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- #1

- Feb 5, 2012

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Here's a question which I am not sure whether my approach is correct. My understanding about Bilinear functions and Gramian matrix is limited, so this might be totally wrong. Hope you can provide some insight.

**Question:**

A bilinear function \(f:\Re^3\times \Re^3 \rightarrow \Re \) is given in standard basis \(\{e_1,\,e_2,\,e_3\}\) by the Gram matrix,

\[G_f=\begin{pmatrix}4&3&2\\1&3&5\\3&6&9 \end{pmatrix}\]

Find the left and a right kernel of \(f\).

My Solution:

My Solution:

I would find the left kernel of \(f\) using,

\[\left\{v\in\Re^3:\, \begin{pmatrix}v_1& v_2& v_3\end{pmatrix} \begin{pmatrix} 4&3&2\\1&3&5\\3&6&9 \end{pmatrix} \begin{pmatrix} u_1\\u_2\\u_3 \end{pmatrix}=0\mbox{ for all } \begin{pmatrix} u_1 \\u_2 \\u_3 \end{pmatrix} \in \Re^3 \right\}\]

and the right kernel of \(f\) using,

\[\left\{u\in\Re^3:\, \begin{pmatrix}v_1& v_2& v_3\end{pmatrix} \begin{pmatrix} 4&3&2\\1&3&5\\3&6&9 \end{pmatrix} \begin{pmatrix} u_1\\u_2\\u_3 \end{pmatrix}=0\mbox{ for all } \begin{pmatrix} v_1 &v_2 &v_3 \end{pmatrix} \in \Re^3 \right\}\]

Is this approach correct? Is this how the left and right kernels are given when the bilinear function is represented by the so called Gramian matrix?