University:
Whitman College
Course:
MATH-240 | Linear Algebra
Academic year:

2024
Views:

425

Pages:

1
Author:

beddgell23k

Section 4.6: The Rank of a Matrix We will use the concept of “rank” to develop and study the four fundamental subspaces associated to a matrix A. 1. Definition: The row space of A, Row(A) is the span of the rows of A (which is Col(AT )). 2. Theorem: If two matrices A and B are row equivalent, then there row spaces are equivalent. If B is in Echelon Form, then the nonzeros rows of B form a basis for both Row(A) and Row(B). We can use either the RREF of A or the Echelon Form of A. 3. Remark: Row operations do not preserve the linear dependence relations among the rows (although they do for the columns). This means, for example, that while the first k rows of B may be linearly independent, while the first k rows of A may not be. (In the row reduction process, all the rows of zeros are shuffled to the bottom). 4. Definition: The rank of matrix A is the dimension of the column space (i.e., the number of independent columns). 5. The Rank Theorem: • Rank of A = # of pivots in A • dim(Col(A)) = dim(Row(A)) or Rank of A = Rank of AT . • rank(A) + dim(Null(A)) = n or the number of pivots + number of free vars = n 6. Let A be an m × n matrix with rank k. Let B be the RREF of A, and D be the RREF of AT . The four subspaces and their dimensions are: Symbol Col(A) Basis found by Look for pivots in B, use cols of A Space IR k n n−k Null(A) Use B to solve Ax = 0 IR Row(A) Use the nonzero rows of B IRn T Null(A ) T Use D to solve A x = 0 7. We have the following additions to the Invertible Matrix Theorem: (m.) The cols of A form a basis for IRn (n.) Col(A) = IRn (o.) dim(Col(A)) = n (p.) Null(A) = {0} (q.) dim(Null(A)) = 0 8. If u, v ∈ IRn , then show that the n × n matrix uv T has rank 0 or 1. 1 IR Dim m m k m−k

Linear Algebra Section 4.6 The Rank of a Matrix

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