watch: LA - 高山 04 | Rank of matrix

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Source video: 【俗说矩阵】初识矩阵的秩,从最简单的例子讲起!

Find the rank

Rank of a matrix 𝐀 equals to the number of the non-zero rows of a matrix in row echelon form (after performing elementary row operations).

Definition of the rank

determinant

Properties

  • The matrices appearing in the transforming process are able to reach the same row echelon form, so they have the same rank.
  • In another word, the elementary row operations don’t change the rank of matrices.
  • The rank of a matrix is no bigger than the number of rows or columns.

Thus, the solution of a linear equation system can be represented as its rank.

Solution judged by rank

Homogeneous linear equation system:

  1. Only zero solution: r(𝐀) = n, the rank of the coefficient matrix equals to the number of unknowns
  2. Exist non-zero solution: r(𝐀) < n,

Non-homogeneous linear equation system:

  1. Single unique solution: r(𝐀|𝐛) = r(𝐀) = n, the rank of the augmented matrix = the rank of the coefficient matrix = the number of the unknowns;
  2. Inifinitely many solutions: r(𝐀|𝐛) = r(𝐀) < n;
  3. No solution: r(𝐀|𝐛) ≠ r(𝐀)

r(𝐀) = 0

𝐀 = $[^{\_{0\ 0\ 0}} \_{^{0\ 0\ 0} \_{0\ 0\ 0}}]$

All of its elements are 0. There is no non-zero rows. So r(𝐀) = 0.

This is the so-called zero matrix noted as 𝐀 = 𝟎

r(𝐀) = 0 and 𝐀=𝟎 are equivalent, because if any one of elements is not 0, then there is a non-zero row resulting the r(𝐀) ≠ 0.

r(𝐀) = 1

𝐀 = $[^{\_{1\ 2\ 3}} \_{^{2\ 4\ 6}\_{3\ 6\ 9}}]$ ➔ $[^{\_{1\ 2\ 3}} \_{^{0\ 0\ 0}\_{0\ 0\ 0}}]$

  • The rows in the matrix of r(𝐀) = 1 are proportional to each other. In fact, the columns are also proportional.
  • Their ratio factor can be 0, and also there must be at least 1 element which is not zero in the matrix. Otherwise, it’s the zero matrix.

𝐚= [1 2 3] and 𝐛 = $[^{_1} \_{^2\_3}]$ are 1x1 matrix with rank = 1.

  • Vectors are the matrix with only 1 row or 1 column.

  • Row vector with at least 1 non-zero value is of rank=1.

  • Column vector has multiple rows with a single value. If it’s not a zero vector, the later rows can be reduced to 0 by adding the row1 multiplying with different factors. 𝐛 = $[^{_1} \_{^2\_3}]$ ➔ $[^{_1} \_{^1\_1}]$

  • For all the non-zero vector, no matter row vector or column vector, their rank = 1.

  • Vector 𝐚 ≠ 0 and r(𝐚) = 1 are equivalent.

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