In order to verify that a subset of R n is in fact a subspace, one has to check the three defining properties. That is, unless the subset has already been verified to be a subspace: see this important note below. Now let V be a subspace of R n. If V is the zero subspace, then it is the span of the empty set, so we may assume V is nonzero. Choose a nonzero vector v 1 in V. This process terminates after at most n steps by this important note in Section 2. The column space is defined to be a span, so it is a subspace by the above theorem.
We need to verify that the null space is really a subspace. In Section 2. We have to verify the three defining properties. Since Nul A satisfies the three defining properties of a subspace, it is a subspace. The column space and the null space of a matrix are both subspaces, so they are both spans.
The column space of a matrix A is defined to be the span of the columns of A. In other words, it is easier to show that the null space is a subspace than to show it is a span—see the proof above.
In order to do computations, however, it is usually necessary to find a spanning set. The null space of a matrix is the solution set of a homogeneous system of equations. For example, the null space of the matrix. Conversely, the solution set of any homogeneous system of equations is precisely the null space of the corresponding coefficient matrix. To find a spanning set for the null space, one has to solve a system of homogeneous equations.
The vectors attached to the free variables form a spanning set for Nul A. Email Address. Linear Algebra. Group Theory. Problem Each of the following sets are not a subspace of the specified vector space. Contents Problem Solution. Solution 1. Solution True or False problems of vector spaces and linear transformations — Problems in Mathematics. Basis of span in vector space of polynomials of degree 2 or less — Problems in Mathematics.
Leave a Reply Cancel reply Your email address will not be published. This website is no longer maintained by Yu. It only takes a minute to sign up. Connect and share knowledge within a single location that is structured and easy to search. That should be enough to get you started; see if you can check for closure under scalar multiplication on your own. Yes, in each case you need to determine if all three properties hold.
Property 1 fails. There can be no zero vector in the set. Property 2 fails. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams?
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