MHB How Do You Convert Expressions into Bilinear Form?

[Sudharaka]

Hi everyone, :)

I don't quite get what this question means. How do we convert the expression into the bilinear form first? Hope you people can give me some insight. :)

Question:

Find the rank of the bilinear function \((e^1+e^3)\otimes (e^2+e^4) - (e^2-e^4)\otimes (e_1-e_3)\).

 

[Sudharaka]

Hi everyone, :)

I don't quite get what this question means. How do we convert the expression into the bilinear form first? Hope you people can give me some insight. :)

Question:

Find the rank of the bilinear function \((e^1+e^3)\otimes (e^2+e^4) - (e^2-e^4)\otimes (e_1-e_3)\).

If we simplify the tensor product expression we get has,

\((e^1\otimes e^2)+(e^1\otimes e^4)+(e^3\otimes e^2)+(e^3\otimes e^4)\)

as the fist four terms. It's these four terms that I have trouble. I know that if we have all the terms as \(e^i\otimes e_j\) then we can easily find out the rank of the bilinear function. But with terms of the form \(e^i\otimes e^j\) I don't understand how we can get the rank.

Thinking further about this question I have the feeling that there maybe a typo. The rank of the bilinear function can be found out if it's given as \((e^1+e^3)\otimes (e_\color{red}{2}+e_\color{red}{4}) - (e^2-e^4)\otimes (e_1-e_3)\).

 

[Opalg]

I have the feeling that there maybe a typo. The rank of the bilinear function can be found out if it's given as \((e^1+e^3)\otimes (e_\color{red}{2}+e_\color{red}{4}) - (e^2-e^4)\otimes (e_1-e_3)\).

That seems the likeliest explanation to me. If the function is $(e^1+e^3)\otimes (e^{2}+e^{4}) - (e^2-e^4)\otimes (e_1-e_3)$ then it is not clear what its domain could be, since the two terms are operating on different spaces.

 

[Sudharaka]

That seems the likeliest explanation to me. If the function is $(e^1+e^3)\otimes (e^{2}+e^{4}) - (e^2-e^4)\otimes (e_1-e_3)$ then it is not clear what its domain could be, since the two terms are operating on different spaces.

Thanks so much for the confirmation. I changed the problem and I get the answer as \(8\). Am I correct? :)

 

[Opalg]

I changed the problem and I get the answer as \(8\). Am I correct? :)

As in a previous question, that depends on what is meant by "rank" here. I would take it to mean the rank of the associated matrix. I would associate each basic tensor of the form $e^i\otimes e_j$ with the $4\times4$ matrix having a $1$ in the $(i,j)$-position and zeros elsewhere. Then the bilinear form $(e^1+e^3)\otimes (e_{2}+e_{4}) - (e^2-e^4)\otimes (e_1-e_3)$ corresponds to the matrix $\begin{bmatrix}0&1&0&1 \\ -1&0&1&0 \\ 0&1&0&1 \\ 1&0&-1&0\end{bmatrix}$, which has rank $2.$

 

[Sudharaka]

As in a previous question, that depends on what is meant by "rank" here. I would take it to mean the rank of the associated matrix. I would associate each basic tensor of the form $e^i\otimes e_j$ with the $4\times4$ matrix having a $1$ in the $(i,j)$-position and zeros elsewhere. Then the bilinear form $(e^1+e^3)\otimes (e_{2}+e_{4}) - (e^2-e^4)\otimes (e_1-e_3)$ corresponds to the matrix $\begin{bmatrix}0&1&0&1 \\ -1&0&1&0 \\ 0&1&0&1 \\ 1&0&-1&0\end{bmatrix}$, which has rank $2.$

I think what you meant by rank is what is meant to find in the question. Of course we used the word rank to mean different things in our linear algebra course, but in this case I think your answer makes more sense and I guess rank refers to the matrix rank. Thanks so much for helping me out. :)

 

[Sudharaka]

Hi again, :)

Our professor changed the question so as all the indices were taken as superscripts. So the the question is,

Question:

Find the rank of the bilinear function \[(e^1+e^3)\otimes (e^2+e^4) - (e^2-e^4)\otimes (e^1-e^3)\]

Now can you give me any idea as to how this can me converted into it's matrix form. :)

 

[Opalg]

Hi again, :)

Our professor changed the question so as all the indices were taken as superscripts. So the the question is,

Question:

Find the rank of the bilinear function \[(e^1+e^3)\otimes (e^2+e^4) - (e^2-e^4)\otimes (e^1-e^3)\]

Now can you give me any idea as to how this can me converted into it's matrix form. :)

It should make no difference. This time, the matrix represents a map from $V^*\otimes V^*$ to itself rather than a map from $V^*\otimes V^*$ to $V\otimes V$. But the matrix entries, and therefore the rank, are the same as before.

 

[Sudharaka]

It should make no difference. This time, the matrix represents a map from $V^*\otimes V^*$ to itself rather than a map from $V^*\otimes V^*$ to $V\otimes V$. But the matrix entries, and therefore the rank, are the same as before.

Thanks very much for the reply. So that means, for example, the matrix representation of \(e^1\otimes e_2\) and the matrix representation of \(e^1\otimes e^2\) is the same, given by,

\begin{pmatrix}0&1&0&0\\0&0&0&0 \\0&0&0&0\\0&0&0&0\end{pmatrix}

Am I correct? Furthermore is this same matrix used to represent \(e_1\otimes e_2\) ?