Let a be an invertible real matrix of order. It is proved that there exists a positive definite symmetric matrix s and an orthogonal matrix U such that

Let a be an invertible real matrix of order. It is proved that there exists a positive definite symmetric matrix s and an orthogonal matrix U such that

Note: it is a positive definite symmetric matrix. So there is a positive definite matrix s in Exercise 2, such that =. Let's see how to take U.]

It is proved that a real symmetric matrix A of order n is positive definite if and only if there exists an invertible real symmetric matrix B, satisfying a = b * B

If a is positive definite, then there is an orthogonal matrix T, a = T ^ (- 1) Pt, where p = diag (A1 A) is the standard form of a, AI > 0. Mark q = diag (√ A1 If a = B ^ 2, B is real symmetric, similar to the above idea, there is an orthogonal matrix T, B = T ^ (- 1) RT, where r = diag (B1 B N) is the standard form of B

Let a and B be positive semidefinite matrices of order n, and prove that all eigenvalues of AB are nonnegative real numbers

Firstly, if a is positive definite, then AB is similar to a ^ {- 1 / 2} ABA ^ {1 / 2} = a ^ {1 / 2} Ba ^ {1 / 2}. According to the inertia theorem, the latter is semi positive definite and the eigenvalue is nonnegative
If a is positive semidefinite, then a + Ti is positive definite when t > 0, and the eigenvalue of (a + Ti) B is nonnegative, then let t - > 0 +, and the conclusion is obtained from the continuity of eigenvalue

If the a-th power of (3-A) x + 3 = 0 is a linear equation of one variable with respect to x, find the value of a-6

Because the equation is a linear equation of one variable
So a = 1
So a - 6 = 1 - 6 = - 5
A: the value of a - 6 is - 5

It is known that the n-th power of (m-1) X-2 = 1 is a linear equation of one variable with respect to X. what are the conditions that m and n should satisfy?

1219354638 ：
∵ (m-1) x ^ (n-2) = 1 is a linear equation of one variable with respect to X
∴m-1≠0
n-2=1
The solution is m ≠ 1, n = 3
The condition that m, n should satisfy is m ≠ 1, n = 3

If the m-2 power of X + 2 = 0 is a linear equation with one variable, then what is the M = value

m-2=1
m=3

(1)（X³-2x²+x-4)-(2x³-5x-4) （2)3(a²-2a-3)-5(-5a²+a-2)

（X³-2x²+x-4)-(2x³-5x-4) =x³-2x²+x-4-2x³+5x+4=x³-2x³-2x²-4+4=-x³-2x²3(a²-2a-3)-5(-5a²+a-2)=3a²-6a-9+25a²-5a+10=3a²+25a&#...

The remainder of 90.4 divided by 2.7 is ()
A.0 B.13 C.1.3 D.0.13

90.4 / 2.7 = 33 + 1.3
Choose C

10 times of 10 times of 10 times divided by the remainder of 7

1 2 3 4 5 6 7 8 9 10 11 12 ...
3 2 6 4 5 1 3 2 6 4 5 1 ...
For numbers like 1000... 00
The first row is the number of 0, and the second row is the remainder of 7
There are 10 zeros in 10 times of 10, and 100 zeros in 10 times
Six 0-cycles are the same as the remainder of 10000, i.e. 4

Look at the following sequence to find out what is the remainder of the 90th number divided by 3? 10，13，23，36，59，95，154，…

From the first number, each number divided by the remainder of 3 will get a rule: 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 2, 1, 0 There is a cycle for every 8 numbers, so & nbsp; 90 △ 8 = 11 2, that is, the remainder of the 90th number divided by 3 is the second number of the cycle, that is, the remainder of the 90th number divided by 3 is 1