = delta(ij)
means the basis vectors are their own reciprocal vectors
x^-1 = x^+
Spectral resolution ANS✔✔ perfect matrix:
nxn matrix with n linearly independent eigenvectors
use eigenvectors as bases vectors
A = sum(lambdai*xi*zi^+)
, Solution 2024/2025
Pepper
Ay = sum(lambdai*xi*<zi,y>
spectral resolution of a polynomial function ANS✔✔ A^2 =
sum(lambdai^2*xi*zi^+)
integer k is greater than or equal to zero
A^k = sum(lambdai^k*xi*zi^+)
if A is nonsingular any integer k applies
A^(-k) = sum(lambdai^(-k)*xi*zi^+)
Spectral resolution of inverse functions ANS✔✔ A^-1 has the same
eigenvalues as A but the eigenvalues of A^-1 are 1/lambdai (lambdai are
the eigenvalues of A)
spectral resolution of exp ANS✔✔ exp(A) = sum(exp(lambdai)*xi*zi^+)
Spectral resolution: connection to differential equations ANS✔✔ any function
f(A)
f(A) = sum(f(lambdai)*xi*zi^+)
dy/dt = sum(exp(lambdai*t)*xi*zi^+*y)
similarity of matrices ANS✔✔ A and B are similar if S exists such that AS=SB
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