How can e^{Diag Matrix} not be an infinite series?

[kostoglotov]

So, in a section on applying Eigenvectors to Differential Equations (what a jump in the learning curve), I've encountered

e^{At} \vec{u}(0) = \vec{u}(t)

as a solution to certain differential equations, if we are considering the trial substitution y = e^{\lambda t} and solving for constant coefficients. I understand the idea more or less, but the math gets quite involved, particularly with complex eigenvalues and eigenvectors.

I can see how (or at least I can follow and accept the explanation for) e^{At} being an infinite series.

But how does e^{\Lambda t} = a single matrix with e^{\lambda_i t} on it's diagonal?

 

[Geofleur]

If we write the series expansion of the exponential, we get

$e^{\Lambda t} = \mathbf{1} + \Lambda t + \frac{1}{2}t^2 \Lambda^2 + \cdots$,

where $\mathbf{1}$ is the identity matrix. Now $\Lambda$ is a diagonal matrix, so the result of multiplying it by itself is simple, e.g., $\Lambda^2 = diag(\lambda_1^2, ..., \lambda_n^2)$. Therefore,

$e^{\Lambda t} = \mathbf{1} + diag(\lambda_1, ...,\lambda_n) t + \frac{1}{2}t^2 diag(\lambda_1^2,...,\lambda_n^2) + \cdots$.

If you write this out in matrix form, summing to form a single matrix, you will see that each entry in the resulting diagonal is an infinite series that itself equals $e^{t \lambda_j }$ for the $jj$ matrix entry.

 

[DrClaude]

It works because for $D = \mathrm{diag}(d_1, d_2, \dots d_N)$, $D^n = \mathrm{diag}(d_1^n, d_2^n, \dots d_N^n)$.

Looks like Geofleur beat me to it