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Let $V = Mat_{2x2}(\mathbb{R})$. Define $\varphi(A,B) := det(A+B) - det(A) - det(B)$.

I have to show that $\varphi(A,B)$ is a symmetric bilinear form. It is easy to see that $\varphi$ is symmetric. My problem is to show the bilinearity. I know i have to show that:

$\varphi(\lambda(A)+\mu(C),B) = \lambda \varphi(A,B) + \mu \varphi(C,B)$.

And the same for the second argument.

But I don't know how to get to the equality.

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We can do the computaion in coordinate. Using the notation below \begin{equation} X= \left[ \begin{matrix} x_1 & x_2\\ x_3 & x_4 \end{matrix} \right] \end{equation} we have: \begin{multline} \varphi(A,B) = [(a_1+b_1)(a_4+b_4)-(a_2+b_2)(a_3+b_3)] - [a_1a_4 -a_2a_3] - [b_1b_4-b_2b_3] =\\ a_1b_4+b_1a_4-a_2b_3-b_2a_3 \end{multline} At this step you have only to do the computation in order to find the bilinearity.

But I prefer another approach. Under the isomorphism of coordinates: \begin{equation} f:Mat_{2\times 2}(\mathbb R)\longrightarrow \mathbb R^4, \quad f(X) = (x_1,x_2,x_3,x_4)^T \end{equation} you can see that $\varphi$ corresponds to the function: \begin{gather} \tilde \varphi:\mathbb R^4 \times \mathbb R^4 \longrightarrow \mathbb R\\ (a,b) \longmapsto a^T M b \end{gather} where $a=(a_1,a_2,a_3,a_4)^T$, $b=(b_1,b_2,b_3,b_4)^T$ and \begin{equation} M= \left[ \begin{matrix} 0 & 0 & 0 &1\\ 0 & 0 & -1 & 0\\ 0 & -1 & 0 & 0\\ 1 & 0 & 0 &0 \end{matrix} \right] \end{equation} and this is obviously a scalar product.

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  • $\begingroup$ So this M is the matrix associated to $\varphi$ with respect to the standard Basis of $V$, if I understand correctly? $\endgroup$ – MathematicalMoose Apr 2 '20 at 8:37
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    $\begingroup$ Yes, it is correct. $\endgroup$ – Menezio Apr 2 '20 at 8:42
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    $\begingroup$ Thank you very much. $\endgroup$ – MathematicalMoose Apr 2 '20 at 8:42
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    $\begingroup$ It's a pleasure! $\endgroup$ – Menezio Apr 2 '20 at 8:43

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