Question

show that if an \(n*n\) matrix has eigenvalues \(\lambda_1, \lambda_2,...,\lambda_n\) (not necessarily distinct) with linearly independent eigenvectors\[x_1,x_2,...,x_n\]respectively, then a general solution of \(x'=Ax\) is \(x(t)=c_{1}e^{\lambda_{1}t}x_{1}+...+c_{n}e^{\lambda_{n}t}x_{n}\)

Answer 1

x' is a derivative or just some other vector?

Answer 2

x' is the derivative of the eigenvector

Answer 3

A is the nxn matrix, right?

Answer 4

yes. sorry I was trying to look for other resources

Answer 5

Since A has n linearly independent eigenvectors then x can be written as a linear combination of these.
That is
\[x = k_1x_1+k_2x_2+\cdots +k_nx_n\]
Thus
\[Ax = A(k_1x_1+k_2x_2+\cdots +k_nx_n)= A(k_1x_1)+A(k_2x_2)+\cdots +A(k_nx_n)\]\[=k_1(A x_1)+k_2(Ax_2)+\cdots +k_n(Ax_n)\]
\[\Rightarrow Ax = k_1\lambda_1 x_1+k_2\lambda_2x_2+\cdots +k_n\lambda_n x_n \]
However, x' can also be written as a linear combination of these same eigenvectors, that is
\[x' = l_1x_1+l_2x_2+\cdots +l_n x_n\]

Answer 6

So
\[\frac{ d }{ dt }x(t) = k_1 \lambda_1x_1+k_2 \lambda_2x_2+\cdots+k_n \lambda_nx_n \Rightarrow x(t) = k_1e^{\lambda_1t}x_1+k_2e^{\lambda_2t}x_2+\cdots+k_n e^{\lambda_nt}x_n\]

Answer 7

Something like that...