- #1
A.Magnus
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I am working on this problem: Show that the index of positivity is $n$ for this quadratic form:
$$f: \mathbb R^n \rightarrow \mathbb R, \ f(X) = \sum^{n}_{i=1} x_{2}^{i} + \frac{1}{n} \sum_{i \neq j}x_ix_j.$$
Here is a solution I got from other sources: Since the index of quadratic form is the number of positive terms (or square terms) of its canonical form, and since the canonical form of $f$ is
$$\big(x_1^2 + x_2^2 + ... + x_n^2 \big)+ \frac{1}{n}\big((x_1x_2 + x_1x_3 + ... + x_1x_n) + (x_2x_3 + x_2x_4 + ... + x_2x_n) + ... \big),$$
therefore the index is $n$.
I have strong doubt that this solution is correct. Is this correct? How do I have to go about if this is wrong? Thank you for your gracious help and time. ~MA
$$f: \mathbb R^n \rightarrow \mathbb R, \ f(X) = \sum^{n}_{i=1} x_{2}^{i} + \frac{1}{n} \sum_{i \neq j}x_ix_j.$$
Here is a solution I got from other sources: Since the index of quadratic form is the number of positive terms (or square terms) of its canonical form, and since the canonical form of $f$ is
$$\big(x_1^2 + x_2^2 + ... + x_n^2 \big)+ \frac{1}{n}\big((x_1x_2 + x_1x_3 + ... + x_1x_n) + (x_2x_3 + x_2x_4 + ... + x_2x_n) + ... \big),$$
therefore the index is $n$.
I have strong doubt that this solution is correct. Is this correct? How do I have to go about if this is wrong? Thank you for your gracious help and time. ~MA
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