Let $V$ be the vector space of symmetric $2 \times 2$ real matrices and
$$ q: A \in V \mapsto det(A) \in \mathbb{R} $$
be a functional on $V$. How can I show that $q$ is a quadratic form? I have no clue.
Quadratic forms on the space of symmetric matrices
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$\begingroup$
linear-algebra
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0I can´t see how the trace of a matirx can be a quadratic in something... – 2017-01-18
1 Answers
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If $A=\pmatrix{a & b\\b & c}$, $\det(A)=ac-b^2$ is a quadratic form in 3 variables $a,b,c$ because the (global) degree of its two terms is 2.
Said otherwise: $det(\lambda A)=\lambda^2 \det(A)$.
(homogeneous with degree 2, which is another way to characterize quadratic forms).
The bilinear form associated with this quadratic form is:
$$B((a,b,c),(a',b',c'))=\frac12(ac'+a'c)-bb'$$
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0Can you show that there exists a bilinear form whose quadratic form is exactly $q$? – 2017-01-18