# Dual space

• Oct 12th 2009, 12:28 PM
Krizalid
Dual space
okay, i dislike linear algebra!!! i hope you guys may help me... hahah.

Let $\displaystyle \alpha=\left\{ \overbrace{\left( \begin{matrix} 1 & 0 \\ 0 & 0 \\ \end{matrix} \right)}^{v_{1}},\overbrace{\left( \begin{matrix} 1 & 1 \\ 1 & 0 \\ \end{matrix} \right)}^{v_{2}},\overbrace{\left( \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right)}^{v_{3}} \right\}$ be a basis for the vector space $\displaystyle \mathbb W=\{A\in\mathbb M_{2}(\mathbb R):A=A^t\}.$

a) Find $\displaystyle \alpha^*,$ corresponding basis of the dual space $\displaystyle \mathbb W^*.$

b) Find $\displaystyle [f]_{\alpha^*},$ if $\displaystyle f\in\mathbb W^*$ so that $\displaystyle f\left( \left( \begin{matrix} x & y \\ y & z \\ \end{matrix} \right) \right)=x+y-z.$

okay, i have some ideas to do a): put $\displaystyle \alpha^*=\{\varphi_1,\varphi_2,\varphi_3\}$ then we have $\displaystyle \varphi_1(v_1)=1,$ and $\displaystyle \varphi_1(v_2)=\varphi_1(v_3)=0.$ In the same fashion for $\displaystyle \varphi_2$ and $\displaystyle \varphi_3.$

Now, assume $\displaystyle \exists\,\alpha_1,\alpha_2,\alpha_3\in\mathbb R$ so that $\displaystyle \varphi _{1}\left( \left( \begin{matrix} x & y \\ z & w \\ \end{matrix} \right) \right)=\varphi_1\big(\alpha_1v_1+\alpha_2v_2+\alp ha_3v_3\big),$ and now i must find those $\displaystyle \alpha_1,\alpha_2,\alpha_3$ and i'll find $\displaystyle \varphi _{1}\left( \left( \begin{matrix} x & y \\ z & w \\ \end{matrix} \right) \right)$ since this equals $\displaystyle \alpha_1\varphi_1(v_1)$ ('cause the others are zero), so, is this the procedure?

As for b), no idea. :(
• Oct 12th 2009, 02:24 PM
NonCommAlg
Quote:

Originally Posted by Krizalid
okay, i dislike linear algebra!!! i hope you guys may help me... hahah.

Let $\displaystyle \alpha=\left\{ \overbrace{\left( \begin{matrix} 1 & 0 \\ 0 & 0 \\ \end{matrix} \right)}^{v_{1}},\overbrace{\left( \begin{matrix} 1 & 1 \\ 1 & 0 \\ \end{matrix} \right)}^{v_{2}},\overbrace{\left( \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right)}^{v_{3}} \right\}$ be a basis for the vector space $\displaystyle \mathbb W=\{A\in\mathbb M_{2}(\mathbb R):A=A^t\}.$

a) Find $\displaystyle \alpha^*,$ corresponding basis of the dual space $\displaystyle \mathbb W^*.$

$\displaystyle \alpha^*=\{v_1^*, v_2^*, v_3^* \},$ where for every $\displaystyle 1 \leq i \leq 3$ we define $\displaystyle v_i^*(c_1v_1+c_2v_2+c_3v_3)=c_1\delta_{i1} + c_2 \delta_{i2} + c_3 \delta_{i3}.$ (here $\displaystyle \delta_{ij}$ is the Kronecker's delta.)

Quote:

b) Find $\displaystyle [f]_{\alpha^*},$ if $\displaystyle f\in\mathbb W^*$ so that $\displaystyle f\left( \left( \begin{matrix} x & y \\ y & z \\ \end{matrix} \right) \right)=x+y-z.$

let $\displaystyle f=c_1v_1^* + c_2 v_2^* + c_3 v_3^*.$ then since $\displaystyle v_i^*(v_j)=\delta_{ij},$ we have $\displaystyle f(v_1)=c_1, \ f(v_2)=c_2, \ f(v_3)=c_3.$ thus $\displaystyle [f]_{\alpha^*}=\begin{bmatrix}f(v_1) \\ f(v_2) \\ f(v_3) \end{bmatrix}=\begin{bmatrix}1 \\ 2 \\ 1 \end{bmatrix}.$
• Oct 12th 2009, 02:35 PM
Krizalid
Quote:

Originally Posted by NonCommAlg
$\displaystyle \alpha^*=\{v_1^*, v_2^*, v_3^* \},$ where for every $\displaystyle 1 \leq i \leq 3$ we define $\displaystyle v_i^*(c_1v_1+c_2v_2+c_3v_3)=c_1\delta_{i1} + c_2 \delta_{i2} + c_3 \delta_{i3}.$ (here $\displaystyle \delta_{ij}$ is the Kronecker's delta.)

okay but, is my procedure the way to solve this? because what you wrote is what i stated above.