"Every quadratic form ${x}^{T}Ax$ with $A$ an invertible matrix is either positive definite, negative definite, or indefinite."

Is this true or false?

Is this true or false?

pominjaneh6
2022-08-12
Answered

"Every quadratic form ${x}^{T}Ax$ with $A$ an invertible matrix is either positive definite, negative definite, or indefinite."

Is this true or false?

Is this true or false?

You can still ask an expert for help

Ashlynn Stephens

Answered 2022-08-13
Author has **25** answers

Quadratic firms are associated to symmetric matrices, and you can easily show symmetric matrices over the reals have real eigenvalues.

As for your main question: since the matrix is invertible, it can't have any zero eigenvalues. Thus if it isn't definite one way, it must be indefinite. That is, it'll have both positive and negative values.

As for your main question: since the matrix is invertible, it can't have any zero eigenvalues. Thus if it isn't definite one way, it must be indefinite. That is, it'll have both positive and negative values.

crazygbyo

Answered 2022-08-14
Author has **3** answers

Since this is a quadratic form, matrix $A$ must be symmetric. You can prove (this is a very easy fact) that any symmetric matrix has only real eigenvalues.

asked 2022-04-06

What is the rationale behind ROC curves?

I am not sure how ROC curves work. I see that the X-Axis is the false positive rate while the Y axis is the true positive rate.

1) I don't understand how for a given statistical learning model, you could have the true positive and false positive rate to vary from 0 to

1. Are you changing parameters in the model to make it so?

2) What about true negatives and false negatives? How are these represented in the curve?

I am not sure how ROC curves work. I see that the X-Axis is the false positive rate while the Y axis is the true positive rate.

1) I don't understand how for a given statistical learning model, you could have the true positive and false positive rate to vary from 0 to

1. Are you changing parameters in the model to make it so?

2) What about true negatives and false negatives? How are these represented in the curve?

asked 2022-06-02

A hospital identifies patients with an average false positive rate of 0.01. If it checks 31333 patients and returns a total of 357 positive results, then what's the expected number of true positives?

asked 2022-07-11

Simplification step by step (exponential and logarithm) - regarding BLOOM FILTER - FALSE POSITIVE probability

$(1-\frac{1}{m}{)}^{kn}\approx {e}^{-kn/m}$

I don't get how to reach ${e}^{-kn/m}$ from $(1-\frac{1}{m}{)}^{kn}$

I have reviewed logarithm and exponent rules but I always get stuck, here is what I have tried; Starting from $(1-\frac{1}{m}{)}^{kn}$, I can write:

- ${e}^{(ln(1-\frac{1}{m}{)}^{kn})}$

- ${e}^{(kn\times ln(1-\frac{1}{m}))}$

Or the other way around:

$ln({e}^{(1-\frac{1}{m}{)}^{kn}})$

$ln({e}^{(1-\frac{1}{m})\times kn})$

$ln({e}^{(1-\frac{1}{m})})+ln(kn)$

$(1-\frac{1}{m})+ln(kn)$

After few steps I fall in an unsolvable loop hole playing with $ln$ and $e$ trying to reach ${e}^{-kn/m}$.

$(1-\frac{1}{m}{)}^{kn}\approx {e}^{-kn/m}$

I don't get how to reach ${e}^{-kn/m}$ from $(1-\frac{1}{m}{)}^{kn}$

I have reviewed logarithm and exponent rules but I always get stuck, here is what I have tried; Starting from $(1-\frac{1}{m}{)}^{kn}$, I can write:

- ${e}^{(ln(1-\frac{1}{m}{)}^{kn})}$

- ${e}^{(kn\times ln(1-\frac{1}{m}))}$

Or the other way around:

$ln({e}^{(1-\frac{1}{m}{)}^{kn}})$

$ln({e}^{(1-\frac{1}{m})\times kn})$

$ln({e}^{(1-\frac{1}{m})})+ln(kn)$

$(1-\frac{1}{m})+ln(kn)$

After few steps I fall in an unsolvable loop hole playing with $ln$ and $e$ trying to reach ${e}^{-kn/m}$.

asked 2022-07-23

Test for HIV. There's a false positive rate of 0.025 and a false negative rate of 0.08. Let's say a particular patient has a probability of testing positive for HIV of 0.005. The patient gets tested and it's positive. What are the chances that the patient actually has HIV?

asked 2022-07-02

If $m$ is the number of bits in the array, $k$ is the number of hash functions and we have $n$ entries, then the false positive probability upper bound can be approximated as:

$(1-{e}^{\frac{-kn}{m}}{)}^{k}$

It further states in the wiki article that:

"The number of hash functions, $k$, must be a positive integer. Putting this constraint aside, for a given $m$ and $n$, the value of $k$ that minimizes the false positive probability is:"

$k=\frac{m}{n}\mathrm{ln}2$

How can I find the $k$ value that minimizes the false positive function ?

$(1-{e}^{\frac{-kn}{m}}{)}^{k}$

It further states in the wiki article that:

"The number of hash functions, $k$, must be a positive integer. Putting this constraint aside, for a given $m$ and $n$, the value of $k$ that minimizes the false positive probability is:"

$k=\frac{m}{n}\mathrm{ln}2$

How can I find the $k$ value that minimizes the false positive function ?

asked 2022-05-10

"A certain disease has an incidence rate of 2%. If the false negative rate is 10% and the false positive rate is 1%, compute the probability that a person who tests positive actually has the disease."

Why do we need to use Bayes' Theorem for this question?

Why do we need to use Bayes' Theorem for this question?

asked 2022-06-30

Is there some indicator for false positives on the Legendre symbol?

For example,

$\left(\frac{35}{13}\right)=1$

yet ${x}^{2}-35{y}^{2}=13$ has no solutions in integers. For a non-UFD like this, is there some way to take the Legendre symbol, add to it or multiply it by something else, like maybe something to the power of the class number to get the $-1$ meaning that $13$ is irreducible in this particular domain of algebraic integers?

For example,

$\left(\frac{35}{13}\right)=1$

yet ${x}^{2}-35{y}^{2}=13$ has no solutions in integers. For a non-UFD like this, is there some way to take the Legendre symbol, add to it or multiply it by something else, like maybe something to the power of the class number to get the $-1$ meaning that $13$ is irreducible in this particular domain of algebraic integers?