# Can we use the singular value decomposition to compute the matrix exponential for a non-diagonalisable matrix?

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?

• The matrix exponential formula for the diagonalizable case works out because when you take powers of $A=U\Lambda U^{-1}$, you only collect powers on $\Lambda$. Why do you think that happens? Can you now see why this does work out for the SVD case? Commented Mar 6, 2023 at 6:25
• I can see that $A^n = U \Lambda^n U^{-1}$, but I don't see why we can't also use $A^n = U \Sigma^n V^T$ by the same logic. Commented Mar 6, 2023 at 6:33
• Actually if we do $A^n = (U \Sigma V^{-1})^n = (U \Sigma V^{-1}) (U \Sigma V^{-1}) (U \Sigma V^{-1}) ... (U \Sigma V^{-1})$ then I think it doesn't work because we can't simplify $V^{-1} U$ into $I$? Is that right? Commented Mar 6, 2023 at 6:35
• That's right, the formula only works if $V^TU=I$. But then $U=V$ anyways. Commented Mar 6, 2023 at 6:41
• @QuartelQuartz: I think the problem here was that above you wrote "why this does work out" but actually meant "why this doesn't work out"? Commented Mar 6, 2023 at 8:34

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 $$(U\Sigma V^\top)^n$$.