# A The product of a matrix exponential and a vector

#### Aleolomorfo

Hello everybody!
I was studying the Glashow-Weinberg-Salam theory and I have found this relation:
$$e^{\frac{i\beta}{2}}\,e^{\frac{i\alpha_3}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}}\, \frac{1}{\sqrt{2}}\begin{pmatrix} 0\\ v \\ \end{pmatrix} = e^{\frac{i\beta}{2}}\,e^{-\frac{i\alpha_3}{2}}\, \frac{1}{\sqrt{2}}\begin{pmatrix} 0\\ v \\ \end{pmatrix}$$
I do not know how the exponential matrix acts on the vector.

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#### DrClaude

Mentor
The exponential of a diagonal matrix is diagonal matrix containing the exponential of the diagonal elements:
$$\exp \left[ \begin{pmatrix} d_1 & 0 & 0 \\ 0 & d_2 & 0 \\ 0 & 0 & d_3 \end{pmatrix} \right] = \begin{pmatrix} e^{d_1} & 0 & 0 \\ 0 & e^{d_2} & 0 \\ 0 & 0 & e^{d_3} \end{pmatrix}$$
which can easily be shown using the series expansion of the exponential.

I guess that the right-hand side of the equation you wrote only has $v$, not the original vector.

#### DaveE

More here: https://en.wikipedia.org/wiki/Matrix_exponential
But you got what you need from DrClaude.
The easy way is to transform the matrix into a diagonal with eigen vectors and such. Do the exponent thing above and, if you want, transform it back.

Last edited:

#### Ray Vickson

Homework Helper
Dearly Missed
Hello everybody!
I was studying the Glashow-Weinberg-Salam theory and I have found this relation:
$$e^{\frac{i\beta}{2}}\,e^{\frac{i\alpha_3}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}}\, \frac{1}{\sqrt{2}}\begin{pmatrix} 0\\ v \\ \end{pmatrix} = e^{\frac{i\beta}{2}}\,e^{-\frac{i\alpha_3}{2}}\, \frac{1}{\sqrt{2}}\begin{pmatrix} 0\\ v \\ \end{pmatrix}$$
I do not know how the exponential matrix acts on the vector.
The exponential of an $m \times m$ matrix is another $m \times m$ matrix:
$$e^A \equiv I + A + \frac{1}{2!} A^2 + \frac{1}{3!} A^3 + \cdots + \frac{1}{n!} A^n + \cdots,$$ where $I$ is the identity matrix. This series converges for any finite matrix $A$.

The series is usually avoided when we actually want to compute $e^A$. Instead, there are several quite good algorithms available to do the computation, mostly based on the use of eigenvalues.

#### FactChecker

Gold Member
2018 Award
The exponential of an $m \times m$ matrix is another $m \times m$ matrix:
$$e^A \equiv I + A + \frac{1}{2!} A^2 + \frac{1}{3!} A^3 + \cdots + \frac{1}{n!} A^n + \cdots,$$ where $I$ is the identity matrix. This series converges for any finite matrix $A$.

The series is usually avoided when we actually want to compute $e^A$. Instead, there are several quite good algorithms available to do the computation, mostly based on the use of eigenvalues.
And one can use the characteristic polynomial of the matrix to convert the infinite series to a finite-order polynomial.

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