- #1
Joelly
- 3
- 0
Click For Summary
The discussion focuses on determining if the mapping from the set of numbers of the form \(a + b\sqrt{2}\) to matrices of the form \(\begin{bmatrix} a & 2b \\ b & a \end{bmatrix}\) is an isomorphism. It confirms that the mapping is both one-to-one and onto. To establish that it preserves multiplication, the products of two numbers and their corresponding matrices must be compared. Additionally, the discussion emphasizes the need to check if the mapping preserves addition by comparing the sums of the pairs. The conclusion is that both operations should be verified to confirm the isomorphism.
- #2
HOI
- 921
- 2
An obvious "guess", practically forced by the way the problem is stated, is the function that maps $a+ b\sqrt{2}$ to $\begin{bmatrix} a & 2b \\ b & a \end{bmatrix}$. Is that an isomorphism? That is, is it "one to one", is it "onto", and does it "preserve the operation".
The first two are simple. To prove that this "preserves the operation" you need to take two members of the first set, $a+ b\sqrt{2}$ and $x+ y\sqrt{2}$, which map to $\begin{bmatrix}a & 2b \\ b & a\end{bmatrix}$ and $\begin{bmatrix}x & 2y \\ y & a\end{bmatrix}$ and show that the product $(a+ b\sqrt{2})(x+ y\sqrt{2})$ maps to the product $\begin{bmatrix}a & 2b \\ b & a \end{bmatrix}\begin{bmatrix}x & 2y \\ y & x\end{bmatrix}$.
Do those two multiplications and compare them.
To determine whether they are also isomorphic under addition, add those pairs and compare them.
The first two are simple. To prove that this "preserves the operation" you need to take two members of the first set, $a+ b\sqrt{2}$ and $x+ y\sqrt{2}$, which map to $\begin{bmatrix}a & 2b \\ b & a\end{bmatrix}$ and $\begin{bmatrix}x & 2y \\ y & a\end{bmatrix}$ and show that the product $(a+ b\sqrt{2})(x+ y\sqrt{2})$ maps to the product $\begin{bmatrix}a & 2b \\ b & a \end{bmatrix}\begin{bmatrix}x & 2y \\ y & x\end{bmatrix}$.
Do those two multiplications and compare them.
To determine whether they are also isomorphic under addition, add those pairs and compare them.
- #3
Joelly
- 3
- 0
Hello!
In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way:
I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true?
And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...
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