MHB Bilinear Function and Gramian Matrix

[Sudharaka]

Hi everyone, :)

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:

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? :)

 

[I like Serena]

Hi everyone, :)

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:

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? :)

Sure.
Looks good.

You can simplify it a bit though.
You can find the left kernel of \(f\) using:
\[\left\{v\in \mathbb R^3:\, v^T \begin{pmatrix} 4&3&2\\1&3&5\\3&6&9 \end{pmatrix}=0 \right\}\]
That amounts to the same thing.
Do you see why?

Note that these are the eigenvectors of the transpose for the eigenvalue 0.

 

[Sudharaka]

Sure.
Looks good.

You can simplify it a bit though.
You can find the left kernel of \(f\) using:
\[\left\{v\in \mathbb R^3:\, v^T \begin{pmatrix} 4&3&2\\1&3&5\\3&6&9 \end{pmatrix}=0 \right\}\]
That amounts to the same thing.
Do you see why?

Note that these are the eigenvectors of the transpose for the eigenvalue 0.

Thanks very much for the confirmation. Of course I see it. Since,

\[\begin{pmatrix}v_1& v_2& v_3\end{pmatrix} \begin{pmatrix} 4&3&2\\1&3&5\\3&6&9 \end{pmatrix} \begin{pmatrix} u_1\\u_2\\u_3 \end{pmatrix}=0\]

holds for each vector \(\begin{pmatrix} u_1\\u_2\\u_3 \end{pmatrix}\) we can choose any two vectors \(\begin{pmatrix} u^1_1\\u^1_2\\u^1_3 \end{pmatrix}\) and \(\begin{pmatrix} u^2_1\\u^2_2\\u^2_3 \end{pmatrix}\) such that,

\[\begin{pmatrix} u^1_1\\u^1_2\\u^1_3 \end{pmatrix}\neq\begin{pmatrix} u^2_1\\u^2_2\\u^2_3 \end{pmatrix}\]

Then,

\[ \begin{pmatrix}v_1& v_2& v_3\end{pmatrix} \begin{pmatrix} 4&3&2\\1&3&5\\3&6&9 \end{pmatrix} \begin{pmatrix} u^1_1\\u^1_2\\u^1_3 \end{pmatrix} = \begin{pmatrix}v_1& v_2& v_3\end{pmatrix} \begin{pmatrix} 4&3&2\\1&3&5\\3&6&9 \end{pmatrix} \begin{pmatrix} u^2_1\\u^2_2\\u^2_3\end{pmatrix}=0\]

Hence,

\[\begin{pmatrix}v_1& v_2& v_3\end{pmatrix} \begin{pmatrix} 4&3&2\\1&3&5\\3&6&9 \end{pmatrix}\left( \begin{pmatrix} u^1_1\\u^1_2\\u^1_3 \end{pmatrix} -\begin{pmatrix}u^2_1\\u^2_2\\u^2_3\end{pmatrix} \right)=0\]

Since,

\[ \begin{pmatrix} u^1_1\\u^1_2\\u^1_3 \end{pmatrix} -\begin{pmatrix}u^2_1\\u^2_2\\u^2_3\end{pmatrix} \neq 0\]

\[\begin{pmatrix}v_1& v_2& v_3\end{pmatrix} \begin{pmatrix} 4&3&2\\1&3&5\\3&6&9 \end{pmatrix}=0\]

Is there a shorter version for this? :)

 

[I like Serena]

Looks good! :)

Is there a shorter version for this?

Hmm, a shorter version?

How about:

Let A be the matrix.

Suppose we have a $v$ in the left kernel such that $v^T A \ne 0$.
Then there is a vector $u$ such that $v^T A u \ne 0$
This is a contradiction.

If on the other hand we pick any $v$ such that $v^T A = 0$, then for any $u$ we have $v^T A u = 0$.

Therefore the left kernel is the set of all $v$ with $v^T A = 0$. $\qquad \blacksquare$

 

[Sudharaka]

Looks good! :)
Hmm, a shorter version?

How about:

Let A be the matrix.

Suppose we have a $v$ in the left kernel such that $v^T A \ne 0$.
Then there is a vector $u$ such that $v^T A u \ne 0$
This is a contradiction.

If on the other hand we pick any $v$ such that $v^T A = 0$, then for any $u$ we have $v^T A u = 0$.

Therefore the left kernel is the set of all $v$ with $v^T A = 0$. $\qquad \blacksquare$

Thanks very much Serena. :) Indeed this is something much more simpler.