Choosing a career means you also need to think about your .

Mathematics

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Sever21 [200]3 years ago

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Answer:

Life Path

Step-by-step explanation:

It's two words, and they are both of equal length. I hope that helps.

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Let g' be the group of real matricies of the form [1 x 0 1]. Is the map that sends x to this matrix an isomorphism?

aliina [53]

Yes. Conceptually, all the matrices in the group have the same structure, except for the variable component $x$ . So, each matrix is identified by its top-right coefficient, since the other three entries remain constant.

However, let's prove in a more formal way that

$\phi:\ \mathbb{R} \to G,\quad \phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]$

is an isomorphism.

First of all, it is injective: suppose $x \neq y$ . Then, you trivially have $\phi(x) \neq \phi(y)$ , because they are two different matrices:

$\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right],\quad \phi(y) = \left[\begin{array}{cc}1&y\\0&1\end{array}\right]$

Secondly, it is trivially surjective: the matrix

$\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]$

is clearly the image of the real number x.

Finally, $\phi$ and its inverse are both homomorphisms: if we consider the usual product between matrices to be the operation for the group G and the real numbers to be an additive group, we have

$\phi (x+y) = \left[\begin{array}{cc}1&x+y\\0&1\end{array}\right] = \left[\begin{array}{cc}1&x\\0&1\end{array}\right] \cdot \left[\begin{array}{cc}1&y\\0&1\end{array}\right] = \phi(x) \cdot \phi(y)$