University:
Massachusetts Institute of Technology
Course:
18.06SC | Linear Algebra
Academic year:

2024
Views:

367

Pages:

3
Author:

Lorena Carter

The four fundamental subspaces In this lecture we discuss the four fundamental spaces associated with a matrix and the relations between them. Four subspaces Any m by n matrix A determines four subspaces (possibly containing only the zero vector): Column space, C ( A) C ( A) consists of all combinations of the columns of A and is a vector space in Rm . Nullspace, N ( A) This consists of all solutions x of the equation Ax = 0 and lies in Rn . Row space, C ( A T ) The combinations of the row vectors of A form a subspace of Rn . We equate this with C ( A T ), the column space of the transpose of A. Left nullspace, N ( A T ) We call the nullspace of A T the left nullspace of A. This is a subspace of Rm . Basis and Dimension Column space The r pivot columns form a basis for C ( A) dim C ( A) = r. Nullspace The special solutions to Ax = 0 correspond to free variables and form a basis for N ( A). An m by n matrix has n − r free variables: dim N ( A) = n − r. Row space We could perform row reduction on A T , but instead we make use of R, the row reduced echelon form of A. ⎡ ⎤ ⎡ ⎤ � � 1 2 3 1 1 0 1 1 I F A = ⎣ 1 1 2 1 ⎦ → ··· → ⎣ 0 1 1 0 ⎦ = =R 0 0 1 2 3 1 0 0 0 0 Although the column spaces of A and R are different, the row space of R is the same as the row space of A. The rows of R are combinations of the rows of A, and because reduction is reversible the rows of A are combinations of the rows of R. The ﬁrst r rows of R are the ”echelon” basis for the row space of A: dim C ( A T ) = r. Left nullspace The matrix A T has m columns. We just saw that r is the rank of A T , so the number of free columns of A T must be m − r: dim N ( A T ) = m − r. The left nullspace is the collection of vectors y for which A T y = 0. Equiva lently, y T A = 0; here y and 0 are row vectors. We say “left nullspace” because y T is on the left of A in this equation. To ﬁnd a basis for the left nullspace we reduce an augmented version of A: � � � � Am×n Im×n −→ Rm×n Em×n . From this we get the matrix E for which EA = R. (If A is a square, invertible matrix then E = A−1 .) In our example, ⎡ ⎤⎡ ⎤ ⎡ ⎤ −1 2 0 1 2 3 1 1 0 1 1 EA = ⎣ 1 −1 0 ⎦ ⎣ 1 1 2 1 ⎦ = ⎣ 0 1 1 0 ⎦ = R. −1 0 1 1 2 3 1 0 0 0 0 The bottom m − r rows of E describe linear dependencies of rows of A, because the bottom m − r rows of R are zero. Here m − r = 1 (one zero row in R). The bottom m − r rows of E satisfy the equation y T A = 0 and form a basis for the left nullspace of A. New vector space The collection of all 3 × 3 matrices forms a vector space; call it M. We can add matrices and multiply them by scalars and there’s a zero matrix (additive identity). If we ignore the fact that we can multiply matrices by each other, they behave just like vectors. Some subspaces of M include: • all upper triangular matrices • all symmetric matrices • D, all diagonal matrices D is the intersection of the ﬁrst two spaces. Its dimension is 3; one basis for D is: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 0 1 0 0 0 0 0 ⎣ 0 0 0 ⎦,⎣ 0 3 0 ⎦,⎣ 0 0 0 ⎦. 0 0 0 0 0 0 0 0 7