Subspace Basis and Vector Spaces

• A basis for a subspace S of Rⁿ is a set of vectors in S that:
• Spans S.
• Is linearly independent.

Standard Basis

• The standard unit vectors e₁, e₂, ..., eₙ in Rⁿ are linearly independent and span Rⁿ.
• They form a basis for Rⁿ, called the standard basis.

Subspace Bases

• A subspace can have more than one basis.
• For example, R² has the standard basis {[1 0], [0 1]} and the basis {[2 -1], [1 3]}.
• The number of vectors in a basis for a given subspace is always the same.

Finding a Basis for Span(u, v, w)

• Given vectors u, v, and w that span S, they form a basis for S if they are linearly independent.
• If w can be expressed as a linear combination of u and v (e.g., w = 2u - 3v), then w can be ignored.
• S = span(u, v), and if u and v are linearly independent, they form a basis for S.
• Geometrically, u, v, and w lie in the same plane, and u and v can serve as direction vectors for this plane.

Finding a Basis for the Row Space of a Matrix

• Finding a basis for the row space of a matrix involves finding the reduced row echelon form of A.
• The row space of A is equal to the row space of R {row(A) = row(R)}.
• row(R) is spanned by its nonzero rows, making it easy to identify a basis.
• The staircase pattern forces the first three rows of R to be linearly independent.
• Basis for the row space of A is {[1 0 1 0 -1], [0 1 2 0 3], [0 0 0 1 4]}.

Finding a Basis by Transposing Vectors

• Transpose vectors u, v, and w to get row vectors, and form a matrix with these vectors as its rows.
• Reduce B to its reduced row echelon form.
• Use the nonzero row vectors as a basis for the row space.
• Since you started with column vectors, you must transpose again.
• Thus, a basis for span (u, v, w) is {[1 2 0], [0 1 -5/2]}.

Row Echelon Form

• Do not need to go all the way to reduced row echelon form, row echelon form is far enough.
• If U is a row echelon form of A, then the nonzero row vectors of U will form a basis for row(A).

Finding a Basis for the Column Space of a Matrix

• Transpose the matrix.
• The column vectors of A become the row vectors of Aᵀ, and apply the method of Example 3.45 to find a basis for row(Aᵀ).
• Transposing these vectors then gives a basis for col(A).
• A product Ax of a matrix and a vector corresponds to a linear combination of the columns of A with the entries of x as coefficients.
• A nontrivial solution to Ax = 0 represents a dependence relation among the columns of A.
• Elementary row operations do not affect the solution set, if A is row equivalent to R, the columns of A have the same dependence relationships as the columns of R.
• Columns of A have the same dependence relationships.
• What is the fastest way to find this basis? Use the columns of A that correspond to the columns of R containing the leading 1s.

Finding a Basis for the Null Space of a Matrix

• Find the solutions of the homogeneous system Ax = 0.
• Use the reduced row echelon form R of A, so all that remains to be done in Gauss-Jordan elimination is to solve for the leading variables in terms of the free variables.
• If the leading 1s are in columns 1, 2, and 4, solve for x₁, x₂, and x₄ in terms of the free variables x₃ and x₅ we get x₁ = -x₃ + x₅, x₂ = -2x₃ - 3x₅, and x₄ = -4x₅.
• Set x₃ = s and x₅ = t.
• Basis null(A) u and v.

Finding Bases

• Find the reduced row echelon form R of A.
• Use the nonzero row vectors of R (containing the leading 1s) to form a basis for row(A).
• Use the column vectors of A that correspond to the columns of R containing the leading 1s (the pivot columns) to form a basis for col(A).
• Solve for the leading variables of R x = 0 in terms of the free variables, set the free variables equal to parameters, substitute back into x, and write the result as a linear combination of f vectors (where f is the number of free variables). These f vectors form a basis for null(A).
• If we do not need to find the null space, then it is faster to simply reduce A to row echelon form to find bases for the row and column spaces. Steps 2 and 3 above remain valid (with the substitution of the word “pivots” for “leading 1s").

Dimension and Rank

• A subspace will has different bases, each basis has the same number of vectors.

The Basis Theorem

• Let S be a subspace of Rⁿ. Then any two bases for S have the same number of vectors.

Definition of Dimension

• If S is a subspace of Rⁿ, then the number of vectors in a basis for S is called the dimension of S, denoted dim S.
• The zero vector 0 by itself is always a subspace of Rⁿ.
• Any set containing the zero vector (and, in particular, {0}) is linearly dependent, so {0} cannot have a basis.
• Define dim {0} to be 0.

Dimensionality

• The standard basis for Rn has n vectors, dim Rⁿ = n.

Theorems

• The row and column spaces of a matrix A have the same dimension.
• The rank of a matrix A is the dimension of its row and column spaces and is denoted by rank(A).

Remarks

• The preceding definition agrees with the more informal definition of rank that was introduced in Chapter 2.
• The advantage of new definition is that it is much more flexible.
• The rank of a matrix gives us information about linear dependence among the row vectors of the matrix and among its column vectors.
• It tells us the number of rows and columns that are linearly independent (and this number is the same in each case!).
• Since the row vectors of A are the column vectors of Aᵀ, Theorem 3.24 has the following immediate corollary.

Nullity of a Matrix

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

In Other Words

• Nullity(A) is the dimension of the solution space of Ax = 0, which is the same as the number of free variables in the solution.
• We can now revisit the Rank Theorem (Theorem 2.2), rephrasing it in terms of our new definitions.

The Rank Theorem

• If A is an m × n matrix, then: rank(A) + nullity(A) = n
• If A is an m x n matrix, then: rank(Aᵀ) = rank(A)

Vector as a Basis

• According to the Fundamental Theorem, the vectors will form a basis for R³ if and only if a matrix with these vectors as its columns (or rows) has rank 3.

Show that the Vectors

• Given 3 vectors as columns of matrix and perform row operations on a matrix with vectors as column
• We see that A has rank 3, so the given vectors are a basis for R³ by (f) and (j).

A Theorem

• Let A be an m × n matrix.
• rank(AᵀA) = rank(A).
• The n × n matrix AᵀA is invertible if and only if rank(A) = n.

Coordinates

• A plane through the origin is a two-dimensional subspace of R³, with any set of two direction vectors serving as a basis.
• Basis vectors locate coordinate axes in the plane/subspace, in turn allowing us to view the plane as a “copy” of R².

Theorem Guaranteeing Unique Coordinates

• Let S be a subspace of Rⁿ and let B = {v₁, v₂, ..., vk} be a basis for S.
• For every vector v in S, there is exactly one way to write v as a linear combination of the basis vectors in B: v = c₁v₁ + c₂v₂ + ... + ckVk

Definition

• Let S be a subspace of Rⁿ and let B = {v₁, v₂, ..., vₖ} be a basis for S.
• Let v be a vector in S, and write v = c₁v₁ + c₂v₂ + ... + cₖvₖ.
• Then c₁, c₂, ..., cₖ are called the coordinates of v with respect to B, and the column vector [v]B = [c₁ c₂ ... cₖ]ᵀ is called the coordinate vector of v with respect to B.

What if

• Let E = {e₁, e₂, e₃} be the standard basis for R³.
• Find the coordinate vector of v = [2 7 4]T with respect to E.
• Solution Since v = 2e₁ + 7e₂ + 4e₃, [v]E = [2 7 4]T.

What Happens

• It should be clear that the coordinate vector of every (column) vector in Rⁿ with respect to the standard basis is just the vector itself.

Same Subspace

• In Example 3.44, we saw that u = [3 -1 5]T, v = [2 1 3]T, and w = [0 -5 1]T are three vectors in the same subspace (plane through the origin) S of R³ and that B = {u, v} is a basis for S.
• Since w = 2u - 3v, we have [w]B = [2 -3]T.

Linear Transformations:

• A transformation (or mapping or function) T from Rⁿ to Rᵐ is a rule that assigns to each vector v in Rⁿ a unique vector T(v) in Rᵐ.
• The domain of T is Rⁿ, and the codomain of T is Rᵐ. Indicate by T : Rⁿ → Rᵐ.
• For a vector v in the domain of T, the vector T(v) in the codomain is called the image of v under (the action of) T.
• The set of all possible images T(v) (as v varies throughout the domain of T) is called the range of T.

Transformation Notation

• Transformation is TA: R² → R³.
• The image of v = [1 1]T is w = [1 3 -1]T.

Transformation to Linear Transformation

• A transformation T : Rⁿ → Rᵐ is called a linear transformation if (1) T(u + v) = T(u) + T(v) for all u and v in Rⁿ and (2) T(cv) = cT(v) for all v in Rⁿ and all scalars c.

Exmaple

• Let's check that T is a linear transformation. To verify (1), we let u = [x₁ y₁]T and v = [x₂ y₂]T.
• Therefore, with T(u + v) = T([x₁ y₁]T + [x₂ y₂]T) = T([x₁ + x₂ y₁ + y₂]T) = [2(x₁ + x₂) - (y₁ + y₂) 3(x₁ + x₂) + 4(y₁ + y₂)] = = [(2x₁ - j₁) + (2x₂ - y₂) (3x₁ + 4₁) + (3x₂ + 4y₂)] = [3x₁ + 4y1] T