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

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1. Nov 12, 2015

### 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?

2. Nov 12, 2015

### 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.

3. Nov 12, 2015

### Staff: Mentor

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

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