Calculate LiMn → 2 (2x ^ 2-3x + 1)

Calculate LiMn → 2 (2x ^ 2-3x + 1)

It should be X - > 2
The original formula is 3

limit of sequence
For the sequence {xn}, the limit of XN is a. prove that the limit of X2N is a, and the limit of X2N + 1 is a

From the title we know LIM (n →∞) xn = a
That is, xn is a convergent sequence
According to the theorem: any subsequence of convergent sequence is convergent, and the limit is the same
We can see that:
Both X (2n) and X (2n + 1) converge and the limit is a
This is the fastest way to prove, using a theorem
It is also possible to prove strictly, and it is also very simple:
Because LIM (n →∞) xn = a
Defined by:
When n > N, there are | xn-a | 0 and N > 0. When n > N, there are 2n > 2n > N, 2n + 1 > 2n + 1 > n
So immediately (according to the definition written above, as long as the subscript is larger than N, the following inequality holds)
|X(2n)-a|

Let a be an invertible real matrix of order n, and prove that the eigenvalue of a (at) is greater than 0. At is the transpose matrix of A
Let a be an invertible real matrix of order n, and prove that the eigenvalue of a (at) is greater than 0
At is the transpose matrix of A

That is to prove that AA ^ t is a positive definite matrix
Because for any n-dimensional column vector x, x ^ t (AA ^ t) x = (a ^ TX) ^ t (a ^ TX) > = 0,
If and only if and only if a ^ TX = 0 and a is invertible, that is, a ^ t is invertible
If and only if x = 0, then AA ^ t is positive definite and the eigenvalues are greater than 0

An advanced problem: A is a matrix of order n, R (a) = R

Consider the matrix P
0 I(n-1)
0 0
Fill in the next row with the lowest a, and then delete the lowest a
Matrix Q
0 0
I(n-1) 0
AQ removes the first column of a and adds a 0 column at the end
Obviously, PAQ moves a one space to the left and up, and then adds 0 to the right side and 0 to the bottom side
Consider the matrix A, since R (a) = R, there must be transformation matrices m, n such that
M ar n = a, where AR is its canonical form, the upper left corner is an identity matrix of order r, and all others are 0,
It can be seen from the front that ar = P P P... P e Q... Q, where the number of P and Q is n-r
MP, n-r-2 PQ, and QN are all rank n-1 matrices, and their product is a

To solve an advanced algebra problem: prove that the N-level matrix A is commutative with all N-level matrices, then a must be a quantity matrix

In this paper, we prove that the matrices commutative with all diagonal matrices are diagonal matrices, so a must be diagonal matrix
It is proved that a and all matrices (E (I, J)) with only one element 1 are commutative

A is an invertible matrix. It is proved that the inverse of the adjoint matrix of a is equal to the inverse of the adjoint matrix of A

Because | a | a inverse = a*
Then (a inverse) * = | a inverse | (a inverse) inverse = A / | a|
And (a *) inverse = (|a|a inverse) inverse = (a inverse) inverse / |a| = A / |a|
(the second one uses the formula (AA) inverse = a inverse / a)
So the two are equal

If a (n * n) is invertible, it is proved that the adjoint matrix A * is also invertible

First of all, no matter what
A (a *) = (a *) a = | a | I is inevitable
Now a is reversible, so | a | is not zero
So (A / | a |) (a *) = (a *) (A / | a |) = I
By definition, a * is reversible and its inverse is a / | a|

It is proved that if the adjoint matrix A * of square matrix A of order n is invertible, then a is invertible

To the contrary
Suppose a is irreversible, then | a | = 0
All a · a * = | a | e = 0
Because of the inverse of a *, right multiply the inverse of a * on both sides of the equation
Inverse of a = a · a * · a * = inverse of a · a * · a * = inverse of 0 · a * = 0
That is, a = 0
Then a * = 0 (according to the meaning of adjoint matrix)
It is in contradiction with a *
So, the assumption is wrong
So a is reversible
Twenty years of teaching experience, professional trustworthy!
Click "adopt answer" in the upper right corner

It is proved that if the square matrix A is invertible, then the adjoint matrix A * of a is also invertible

The square matrix A of order n is invertible, | a | ≠ 0
A A*=|A|E
A*=|A|A^(-1)
|A*|=|A|^(n-1)≠0
A * reversible

How to prove that transpose matrix of invertible matrix is also invertible
It is proved that the inverse matrix of a transpose is equal to the transpose of the inverse matrix of A

Because
A reversible
therefore
|A|≠0
and
|A|=|A^T|
therefore
|A^T|≠0
therefore
A ^ t is reversible
[A^(-1)]^TA^T
=(AA^(-1))^T
=E^T
=E
therefore
The inverse matrix of the transpose of a is equal to the transpose of the inverse matrix of A