Line generation Summary 1 linear equations in Linear Algebra

Review... Copy linear algebra and Its Application

Linear Equations

1. Similar

X_1-2x_2 =-1

-X_1 + 3x_2 = 3

There are three situations

1. No solution 2. There is a unique solution 3. There is an infinite Solution

Consider two parallel lines, the intersection line, and the exact coincidence line.

Solving Equations

Primary Line Transformation (Multiplier, swap, multiply)

Two Problems of Linear Equations

1. Is there at least one solution?

2. If there is a solution, is it unique?

Line simplification, step matrix, augmented matrix, Principal Component

You can convert any linear equations into a tiered matrix by using a line elementary transformation similar to the process of solving equations.

1 0-5 1

0 1 1 4

0 0 0 0

Corresponding

X_1-5 X_3 = 1

X_2 + X_3 = 4

0 = 0

X_1 = 1 + 5x_3

X_2 = 4-X_3

X_3 is a free variable.

1. Is there at least one solution?

The last column of the augmented matrix is not the principal component column, that is, there is no [0... 0 B] B! = 0

2. If there is a solution, is it unique?

If 1 is compatible, there is a unique solution if there is no free variable, and if yes, there is an infinite solution.

 

Vector Equation

The above equation can be regarded as a solution [X_1... X_n] acts on a group of vectors to obtain the final result solution vector.

1 x_1 + 2 X_2 = 7

-2 x_1 + 5 X_2 = 4

-5 x_1 + 6 X_2 =-3

Can be viewedA_1= [1,-2,-5] ^ tA_2= [2, 5, 6] ^ tB= [7, 4,-3] ^ t

A_1X_1 +A_2X_2 =B

A major idea of linear algebra is that research can be expressed as a fixedVectorSet {v_1, V_2... V_p}Linear CombinationAll vectors.

Determining vector equation X_1V_1+ X_2V_2+... X_pV_p=Does B have a solution?

Equivalent to the judgment VectorBBelongSPAN {v_1... V_p}

This is equivalent to determining the Augmented Matrix[V_1 V_2... V_p B]Is there any solution to the linear equations.

 

Matrix equation Ax = B

The third perspective is the angle of matrix equations. Coming soonLinear Combination of VectorsSeeMatrix and Vector Product.

 

Ax = [A1 A2... An] X1 = x1A1+ X2A2+... + XnAn

X2

...

XN

SoFor linear algebra, solutions to linear equations can be viewed from three different perspectives.

1. Matrix Equation

2. vector equation

3. Linear Equations

 

The following propositions are equivalent:

A m * n matrix

In A. R ^ m, each vector B and Ax = B Have solutions.

In B. R ^ m, each B is a linear combination of column.

C. Generate R ^ m for each column of

D. Each column of A has a principal component position.

(Similar to a = 1 0 0

0 1 0

0 0 1

A_1 = [1 0 0] ^ t A_2 = [0 1 0] ^ t A_3 = [0 0 1] ^ t

R ^ 3 can be generated for each A_1, A_2, and A_3 columns)

 

Solution Set of Linear Equations

1. The homogeneous equation is similar to the Ax = 0 vector, that is, the ordinary solution must be a solution. Whether there are other solutions depends on whether the equation has free variables. Therefore, if there is an extraordinary solution, there must be more than one.
2. For non-homogeneous equations, The Ax = B solution can be expressed as W = P + V_H P is a special solution of Ax = B, and V_H is any solution of homogeneous equation AX = 0.

 

Linear Independence

Define a set of vectors {v_1 ,... V_p} linear independence is equivalent

X_1V_1+ X_2V_2... + X_pV_p=0Only ordinary solution set

If there is a permission C_1… which is not all 0... C_p makes C_1V_1+... C_pV_p=0The {v_1... V_p} vector group is linearly related.

 

Linear Independence of each column in matrix,If and only when AX = 0, There is only an ordinary solution.

 

Linear Correlation actually indicates that vector groups have redundancy, that is, at least one vector can be expressed by Linear Combinations of other vectors...

3 6

1 2*2 (* 1/2) of linear correlation between two columns of this matrix can be represented by one vector of the other.

If a group of vectors are linearly correlated, there is a non-zero weight. C_x corresponds to C_x v_x. v_x can be expressed by a linear combination of other vectors (move the other vectors to the right of the equation and divide them by C_x)

 

Linear transformation (ing)

AX=BIt can be seen that matrix A is an object and acts on the vector through MultiplicationX maps to a new vector B.

A = 1 0 0

0 1 0

0 0 0

Ax, for example, x = [1 1] ^ t can be seen as the point (, 1) in R ^ 3 projected to the x_1, x_2 coordinate plane (, 0)

Linear transformation matrix

The conversion of the unit vector (0 0 .. 1) of each dimension actually forms the transformation matrix. Refer to p71, p72.

Existence and Uniqueness

Full shot (value vectors all have mappings from the defined domain), single shot (one-to-one ing)

 

A = 1-4 8 1

0 2-1 3

0 0 0 5

This matrix is full-rank,Each row has a principal position, which means it can generate R ^ 3,For each vector in R ^ 3, The ing exists (full shot),Because there is a free variable, the ing from R ^ 4 to R ^ 3 is not a single shot., (That is mandatory ...)

 

T: R ^ N-> r ^ m linear transformation, T is a single shot and only has an ordinary solution when AX = 0That is to sayLinear Independence of columnAs shown in the preceding figure, the number of a 3*4 rows and 3 <Number of columns 4 (the number of vectors exceeds the number of each vector element) must be linearly related.

Because it exists if there is only an ordinary solution and it is not a single shot.X! =YTX= B tY= B T (X-y) = 0X-YMust be0 Vector, And assume X! = Y conflict

It may be r ^ 3-> r ^ 3, for example

A = 1 0 0

0 1 0

0 0 1

OrNot full shot, but ensure one-to-one

A = 1 0

0 1

0 0

R ^ 2-> r ^ 3 3*2 2*1-> 3*1It is not a single shot because it can only be mapped to a shape such as (A, B, 0), But Ax = 0 only has an ordinary solution, and Vector Linear Independence..

The following are some of my conclusions for your reference only.

If it is a flat-length m * n m <n, it is inevitable that the linear correlation is not a single shot, but it is not guaranteed that each line has a principal component position. If there is a principal component, it is full shot.

If the square is square, if each row has a principal position, full shot, and single shot, because there is no free variable. The opposite is not a full shot or a single shot.

If the vertical and long rows are m> N, it is impossible for each row to have a principal component position, which is not a full shot and R ^ m cannot be generated. If it is linear, the single shot and the non-single shot (not full rank, there are 0 rows in the step matrix, free variables, normal solutions for AX = 0, linear correlation between columns ).

Whether a single shot is determined by linear independence, and whether it is full or not depends on whether it is full rank.