Some Linear Algebra

**funnyinga** · May 6th 2009, 03:49 AM

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

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

(b) $A_{ij} \ \hat{+} \ B_{ij} = A_{ij}B_{ij}$
$k * A_{ij} = {A^k}_{ij}$

**NonCommAlg** · May 6th 2009, 04:14 AM

Originally Posted by funnyinga

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

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

multiplication by scalar is obviously not linear.

(b) $A_{ij} \ \hat{+} \ B_{ij} = A_{ij}B_{ij}$
$k * A_{ij} = {A^k}_{ij}$

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 $A$ be the matrix with 0 in the first row and first column and

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

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