For a diagonalisable matrix $ \bf{A} $ with eigendecomposition $ \bf{A} = \bf{U} \bf{\Lambda} \bf{U}^{-1} $, we know that $ \exp(\bf{A}) = \bf{U} (\exp \bf{\Lambda}) \bf{U}^{-1} $, where $ \exp \bf{\Lambda} $ is a diagonal matrix with $ e^{\lambda} $ terms.
I want to compute $ \exp(\bf{A}) $ where $ \bf{A} $ does not have a full set of eigenvectors (defective) so cannot use this. I tried to use the singular value decomposition by letting $ \bf{A} = \bf{U} \bf{\Sigma} \bf{V}^{T} $ and computing $ \exp(\bf{A}) = \bf{U} (\exp \bf{\Sigma}) \bf{V}^{T} $.
However, trying out a numerical example: (SVD computation, evaluating exp using SVD computation) $$ \mathbf{A} = \begin{bmatrix} 3 & -1\\ 1 & 1 \end{bmatrix} $$ $$ \mathbf{U} = \begin{bmatrix} -0.973249 & -0.229753\\ -0.229753 & 0.973249 \end{bmatrix}, \ \ \ \ \mathbf{\Sigma } = \begin{bmatrix} 3.23607 & 0 \\ 0 & 1.23607 \end{bmatrix}, \ \ \ \ \mathbf{V } = \begin{bmatrix} 0.973249 & 0.229753 \\ 0.229753 & 0.973249 \end{bmatrix} $$ $$ \bf{U} (\exp \bf{\Sigma}) \bf{V}^{T} = \begin{bmatrix} -24.273 & -6.456 \\ -4.917 & 1.9178 \end{bmatrix} $$
but the actual exponential (computation) should be $ \begin{bmatrix} 2e^2 & -e^2 \\ e^2 & 0 \end{bmatrix} = \begin{bmatrix} 14.78 & -7.389 \\ 7.389 & 0 \end{bmatrix} $
so something has gone wrong. Is the SVD a valid method of finding a matrix exponential, or have I just made an error somewhere?
Community wiki answer so the question can be marked as answered:
As discussed in the comments, the reason this works for diagonalization is that all the factors $U^{-1}U$ in $(U\Lambda U^{-1})^n$ cancel, and thus the result is just $U\Lambda^n U^{-1}$; this doesn’t work for singular value decomposition because in that case $V^\top U$ doesn’t generally cancel in $(U\Sigma V^\top)^n$.