# Some Linear Algebra

• May 6th 2009, 03:49 AM
funnyinga
Some Linear Algebra
For the following, show that $\displaystyle M_{m\times n}(\mathbb{R})$ does not form a vector space over $\displaystyle \mathbb{R}$ under the (non standard) operations $\displaystyle \hat{+}$ (addition) and $\displaystyle *$ (multiplication).

(a) $\displaystyle A_{ij} \ \hat{+} \ B_{ij} = A_{ij} + B_{ij}$
$\displaystyle k * A_{ij} = k^2A_{ij}$

(b) $\displaystyle A_{ij} \ \hat{+} \ B_{ij} = A_{ij}B_{ij}$
$\displaystyle k * A_{ij} = {A^k}_{ij}$
• May 6th 2009, 04:14 AM
NonCommAlg
Quote:

Originally Posted by funnyinga
For the following, show that $\displaystyle M_{m\times n}(\mathbb{R})$ does not form a vector space over $\displaystyle \mathbb{R}$ under the (non standard) operations $\displaystyle \hat{+}$ (addition) and $\displaystyle *$ (multiplication).

(a) $\displaystyle A_{ij} \ \hat{+} \ B_{ij} = A_{ij} + B_{ij}$
$\displaystyle k * A_{ij} = k^2A_{ij}$

multiplication by scalar is obviously not linear.

Quote:

(b) $\displaystyle A_{ij} \ \hat{+} \ B_{ij} = A_{ij}B_{ij}$
$\displaystyle k * A_{ij} = {A^k}_{ij}$
the additive inverse doesn't always exist! see that the additive identity, namely 0, here is the matrix with all entries equal 1. now let $\displaystyle A$ be the matrix with 0 in the first row and first column and

whatever you want everywhere else. then $\displaystyle A$ has no additive inverse.