3 years ago

Prove that the eigenvalues of a matrix squared are greater than the matrix eigenvalues

Mathematics

1 answer:

solong [7]3 years ago

6 0

Yes it issssgvrttvnrtbt

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erma4kov [3.2K]

Step-by-step explanation:

$f(x) = ln( \frac{1}{x} )$

To find the derivative, notice we have a function 1/x inside of another function, ln(x)

We use what we call the chain rule,

It states that

derivative of a '

$\frac{d}{dx} f(g(x)) = f '(g(x)) \times \: g '(x)$

Here f is ln(x)

f is 1/x

So first, we know that

$\frac{d}{dx} ( ln(x) = \frac{1}{x}$

$f'(g(x)) = \frac{1}{ \frac{1}{x} }$

We know that

$g'(x) = - \frac{1}{ {x}^{2} }$

So we have

$x \times - \frac{1}{ {x}^{2} } = \frac{ - 1}{x}$

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