How to calculate the power of a matrix? You can't just multiply one by one. Is there a formula? If it is n-order, how to deal with it?

There is no formula, but there is a way to deal with it
To calculate a ^ k, generally speaking, first of all, we change a into Jordan canonical form a = P * J * P ^ {- 1}, and then we have
A^k=P*J^k*P^{-1}
It is quite easy to calculate J ^ k for Jordan canonical form

How to calculate the nth power of matrix (1),
1 1
1 1 is this matrix, which is two rows and two columns

You should be able to use mathematical induction
A=（1 1 ；1 1）
A^2=（2 2 ；2 2）
A^3=（4 4 ；4 4）=（2^2 4 ；4 4）
A^4=（8 8 ；8 8）=（2^3 8 ；8 8）
We should be able to see the law, that is
A ^ n is 2 ^ (n-1)

Matrix A, a = 1 200 1 200 1 to find a ^ n, (the nth power of a)
Matrix A, a = first row 1 0
The second line 0,1,
The third line is 0,0,1
Find a ^ n, (the nth power of a)
When n > = 3, n ^ n=
From the binomial theorem, we get
A^ n = I + n(2N) + ( n(n-1)/2) (2N)^2
= 1 2n 2n(n-1) \ 0 1 2n \ 0 0 1
And binomial theorem

Since a = I + 2 N, where I is the unit matrix, n = 0 10 / 0 01 / 0 00
It is well known that n ^ 2 = 0 01 / 0 00 / 0 00, and n ^ n = 0 when n > = 3
From the binomial theorem, we get
A^ n = I + n(2N) + ( n(n-1)/2) (2N)^2
= 1 2n 2n(n-1) \ 0 1 2n \ 0 0 1
N ^ 3 = 0 can be checked by yourself. Obviously, when k > = 3, n ^ k = n ^ 3, n ^ (K-3) = 0
Expand a ^ n = (I + 2n) ^ n according to binomial theorem, all the items containing (2n) ^ k (k > = 3) are zero, so there are only the first three items
As for what is binomial theorem, I don't need to say, if you don't understand, go back to high school textbooks or Baidu

It is proved that for any real symmetric matrix A, there always exists a sufficiently large real number t such that {Ti (I is the identity matrix) + a} is a positive definite matrix

Let the eigenvalues of a be λ 1, λ 2,..., λ n, then the eigenvalues of te + a are t + λ 1, t + λ 2,..., t + λ n. obviously, no matter how much λ I is, there is always a large enough t to make t + λ I > 0, that is, te + A is a positive definite matrix

A. B is a positive definite matrix and C is an invertible matrix. It is proved that A-B is a symmetric matrix

There's something wrong with your topic. Can't C be used? A and B are positive definite, and their differences are not necessarily symmetrical
A=(10 1; 2 10)
B=(100,4; 1,101)

Let a be a positive definite matrix and prove that there exists a positive definite symmetric matrix s such that a = s ^ 2

Is a positive definite matrix a symmetric matrix?
I read your proof of the following
A. It is proved that AB is a positive definite matrix if and only if AB = ba
Why? Because a and B are positive definite, so a ^ t = a, B ^ t = B?

You see the definition of positive definite matrix, the premise is a symmetric matrix!

A is a real symmetric matrix of order 3, and satisfies the condition that a ^ 2 + a = 0, the rank of a is known to be r (a) = 2, Q: what is the value of K, a + Ke is a positive definite matrix

k> 1. You can think of a as diag (- 1, - 1,0). In fact, it doesn't affect the similarity transformation, because symmetric matrices can always be diagonalized

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

Let m be a symmetric matrix with real coefficients of order n, if for any nonzero vector
X=(x_ 1,...x_ n) If Xmx ′ > 0, we call it positive definite
The positive definite matrix can be transformed into the standard form, i.e. the identity matrix
All symmetric matrices (or Hermitian matrices) with eigenvalues greater than zero are also positive definite matrices
Another definition: a real symmetric matrix. Positive definite quadratic form f (x1, X2,...) The matrix A (a ′) is called positive definite matrix
Some distinguishing methods of positive definite matrix
According to the concept of positive definite matrix, there are the following methods to distinguish positive definite matrix
1. The necessary and sufficient condition for a positive definite symmetric matrix of order n is that all n eigenvalues of a are positive
Proof: If yes, then yes
∴λ＞0
On the contrary, there must be u-cause
That is: A is positive definite
From the above method, it is not difficult to find that a is a positive semidefinite matrix if and only if all the eigenvalues of a are nonnegative
Eigenvalues are calculated on the main diagonal, you know

Must the elements on the diagonal of a symmetric positive definite matrix be the same?

No, for example, all diagonal matrices satisfying that diagonal elements are positive numbers are symmetric and positive definite

How to calculate the power of a matrix? You can't just multiply one by one. Is there a formula? If it is n-order, how to deal with it?