# need help on dual spaces

• Sep 10th 2009, 06:29 PM
Morgan
need help on dual spaces
Let $\displaystyle \alpha = \{v_1,v_2\}$ be a basis for $\displaystyle V$ and $\displaystyle \beta^* = \{(v_1 + v_2)^*,(v_1 - v_2)^*\}$ where $\displaystyle v_1^*,v_2^*$ is the canonical basis of the dual. Prove that $\displaystyle \beta^*$ is basis of $\displaystyle V^*$ and $\displaystyle (v_1 + v_2)^*\ne v_1^* + v_2^*.$
• Sep 10th 2009, 10:24 PM
NonCommAlg
Quote:

Originally Posted by Morgan
Let $\displaystyle \alpha = \{v_1,v_2\}$ be a basis for $\displaystyle V$ and $\displaystyle \beta^* = \{(v_1 + v_2)^*,(v_1 - v_2)^*\}$ where $\displaystyle v_1^*,v_2^*$ is the canonical basis of the dual. Prove that $\displaystyle \beta^*$ is basis of $\displaystyle V^*$ and $\displaystyle (v_1 + v_2)^*\ne v_1^* + v_2^*.$

first of all, the characteristic of your base field $\displaystyle F$ must be $\displaystyle \neq 2$ because otherwise $\displaystyle v_1+v_2=v_1-v_2.$ the first part of your question is trivial because $\displaystyle \{v_1+v_2,v_1-v_2 \}$ is a basis for $\displaystyle V$ and so $\displaystyle \beta^*$ is

a basis for $\displaystyle V^*.$ for the second part, we have $\displaystyle 2(v_1+v_2)^*(v_1)=(v_1+v_2)^*(v_1+v_2 + v_1 - v_2)=1$ but $\displaystyle 2(v_1^* + v_2^*)(v_1)=2.$ so $\displaystyle 2(v_1+v_2)^*(v_1) \neq 2(v_1^* + v_2^*)$ and thus $\displaystyle (v_1+v_2)^* \neq v_1^* + v_2^*$ because

$\displaystyle \text{char}(F) \neq 2.$
• Sep 15th 2009, 07:13 PM
Morgan
"characteristic?"
what do you mean by "characteristic?" also, what does mean char(F) and why is not 2?