# Thread: help with Linear spaces:)

1. ## help with Linear spaces:)

Hey,
Is the group a field in relation to the addition and multiplication of the matrices?

Thanks

2. ## Re: help with Linear spaces:)

A basic property of a field is that every member of a field, other than the additive identitiy (0), has a multiplicative inverse. Such a matrix has a multiplicative inverse if and only if its determinant is $\displaystyle a^2- 2b^2\ne 0$. That is always true for a and b rational numbers.

3. ## Re: help with Linear spaces:)

Should I show the 7 features?
How do I start it?

4. ## Re: help with Linear spaces:)

Can't you use the fact that
1) addition of matrices is commutative.
2) addition of matrices is associative.
3) there exist a "0" matrix, the additive identity.
4) every matrix has an additive inverse.
5) multiplication of matrices is associative.
6) there exist a "1" matrix, the multiplicative identity.
are true of ALL matrices? so the only thing left to show is that the every non-zero matrix is invertible.

(Well, I guess you should show that the 0 matrix, $\displaystyle \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$ is of the given form, with a= b= 0 and that the identity matrix, $\displaystyle \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$, is of that form with a= 1, b= 0.)

5. ## Re: help with Linear spaces:)

Originally Posted by HallsofIvy
Can't you use the fact that
1) addition of matrices is commutative.
2) addition of matrices is associative.
3) there exist a "0" matrix, the additive identity.
4) every matrix has an additive inverse.
5) multiplication of matrices is associative.
6) there exist a "1" matrix, the multiplicative identity.
are true of ALL matrices? so the only thing left to show is that the every non-zero matrix is invertible.

(Well, I guess you should show that the 0 matrix, $\displaystyle \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$ is of the given form, with a= b= 0 and that the identity matrix, $\displaystyle \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$, is of that form with a= 1, b= 0.)

That's the solution?