Let a be a positive definite matrix, and prove that all the elements on the diagonal of a are greater than zero Advanced algebra problems

If a is positive definite, then for any x ≠ 0, x ^ tax > 0
Take x = ε I, the ith component is 1, and the rest components are 0
Then ε I ^ TA, ε I = aii > 0, I = 1,2,..., n
So all the elements on the diagonal of a are greater than zero

What is scalar in linear algebra? Why are the elements on the main diagonal of positive definite matrix greater than 0?

The scalar matrix is a = AE
Where a is a constant and E is the identity matrix
All eigenvalues of a positive definite matrix are greater than zero,
The trace of matrix (i.e. the sum of main diagonal elements) = the sum of all eigenvalues > 0

The sum of two adjacent natural numbers is 97 And ．

(97-1) △ 2, = 96 △ 2, = 48; 48 + 1 = 49; answer: these two natural numbers are 48 and 49

The sum of two continuous natural numbers is 57. What are the two natural numbers respectively? (using equation solution)

Let the first natural number be x and the second X + 1
The equation is: x + (x + 1) = 57
The solution equation is: 2x = 56
X=28

If the sum of the first number and the second number is 47, then the third number is 47 The product of them is , and yes__ ．

The second number is 23 + 1 = 24, the third number is 24 + 1 = 25, the product of the three numbers is 23 × 24 × 25 = 13800, and the sum of the three numbers is 23 + 24 + 25 = 72

The reciprocal sum of two consecutive natural numbers is 1130. These two numbers are And ．

The number of one of the numbers is a, and the other number is a + 1, from the question meaning of & nbsp; 1A + 1A + 1 = 1130, a + (a + 1) a (a + 1) a (a + 1) (a + 1) (a + 1 + 1) = 1130, so a + (a + 1) = 1130, so a + (a + 1 = 11, so a + (a + 1) = 11, so the other number is another number is a + 1, and another number is a + 1, a + 1, from the question is a + 1, by the meaning of the question & nbsp; from the & nbsp; sense of & amp; 1A + 1A + 1A + 1A + 1 = 1130, a + 1, from the question meaning of & nbsp; the & nbsp; by the & nbsp; Question & nbsp; sense & nbsp; the & nbsp; answer is 5 and 6. The answer is 5 and 6. The answer is 5 and 6. The answer is 5 and 6. The answer is the answer is 5 and 6. The answer is 5 and 6. The 6

There is such a kind of number, they can write the square difference of two natural numbers, such as 3 = 2 ^ 2-1 ^ 2, which is called wisdom number, then there are several wisdom numbers in 1 ` 100

50 odd numbers
And then there are multiples of 4, all of which are 25
Oh, four can't. 24
74 in all, no more

Knowledge about wisdom number: in the natural number sequence, which number is the 1990 "wisdom number" starting from 1, and explain the reason,

First of all, your proposition is wrong. 1 is not a wisdom number. The first wisdom number is 3
The 1990 wisdom number is 3980
Because the form of wisdom number must be 2K + 1 or 4K, K ≥ 1
The 1990 "wisdom number" is when k = 1990 / 2 = 995
4k＝4x995＝3980

Proving that the square difference of two adjacent natural numbers must be odd

Because two adjacent natural numbers must be an odd number, an even number, even number square is even number, odd number square is j, the difference between odd number and even number must be odd number

Prove the following proposition: the sequence composed of the square difference of two adjacent natural numbers is continuous odd
Prove the following proposition: the sequence composed of the square difference of two adjacent natural numbers is continuous odd
For example:
1^-0^=1
2^-1^=3
3^-2^=5
4^-3^=7
5^-4^=9
6^-5^=36-25=11
7^-6^=49-36=13
8^-7^=64-49=15
9^-8^=81-64=17
10^-9^=100-81=19
11^-10^=121-100=21
12^-11^=144-121=23
13^-12^=169-144=25
14^-13^=196-169=27
15^-14^=225-196=29
16^-15^=256-225=31
17^-16^=289-256=33
18^-17^=324-289=35
19^-18^=361-324=37
20^-19^=400-361=39
.

(n+1)^-n^=2n+1
n^-(n-1)^=2n-1
You don't need me to prove that the two numbers are odd?
(2n+1)-(2n-1)=2

Let a be a positive definite matrix, and prove that all the elements on the diagonal of a are greater than zero Advanced algebra problems