Of linear algebraic thinking and understanding

Foreword

Just read the "Khan Academy open class - the nature of linear algebra", more than a step on the understanding of linear algebra. Although the university had when it required courses, and after graduation deliberately read the yellow book cover of "Linear Algebra and Its Applications" This more famous foreigners edited books, however, before this course until the video to see, still did not understand a few basic concepts of linear algebra, such as what are linearly related, do not understand why the need to solve the eigenvalues ​​and eigenvectors. Now could not help a bit sorry Why did I teach this class university teachers do not use such popular intuitive way to teach it? Why I did not notice before these high-quality online resources?

This is not directly handling the "essence of linear algebra" The door's open class, but elaborate thinking and understanding after I finish this course in different languages on the same ground. If after reading all readers in this video door open class, look at this article, and you can find resonance in this, then that to some extent you understand the contents of the course. So, before reading this article, take a look at these suggestions to see the door open class b station.

I understand and thinking

1. Why do we prefer to use a feature vector matrix as a base coordinate this matrix it? If the basis vectors are eigenvectors, what happens?

Not all of the eigenvectors of the matrix A has, nor are perpendicular to each other among all the feature vectors for the second point, it does not affect our eigenvector matrix A use as its base vector. Basis vectors need not necessarily perpendicular to each other. According to a characteristic feature vectors, feature vectors always with one corresponding matrix A, so we said feature vector A. Therefore, when the vector v in a linear variation through the matrix A (i.e. Av), all feature vectors of vector A and vector collinear with these variations do not depart from the original line where the occurrence, is likely to change only the telescopic . This characteristic equation from the eigenvectors and eigenvalues ​​Ax = $\lambda$ x can be seen, where x indicates a feature vector, $\lambda$ denotes the current feature vector as a feature value at x. Intuitive understanding of this formula are: role of matrix A x with constant effect $\lambda$ same function of x effects. $\lambda$ variation x changes occurring or stretching direction of a vector, because $\lambda$ is a constant.
If some eigenvectors of the matrix A as we basis vectors, basis vectors are orthogonal to these assumptions, i.e. pairwise orthogonal, they form a new coordinate system Z, then any coordinate system and the coordinate Z is vector collinear axis linear change matrix a are just a telescopic change. Therefore, if the basis vectors are eigenvectors, then, a change in a linear vector v matrix A (Av) in easily calculated feature values for the diagonal matrix composed v is multiplied, if the basis vectors are Orthogonal. As we know, the value of each dimension of a coordinate system of the reference axes are all vectors of the coordinate scale above, for example a value of the first dimension increased by vector b 1, b is a vector corresponding to said first dimension value with respect to a first coordinate axis of a scale increases (reference relationship), the value of each coordinate axis "a scale" with respect to the old coordinate system may be different, if the old coordinate system the new metric Z coordinate system a scale of each coordinate axis lengths, respectively, equal to the characteristic values as a coordinate group corresponding to the feature vectors. So we can say that the new coordinate system values of all the points are based on a new scale of coordinate axes Z coordinate system as the reference. After understanding that, now, a new vector in the coordinate system Z v [v1, v2, v3] varies linearly through the matrix A, the value after the change is equal to the vector v v [v1 * $\lambda 1$ , v2 * $\lambda 2$ , v3 * $\lambda 3$ ] (old scale used to measure the coordinate system).

2. We use the eigenvectors of the matrix A is formed as a new coordinate system Z2 coordinate group, these groups are not necessarily orthogonal vectors. We want the vector v rotated 90 degrees counterclockwise in Z2, the operational thinking what is it?

There are two coordinate systems, we started a coordinate system used to represent the matrix A, which is a standard coordinate system, i.e., a vector group are orthogonal to each metric dimension and each is 1, Z1 represents; second coordinate Department is Z2. It should be noted that the measurement matrix A and vector v are using the Z1, that we are all vectors and matrices are expressed using the Z1. Popular terms using the Z1 "language" expressed. Now we hope matrix A is rotated 90 degrees in the Z2 coordinate system, is the need to translate into Z2 A language, then rotated 90 degrees, then translated back. Why should translate back? Because we represent all of the vector or matrix are using the Z1 language. So there is a formula ( $P^{-1} B P v$), calculated order is right to left, where v is the need to rotate a vector, P is the vector of the new coordinate system, represents a one dimension, B represents a linear variation of the rotation matrix of 90 degrees, $P^{-1}$ represents an inverse matrix of the matrix P. Understand this formula is, after the first v translated into the language Z2, rotated 90 degrees, then we translate it into the current language Z1. In particular, the first column of B $[0, 1]^T$ , the second column $[-1， 0]^T$ . Concluded that the expression $P^{-1}BP$implies a mathematical metastasis.

3. Why can by solving the equation det when solving eigenvalues ​​(A- $\ Lambda I$ ) = 0, where det is the determinant.

From the geometric point to understand determinant, showing that the various dimensions of the sheets into the space where the word "Zhang" is not a parallel vector conceivable on a straight line from the origin of the two formed in two-dimensional space area of ​​the quadrangle; in three space, starting from the origin is conceivable three parallel tetragonal not in the same vector on the plane formed. Parallelogram matrix A two-dimensional vector space formed in the area of ​​DET is (A) values. In the three-dimensional space, three-dimensional volume cube vector matrix A is formed DET value (A) of. More dimensions words, and so on. However, the situation may arise that, in the two-dimensional space, the two vectors A on the same straight line, so that A can not span a parallelogram, so that the area is zero, namely det (A) == 0 ; Similarly, in three dimensions, three if the vector a on the same plane, then they can not span a cube, so its volume is zero, i.e., det (a) == 0. Back problems, according to the calculated characteristic value with (A- $I \ lambda$ ) x = 0, since x is a vector of all zeros can not be, in order that this equation holds, it is necessary to (A- $I \ lambda$ ) is 0, i.e. the matrix (A- $I \ lambda$ amount) of each of the vector space spanned is 0 (the amount represents the area or volume or higher dimensional volume), then this amount is det ((A- $I \ lambda$ )) == 0.

4. Why can put a diagonal matrix A considered: all basis vectors are eigenvectors of the matrix, the matrix of diagonal elements are the eigenvalues ​​do they belong?

The diagonal matrix, each column represents a group vector. Because in this group is the vector coordinate system, all vectors v A linear transformation are only allow a change in size of the value of each dimension of the vector v.

If there is a deviation in understanding, identifying and discussing welcome