Exam 2 Flashcards
To determine if a subset S is a subspace, apply the following steps.
Step 1. Check if 0 is in S. If not, then S is not a subspace.
Step 2. If you can show that S is generated by a set of vectors, then by Theorem 4.2 S is a subspace.
Step 3. Try to verify that conditions (b) and (c) of the definition are met. If so, then S is a subspace. If you cannot show that they hold, then you are likely to uncover a counterexample showing that they do not hold, which demonstrates that S is not a subspace.
Subspace
If A is an n x m matrix, then the solution to Ax=0 is called the null space of A and is denoted by null(A)
Null space of a matrix A
The kernel of T is the set of vectors x such that T(x)=0
Kernel of a linear transformation
A set B={u_1……u_m} is a basis for the subspace S if:
a. B spans s
b. B is linearly independent
Basis
Let S be a subspace of R^n. Then the dimension of S is the number of vectors in any basis of S
Dimension
The nullity of a matrix A is the dimension of the null space of A and is denoted by nullity(A)
Nullity of a matrix
Let A be a nxm matrix:
a. The row space of A is the subspace Rm spanned by the row vectors of A and is denoted by row(a)
b.The column space of A is the subspace of Rn spanned by the column vectors of A and is denoted by Col(A)
Row space and column space a matrix
The rank of a matrix A is the dimension of the row (or column) space of A, and denoted by rank(A)
Rank of a matrix A
Let A be an nxm matrix. Then the nonzero vector u is an eigenvector of A if there exists a scalar(lambda) such at
Au=(lambda)u
The scalar lambda is called the eigenvalue of A
Eigenvalue and Eigenvector
Let A be a square matrix with eigenvalue lambda. The subspace of all eigenvectors associated with lambda, together with the zero vector, is called the eigenspace
Eigenspace
The polynomial that we get from the det(A-lambda(I))
Characteristic Polynomial