Deciding linear independence and computing spans and linear dependencies in a vector space

Let V be an m-dimensional vector space over a field F having basis E = \{e_i\}_{i=1}^m and let \{v_j\}_{j=1}^m be vectors in V. Let A be the m \times n matrix whose jth column is the coordinates of v_j expressed in terms of the ordered basis E. Let A^\prime be the reduced row echelon form of A.

  1. Let B be any matrix that is row equivalent to A, and let \{w_j\}_{i=1}^n be the elements of V whose coordinates (in terms of the ordered basis E) are the columns of B. Prove that if \sum \alpha_i v_i = 0, then \sum \alpha_i w_i = 0.
  2. Prove that the vectors whose E-coordinates are the pivotal columns of A^\prime are linearly independent, and that the vectors whose E-coordinates are given by the nonpivotal columns of A^\prime are linearly dependent on these.
  3. Prove that the vectors \{v_i\}_{i=1}^n are linearly independent if and only if A^\prime has n nonzero rows.
  4. By part (3), the vectors \{v_i\}_{i=1}^n are linearly dependent if and only if A^\prime has nonpivotal columns. In this case, compute all of the linear dependencies among the v_i.
  5. Prove that the dimension of the subspace W spanned by \{v_i\}_{i=1}^n is the row rank of A^\prime. Prove also that the columns of A corresponding to the pivotal columns of A^\prime form a basis for W.

  1. Say A and B are row equivalent via the invertible matrix P– that is, PA = B. Now P[v_1 | \cdots | v_n] = [Pv_1 | \cdots | PV_n] = [w_1 | \cdots | w_n], so that Pv_i = w_i for each i. Now if \sum \alpha_i v_i = 0, then P(\sum \alpha_i v_i) = 0, so that \sum \alpha_i Pv_i = 0, and thus \sum \alpha_i w_i = 0.
  2. Note that the pivotal columns of A^\prime are a subset of E, and thus are linearly independent. Moreover, any nonpivotal column has nonzero entries only in those rows containing a pivot; hence each of the nonpivotal columns of A^\prime is linearly dependent on the pivot columns.
  3. Suppose the v_i are linearly independent. Then by part (1) the columns of A^\prime are linearly independent. By part (2), all n columns of A^\prime are pivot columns, and so the column rank (hence row rank) of A^\prime is n. Using this previous exercise, A^\prime has n nonzero rows.

    Conversely, suppose A^\prime has n nonzero rows. Then the row rank (hence column rank) of A^\prime is n, so that the columns of A^\prime are linearly independent. By part (1), the columns of A are linearly independent.

  4. Using part (1), it suffices to find the linear dependencies among the columns of A^\prime. To that end, let u_i be the columns of A^\prime (in order) and suppose \sum \beta_i u_i = 0. Letting B = [\beta_1 | \cdots | \beta_n]^\mathsf{T}, we have A^\prime B = 0. We computed the solutions B of this matrix equation here; in particular, we may choose the \beta_i corresponding to nonpivotal columns of A^\prime arbitrarily, and the \beta_i corresponding to pivotal columns are then determined uniquely. Every such solution B then gives a linear dependence among the columns of A.
  5. Because the matrix P carrying A to A^\prime is invertible, the span of the v_i is the preimage of the span of the u_i, and the preimage of a basis is a basis. In (2) we saw that the pivotal columns of A^\prime are a maximal linearly independent set of columns of A^\prime, and thus form a basis of the column space of A^\prime. Hence the columns of A corresponding to pivotal columns of A^\prime form a basis of the column space of A. In particular, the dimension of this subspace is the row rank (i.e. number of nonzero rows) of A^\prime.
Post a comment or leave a trackback: Trackback URL.

Comments

  • deedeerr  On November 30, 2011 at 10:16 am

    “Moreover, any nonpivotal column has nonzero entries only in those rows not containing a pivot; hence each of the nonpivotal columns of A^\prime is linearly dependent on the pivot columns.”

    Shouldn’t it be in those rows that do contain a pivot?

    • nbloomf  On December 3, 2011 at 12:39 am

      Indeed. Thanks!

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: