The eigenspace for is spanned by a vector , so its dimension is (the matrix cannot be diagonalizable, since if it was the case would be equal to ). You can compute to see that the computation of the exponential matrix is straightforward.
Hi,
I'm supposed to compute the matrix exponential, e^A, for several matrices (for example,A=
(1 1)
(-1 -1)
I know that when I solve for its eigenvalues i get lambda=0. So then I evaluate the null space of A-Ilambda for lambda=0. Also, my diagonal matrix will be
(e^0 0)
(0 e^0) which is the identity matrix
Solving for (A-O)x=0
(1 1|0)
(0 0|0)
so i get (1,-1)^T twice. So then i should have another matrix X=
( 1 1 )
(-1 -1)
but X doesn't have an inverse . What did I do wrong?
The eigenspace for is spanned by a vector , so its dimension is (the matrix cannot be diagonalizable, since if it was the case would be equal to ). You can compute to see that the computation of the exponential matrix is straightforward.
Well it seems apparent that A^k=O for k> or=2. But I was under the impression that "using the definition of the matrix exponential" involved a diagonalizable matrix, A^k=(X)(D^k)(X^-1). So then how does computing A^2 help me with using (X)(D^k)(X^-1)? Forgive my ignorance.
Your error is in assuming there is a diagonal matrix similar to A. A matrix is diagonalizable if and only if there exist a "complete set of eigenvectors". For a 2 by 2 matrix that would be two independent eigenvectors. If you find two distinct eigenvalues, then the corresponding eigenvectors are independent so the matrix is diagonalizable. But if there is only one (double) eigenvalue, there may or may not be two independent eigenvectors.
You, as you say, in seeking eigenvectors, find only "y= -x" or <1, -1>. You cannot use the same vector (or any multiple) as both columns because, as you saw, that gives a matrix that is not invertible. This matrix is not diagonalizable but can be put in "Jordan Normal Form".
What you need to do is find a generalized eigenvector. It is always true that a matrix satisfies its own characteristic equation: for every vector. For v any multiple of <1, -1>, Av= 0 so . But that is only a one dimensional subspace of two dimensional space. There must exist another one dimensional subspace of vectors u such that but . For such vectors, it is still true that so that Au must be a multiple of <1, -1>. Those are "generalized eigenvectors".
So you need to solve
(any multiple of <1, -1> would do but it is simplest to use <1, -1> itelf).
That is the same as and , both of which are equivalent to . Taking x= 1, that gives y= 0 so the vector is <1, 0> and the corresponding matrix is .
What is ? What is ?
What are the powers of ? And so what is its exponential?