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Math Help - Dual space

  1. #1
    Math Engineering Student
    Krizalid's Avatar
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    Dual space

    okay, i dislike linear algebra!!! i hope you guys may help me... hahah.

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

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

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

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

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

    As for b), no idea.
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  2. #2
    MHF Contributor

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

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

    a) Find \alpha^*, corresponding basis of the dual space \mathbb W^*.
    \alpha^*=\{v_1^*, v_2^*, v_3^* \}, where for every 1 \leq i \leq 3 we define v_i^*(c_1v_1+c_2v_2+c_3v_3)=c_1\delta_{i1} + c_2 \delta_{i2} + c_3 \delta_{i3}. (here \delta_{ij} is the Kronecker's delta.)


    b) Find [f]_{\alpha^*}, if f\in\mathbb W^* so that f\left( \left( \begin{matrix}<br />
x & y \\<br />
y & z \\<br />
\end{matrix} \right) \right)=x+y-z.
    let f=c_1v_1^* + c_2 v_2^* + c_3 v_3^*. then since v_i^*(v_j)=\delta_{ij}, we have f(v_1)=c_1, \ f(v_2)=c_2, \ f(v_3)=c_3. thus [f]_{\alpha^*}=\begin{bmatrix}f(v_1) \\ f(v_2) \\ f(v_3) \end{bmatrix}=\begin{bmatrix}1 \\ 2 \\ 1 \end{bmatrix}.
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    \alpha^*=\{v_1^*, v_2^*, v_3^* \}, where for every 1 \leq i \leq 3 we define v_i^*(c_1v_1+c_2v_2+c_3v_3)=c_1\delta_{i1} + c_2 \delta_{i2} + c_3 \delta_{i3}. (here \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.
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