Exam 2 Flashcards

1
Q

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.

A

Subspace

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2
Q

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)

A

Null space of a matrix A

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3
Q

The kernel of T is the set of vectors x such that T(x)=0

A

Kernel of a linear transformation

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4
Q

A set B={u_1……u_m} is a basis for the subspace S if:

a. B spans s
b. B is linearly independent

A

Basis

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5
Q

Let S be a subspace of R^n. Then the dimension of S is the number of vectors in any basis of S

A

Dimension

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6
Q

The nullity of a matrix A is the dimension of the null space of A and is denoted by nullity(A)

A

Nullity of a matrix

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7
Q

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)

A

Row space and column space a matrix

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8
Q

The rank of a matrix A is the dimension of the row (or column) space of A, and denoted by rank(A)

A

Rank of a matrix A

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9
Q

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

A

Eigenvalue and Eigenvector

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10
Q

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

A

Eigenspace

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11
Q

The polynomial that we get from the det(A-lambda(I))

A

Characteristic Polynomial

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