Suppose N in L(V) is nilpotent. Prove that the minimal polynomial fo N is z^m+1, where m is the length of the longest consecutive strong of 1st that appears on teh line directly aoce the diagonal in teh matrix of N with respect to any jordan basis for N.

your question is poorly written attached below is the complete and well written question

answer : The minimal polynomial has degree m + 1 hence Z^m+1

Step-by-step explanation:

Given that

N ∈ L(V) is nilpotent

attached below is the required prove

the correct answer is semicircle

step-by-step explanation:

the answer is semicircle because the shorter arc joining two points on the circumference of a circle is known as the minor arc, the larger arc is called the major arc, and a semicircle is a half circle, formed by cutting a whole circle along a diameter line.

hope i .

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet