sjeikdom0
2021-03-09
Answered

When is a rank one matrix diagonalizable? Justify your answer. What is one choice of the diagonalizing similarity? What happens when it is not diagonalizable? Justify your answer.

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Tasneem Almond

Answered 2021-03-10
Author has **91** answers

Step 1

Let, the rank of

Step 2

A

Example:

Ay diagonal matrix A is diagonalizable as it is similar to itself.

Step 3

If the matrix is not diagonalizable, the one has to find a matrix with same properties consisting of eigen values on the leading diagonal and either ones or zeros on the super diagonal.

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We are given a ratio:

$\frac{g(x)}{f(x)}$

where:

$g(x)\in {\mathbb{R}}^{+}$

$f(x)\in \mathbb{N}\phantom{\rule{mediummathspace}{0ex}}\cap f(x)\ge 2$

So $g(x)$ returns values in $[0,+\mathrm{\infty}]$ while $f(x)$ returns values in $\{2,3,4,\dots \}$.

I am looking for a confirmation about a very simple question: if I maximize $\frac{g(x)}{f(x)}$, do I also maximize $\frac{g(x)}{f(x)-1}$ in this very particular case?

$\frac{g(x)}{f(x)}$

where:

$g(x)\in {\mathbb{R}}^{+}$

$f(x)\in \mathbb{N}\phantom{\rule{mediummathspace}{0ex}}\cap f(x)\ge 2$

So $g(x)$ returns values in $[0,+\mathrm{\infty}]$ while $f(x)$ returns values in $\{2,3,4,\dots \}$.

I am looking for a confirmation about a very simple question: if I maximize $\frac{g(x)}{f(x)}$, do I also maximize $\frac{g(x)}{f(x)-1}$ in this very particular case?

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Finding surface area and volume of a sphere using only Pappus' Centroid Theorem

I wonder if it is possible to derive surface area and volume of a sphere seperately using techniques involving pappus' theorem.

I did some calculation and found out the ratio of surface area and volume. Here is my work,

1. My key observation is finding out that the centroid of a semidisk of raidus r is also the centroid of a semicircle of raidus $2r/3$ when the centers coincide. (By "centers" I mean the centers of full circle and disc induced by half of them.)

2. I sliced the semidisk identical pieces of triangles(infinitely many) and by locating each triangle's centroid I form a semicircle of radius $2r/3$ which must have the same centroid with semidisk.

3. Then let's say our centroid is located h distance above the center. We still don't know it.

4. Using the theorem, the circular path taken by the centroid of the semidisk times the area of the centroid should give the volume of the sphere of radius r. And similarly, the circular path taken by the same centroid of the semicircle times the arc length of semicircle should give the surface area of a sphere of radius 2r/3.

5. We know that when the radius increases with a proportion, corresponding surface area will also increase with the square of that proportion. Thus, we need to multiply the surface area of sphere of radius 2r/3 by the factor 9/4 to get the surface area of the sphere of radius r.

Here is the calculations,

$V=2\pi h\ast \pi {r}^{2}/2={\pi}^{2}{r}^{2}h$

$S=9/4\ast 2\pi h\ast \pi (2r/3)=3{\pi}^{2}rh$

Since we don't know h, I simply divide them to cancel it out and get, $V/S=r/3.$

I wonder if it is possible to derive surface area and volume of a sphere seperately using techniques involving pappus' theorem.

I did some calculation and found out the ratio of surface area and volume. Here is my work,

1. My key observation is finding out that the centroid of a semidisk of raidus r is also the centroid of a semicircle of raidus $2r/3$ when the centers coincide. (By "centers" I mean the centers of full circle and disc induced by half of them.)

2. I sliced the semidisk identical pieces of triangles(infinitely many) and by locating each triangle's centroid I form a semicircle of radius $2r/3$ which must have the same centroid with semidisk.

3. Then let's say our centroid is located h distance above the center. We still don't know it.

4. Using the theorem, the circular path taken by the centroid of the semidisk times the area of the centroid should give the volume of the sphere of radius r. And similarly, the circular path taken by the same centroid of the semicircle times the arc length of semicircle should give the surface area of a sphere of radius 2r/3.

5. We know that when the radius increases with a proportion, corresponding surface area will also increase with the square of that proportion. Thus, we need to multiply the surface area of sphere of radius 2r/3 by the factor 9/4 to get the surface area of the sphere of radius r.

Here is the calculations,

$V=2\pi h\ast \pi {r}^{2}/2={\pi}^{2}{r}^{2}h$

$S=9/4\ast 2\pi h\ast \pi (2r/3)=3{\pi}^{2}rh$

Since we don't know h, I simply divide them to cancel it out and get, $V/S=r/3.$