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Material Type: Notes; Class: LIN ALG & DIFF EQNS; Subject: Mathematics; University: University of California - Berkeley; Term: Fall 2009;
Typology: Study notes
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Unless otherwise specified, A is an m×n real matrix, and vectors x are column vectors. Row vectors are written as xT^ , where x is the corresponding column vector. The nullspace of A, Null(A) ⊂ Rn, is the space of solutions of the homogeneous system Ax = 0 , the column space Col(A) ⊂ Rm is the span of the columns. There are two more subspaces associated to A, this time consisting of row vectors: the row space Row(A) ⊂ Rn, the span of the rows in A; and the left nullspace LNull(A) ⊂ Rm^ that we’ll meet below. (A quick and dirty definition is as the nullspace of the ‘flipped’ or transposed matrix AT^ : this is the matrix with switched indexing, (AT^ )ij = Aji.) Caution! These four subspaces are distinct in general, with no obvious relation among them. We will learn some subtle relations later. The reduced row echelon form of A is the matrix rref(A) produced from elementary row oper- ations, with the properties that
The columns containing pivots are called pivot columns, the others are the free columns. The nullspace of A agrees with that of rref(A), and can be parametrized as follows: the free variables can be chosen freely, and each equation in the reduced row system can then be used to solve for the corresponding pivot variable. A similar story applies to the inhomogeneous system Ax = b: it has the same general solution as the reduced system rref(A)x = b′, where [rref(A)|b′] is the reduced form of the augmented matrix [A|b]. The general solution can be parametrized by the same procedure. We will learn a strong uniqueness property of rref(A): it is completely determined by nullspace of A. (Similarly, the reduced form of the augmented matrix [A|b] is determined uniquely from the affine space of solutions of Ax = b.) A collection of vectors {v 1 ,... , vr} is linearly independent if the only expression of 0 as a linear combination of the vi is the one with 0 weights: that is,
kz v 1 + · · · + krvr = 0 ⇒ k 1 = k 2 = · · · = kr = 0.
A collection {w 1 ,... , ws} spans the linear subspace L ∈ Rn^ if every wi lies in L and every vector in L can be expressed as a linear combination of the wi. A linearly independent, ordered collection of vectors which spans L is called a basis. Each vector in L can be uniquely expressed as a linear combination of the basis elements (that is, the weights are uniquely determined). The main example is the standard basis e 1 ,... , en of Rn, the unit vectors on the coordinate axes.
c 1 x 1 + · · · + cnxn = 0 (2.1)
holds identically for all x ∈ Null(A). Seeing this in one direction is easy: a vector cT^ in the row space must be a linear combina- tion of the rows of A, therefore the equation (2.1) is a consequence of the equations in the homogeneous system, and must hold for any solution. To see the other direction, start with an equation (2.1), and subtract suitable multiples of the equations in the row-reduced system so as to cancel the coefficients of the pivot variables. The equation we now get involves only the free variables. But the free variables can be chosen freely, so this cannot hold identically on Null(A), unless it is the trivial equation 0 = 0! So, the original equation can only hold if it can be converted to 0 = 0 by subtracting rows of rref(A): that is, if cT^ ∈ Row(A).
d 1 b 1 + · · · + dmbm = 0 (2.2)
holds for all vectors b ∈ Col(A). That is, LNull(A) is the space of homogeneous linear equations which hold on all vectors b for which the system Ax = b is solvable. As above, one can show a converse: every common solution b of all equations in the left nullspace does in fact belong to the column space. So, a basis of LNull(A) allows us to check solvability by testing the linear equations in the basis on b, without performing a row-reduction. (Of course, finding a basis in the first place does require a row-reduction; see below).
The row space and left nullspace are naturally spaces of homogeneous linear equations, and they are best regarded as row vectors.
(3.1) Column space. A basis for Col(A) is given by the pivot columns of A. They form a basis, because any b ∈ Col(A) leads to a solvable system Ax = b, and then to one unique solution with zero-values for the free variables: but this gives a unique expression for b as a combination of the pivot columns.
(3.2) Row space. A basis for the row space of A is given by the non-zero rows in rref(A). Linear independence applies because the pivots are in distinct position: so, in any linear combi- nation, each weight can be read off in the corresponding pivot entry. These vectors span the row space because every row of A can be recovered as a linear combination of rows in rref(A): this just requires tracing back through the row-reduction algorithm.
(4.1) Equality. How can we check whether two subspaces L and L′^ of Rn^ are equal? There are several methods, depending how L and L′^ have been described.
(4.2) Inclusions. The Schubert basis does not help for checking a containment relation, such as L ⊃ L′: even when the latter holds, the Schubert basis for L will usually not contain that for L′. Instead, you can use several methods, depending how L, L′^ have been described.
Homework problem. For each of the following two matrices, find bases of the four subspaces. List all containment or equalities that hold between all pairs of subspaces of the same kind :