applied linear algebra math 5112/6012herronda/linear_algebra/...applied linear algebra subspaces of...

Subspaces of Euclidean Space Rn

Applied Linear AlgebraMATH 5112/6012

Applied Linear Algebra Subspaces of Rn Chapter 3, Section 5 1 / 12

What is a subspace?

Let V be a collection of vectors in Rn. (For example, V could be asolution set to some equation, or it could be all the vectors that have thirdcoordinate -7.)

We say that V closed with respect to scalar multiplication if and only ifwhenever ~v is in V and s is any scalar, then s~v is also in V. For example,if V = Span{~v} (for some ~v in Rn), then V is closed with respect toscalar multiplication. In fact, if V = Span{~v1, ~v2, . . . , ~vp} (for any~v1, ~v2, . . . , ~vp in Rn), then V is closed with respect to scalar multiplication.

We say that V closed with respect to vector addition if and only ifwhenever ~u and ~v are in V, then ~u + ~v is also in V. For example, ifV = Span{~v} (for some ~v in Rn), then V is closed with respect to vectoraddition. In fact, if V = Span{~v1, ~v2, . . . , ~vp} (for any ~v1, ~v2, . . . , ~vp inRn), then V is closed with respect to vector addition.

We call V a vector subspace of Rn if and only if . . .

What is a subspace?

We say that V closed with respect to scalar multiplication if and only ifwhenever ~v is in V and s is any scalar, then s~v is also in V.

For example,if V = Span{~v} (for some ~v in Rn), then V is closed with respect toscalar multiplication. In fact, if V = Span{~v1, ~v2, . . . , ~vp} (for any~v1, ~v2, . . . , ~vp in Rn), then V is closed with respect to scalar multiplication.

What is a subspace?

We say that V closed with respect to scalar multiplication if and only ifwhenever ~v is in V and s is any scalar, then s~v is also in V. For example,if V = Span{~v} (for some ~v in Rn), then V is closed with respect toscalar multiplication.

In fact, if V = Span{~v1, ~v2, . . . , ~vp} (for any~v1, ~v2, . . . , ~vp in Rn), then V is closed with respect to scalar multiplication.

What is a subspace?

We say that V closed with respect to vector addition if and only ifwhenever ~u and ~v are in V, then ~u + ~v is also in V.

For example, ifV = Span{~v} (for some ~v in Rn), then V is closed with respect to vectoraddition. In fact, if V = Span{~v1, ~v2, . . . , ~vp} (for any ~v1, ~v2, . . . , ~vp inRn), then V is closed with respect to vector addition.

What is a subspace?

We say that V closed with respect to vector addition if and only ifwhenever ~u and ~v are in V, then ~u + ~v is also in V. For example, ifV = Span{~v} (for some ~v in Rn), then V is closed with respect to vectoraddition.

In fact, if V = Span{~v1, ~v2, . . . , ~vp} (for any ~v1, ~v2, . . . , ~vp inRn), then V is closed with respect to vector addition.

What is a subspace?

We call V a vector subspace of Rn if and only if . . .Applied Linear Algebra Subspaces of Rn Chapter 3, Section 5 2 / 12

What is a subspace?

Let V be a collection of vectors in Rn.

We call V a vector subspace of Rn if and only if

~0 is in V,

V closed with respect to vector addition, and

V closed with respect to scalar multiplication.

Some simple examples:

V = {~0} is the trivial vector subspace

V = Rn is a vector subspace of itself (also kinda trivial)

V = Span{~v} (for any ~v in Rn)

V = Span{~v1, ~v2, . . . , ~vp} (for any ~v1, ~v2, . . . , ~vp in Rn)

A simple non-example:

V ={all ~v in R4 with third coordinate -7

}is not a subspace

What is a subspace?

Let V be a collection of vectors in Rn.We call V a vector subspace of Rn if and only if

~0 is in V,

}is not a subspace

What is a subspace?

~0 is in V,

}is not a subspace

What is a subspace?

~0 is in V,

}is not a subspace

What is a subspace?

~0 is in V,

}is not a subspace

What is a subspace?

~0 is in V,

}is not a subspace

What is a subspace?

~0 is in V,

}is not a subspace

What is a subspace?

~0 is in V,

}is not a subspace

What is a subspace?

~0 is in V,

}is not a subspace

What is a subspace?

~0 is in V,

}is not a subspace

What is a subspace?

~0 is in V,

}is not a subspace

What is a subspace?

~0 is in V,

}is not a subspace

More Examples—Which are, or are not, vector subspaces?

For each V, decide whether or not V is closed with respect to scalarmultiplication and/or closed with respect to vector addition.

V ={[x

] ∣∣∣ x + y = 1}

~0 not in V, so not VSS

V ={[x

] ∣∣∣ xy ≥ 0}

V not closed wrt vector add

V ={[x

] ∣∣∣ x ≥ 0 and y ≥ 0}

V not closed wrt scalar mult

V ={[x

] ∣∣∣ x + y = 1}

V ={[x

] ∣∣∣ xy ≥ 0}

V ={[x

] ∣∣∣ x ≥ 0 and y ≥ 0}

V ={[x

] ∣∣∣ x + y = 1}

V ={[x

] ∣∣∣ xy ≥ 0}

V ={[x

] ∣∣∣ x ≥ 0 and y ≥ 0}

V ={[x

] ∣∣∣ x + y = 1}

V ={[x

] ∣∣∣ xy ≥ 0}

V ={[x

] ∣∣∣ x ≥ 0 and y ≥ 0}

Figure: x + y = 1

V ={[x

] ∣∣∣ x + y = 1}

V ={[x

] ∣∣∣ xy ≥ 0}

V ={[x

] ∣∣∣ x ≥ 0 and y ≥ 0}

Figure: x + y = 1

V ={[x

] ∣∣∣ x + y = 1}

V ={[x

] ∣∣∣ xy ≥ 0}

V ={[x

] ∣∣∣ x ≥ 0 and y ≥ 0}

V ={[x

] ∣∣∣ x + y = 1}

V ={[x

] ∣∣∣ xy ≥ 0}

V ={[x

] ∣∣∣ x ≥ 0 and y ≥ 0}

~u + ~v

Figure: xy ≥ 0

V ={[x

] ∣∣∣ x + y = 1}

V ={[x

] ∣∣∣ xy ≥ 0}

V ={[x

] ∣∣∣ x ≥ 0 and y ≥ 0}

~u + ~v

Figure: xy ≥ 0

V ={[x

] ∣∣∣ x + y = 1}

V ={[x

] ∣∣∣ xy ≥ 0}

V ={[x

] ∣∣∣ x ≥ 0 and y ≥ 0}

V ={[x

] ∣∣∣ x + y = 1}

V ={[x

] ∣∣∣ xy ≥ 0}

V ={[x

] ∣∣∣ x ≥ 0 and y ≥ 0}

Figure: x ≥ 0 and y ≥ 0

V ={[x

] ∣∣∣ x + y = 1}

V ={[x

] ∣∣∣ xy ≥ 0}

V ={[x

] ∣∣∣ x ≥ 0 and y ≥ 0}

Figure: x ≥ 0 and y ≥ 0

Vector Subspaces—Basic Example

Recall that a collection V of vectors (in Rn) is a vector subspace (of Rn) ifand only if

~0 is in V,V closed with respect to vector addition

(~u, ~v in V =⇒ ~u + ~v in V)

V closed with respect to scalar mult

(s scalar , ~v in V =⇒ s~v in V)

Let V = Span{~v1, ~v2, . . . , ~vp}. Let’s show that V is closed wrt vectoraddition. Let ~u, ~v be vectors in V. This means there are scalarss1, s2, . . . , sp and t1, t2, . . . , tp with

~u = s1~v1 + · · ·+ sp~vp and ~v = t1~v1 + · · ·+ tp~vp

~u + ~v = (s1 + t1)~v1 + (s2 + t2)~v2 + · · ·+ (sp + tp)~vp

which is a vector in V.

Homework: Show that V is closed wrt scalar multiplication.

~0 is in V,

V closed with respect to vector addition

(~u, ~v in V =⇒ ~u + ~v in V)

~u + ~v = (s1 + t1)~v1 + (s2 + t2)~v2 + · · ·+ (sp + tp)~vp

~0 is in V,V closed with respect to vector addition

(~u, ~v in V =⇒ ~u + ~v in V)V closed with respect to scalar mult

~u + ~v = (s1 + t1)~v1 + (s2 + t2)~v2 + · · ·+ (sp + tp)~vp

~0 is in V,V closed with respect to vector addition (~u, ~v in V =⇒ ~u + ~v in V)

~u + ~v = (s1 + t1)~v1 + (s2 + t2)~v2 + · · ·+ (sp + tp)~vp

~0 is in V,V closed with respect to vector addition (~u, ~v in V =⇒ ~u + ~v in V)V closed with respect to scalar mult

(s scalar , ~v in V =⇒ s~v in V)Let V = Span{~v1, ~v2, . . . , ~vp}. Let’s show that V is closed wrt vectoraddition. Let ~u, ~v be vectors in V. This means there are scalarss1, s2, . . . , sp and t1, t2, . . . , tp with

~u + ~v = (s1 + t1)~v1 + (s2 + t2)~v2 + · · ·+ (sp + tp)~vp

~0 is in V,V closed with respect to vector addition (~u, ~v in V =⇒ ~u + ~v in V)V closed with respect to scalar mult (s scalar , ~v in V =⇒ s~v in V)

~u + ~v = (s1 + t1)~v1 + (s2 + t2)~v2 + · · ·+ (sp + tp)~vp

Let V = Span{~v1, ~v2, . . . , ~vp}. Let’s show that V is closed wrt vectoraddition.

Let ~u, ~v be vectors in V. This means there are scalarss1, s2, . . . , sp and t1, t2, . . . , tp with

~u + ~v = (s1 + t1)~v1 + (s2 + t2)~v2 + · · ·+ (sp + tp)~vp

Let V = Span{~v1, ~v2, . . . , ~vp}. Let’s show that V is closed wrt vectoraddition. Let ~u, ~v be vectors in V. This means

there are scalarss1, s2, . . . , sp and t1, t2, . . . , tp with

~u + ~v = (s1 + t1)~v1 + (s2 + t2)~v2 + · · ·+ (sp + tp)~vp

Homework: Show that V is closed wrt scalar multiplication.Applied Linear Algebra Subspaces of Rn Chapter 3, Section 5 5 / 12

Just saw that any V = Span{~v1, ~v2, . . . , ~vp} is both closed wrt vectoraddition and closed wrt scalar multiplication.

Example (Basic Vector SubSpace)

For any ~v1, ~v2, . . . , ~vp in Rn, Span{~v1, ~v2, . . . , ~vp} is a vector subspace.

In fact, every vector subspace can be expressed this way!

Example (Column Space of a Matrix)

The column space CS(A) of a matrix A is the span of the columns of A.

Thus is A is an m × n matrix, then CS(A) is a VSS of

If A =[~a1 ~a2 . . . ~an

], then CS(A) = Span{~a1, ~a2, . . . , ~an}.

If A =[~a1 ~a2 . . . ~an

The column space CS(A) of a matrix A is the span of the columns of A.Thus is A is an m × n matrix, then CS(A) is a VSS of

If A =[~a1 ~a2 . . . ~an

The column space CS(A) of a matrix A is the span of the columns of A.Thus is A is an m × n matrix, then CS(A) is a VSS of Rm.

If A =[~a1 ~a2 . . . ~an

The column space CS(A) of a matrix A is the span of the columns of A.Thus is A is an m × n matrix, then CS(A) is a VSS of Rm.

If A =[~a1 ~a2 . . . ~an

Column Space of a Matrix

Let A =[~a1 ~a2 . . . ~an

]be an m × n matrix; so,

each ~aj is in Rm.

The column space CS(A) of A is the span of the columns of A, i.e.,CS(A) = Span{~a1, ~a2, . . . , ~an}.

Three Ways to View CS(A)The column space CS(A) of A =

[~a1 ~a2 . . . ~an

CS(A) = Span{~a1, ~a2, . . . , ~an}

CS(A) = {~b in Rm∣∣ A~x = ~b has a solution}

CS(A) = Rng(T ) where Rn T−→ Rm is T (~x) = A~x

Let A =[~a1 ~a2 . . . ~an

]be an m × n matrix; so, each ~aj is in

[~a1 ~a2 . . . ~an

CS(A) = Span{~a1, ~a2, . . . , ~an}

Let A =[~a1 ~a2 . . . ~an

]be an m × n matrix; so, each ~aj is in Rm.

[~a1 ~a2 . . . ~an

CS(A) = Span{~a1, ~a2, . . . , ~an}

Let A =[~a1 ~a2 . . . ~an

The column space CS(A) of A is the span of the columns of A, i.e.,

CS(A) = Span{~a1, ~a2, . . . , ~an}.

[~a1 ~a2 . . . ~an

CS(A) = Span{~a1, ~a2, . . . , ~an}

Let A =[~a1 ~a2 . . . ~an

[~a1 ~a2 . . . ~an

CS(A) = Span{~a1, ~a2, . . . , ~an}

Let A =[~a1 ~a2 . . . ~an

[~a1 ~a2 . . . ~an

CS(A) = Span{~a1, ~a2, . . . , ~an}

Let A =[~a1 ~a2 . . . ~an

[~a1 ~a2 . . . ~an

CS(A) = Span{~a1, ~a2, . . . , ~an}

Let A =[~a1 ~a2 . . . ~an

[~a1 ~a2 . . . ~an

CS(A) = Span{~a1, ~a2, . . . , ~an}

Let A =[~a1 ~a2 . . . ~an

[~a1 ~a2 . . . ~an

CS(A) = Span{~a1, ~a2, . . . , ~an}

Three Ways to View the Column Space CS(A)The column space CS(A) of A =

[~a1 ~a2 . . . ~an

CS(A) = Span{~a1, ~a2, . . . , ~an}

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

T (~0) = ~0 ~b = T (~p)

[~a1 ~a2 . . . ~an

CS(A) = Span{~a1, ~a2, . . . , ~an}

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

T (~0) = ~0 ~b = T (~p)

[~a1 ~a2 . . . ~an

CS(A) = Span{~a1, ~a2, . . . , ~an}

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

T (~0) = ~0 ~b = T (~p)

[~a1 ~a2 . . . ~an

CS(A) = Span{~a1, ~a2, . . . , ~an}

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

T (~0) = ~0

~b = T (~p)

[~a1 ~a2 . . . ~an

CS(A) = Span{~a1, ~a2, . . . , ~an}

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

T (~0) = ~0 ~b

= T (~p)

[~a1 ~a2 . . . ~an

CS(A) = Span{~a1, ~a2, . . . , ~an}

~p~y = T (~x)

~y = A~x

CS(A) = Rng(T )

T (~0) = ~0 ~b

= T (~p)

[~a1 ~a2 . . . ~an

CS(A) = Span{~a1, ~a2, . . . , ~an}

~p~y = T (~x)

~y = A~x

CS(A) = Rng(T )

T (~0) = ~0 ~b = T (~p)

Vector Subspaces—Another Basic Example

Again, let A be an m × n matrix.

The null space NS(A) of A is

NS(A) = {~x∣∣ A~x = ~0};

just the solution set for the homogeneous equation A~x = ~0.

This is avector subspace of

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

Again, let A be an m × n matrix. The null space NS(A) of A is

NS(A) = {~x∣∣ A~x = ~0};

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

NS(A) = {~x∣∣ A~x = ~0};

just the solution set for the homogeneous equation A~x = ~0. This is avector subspace of

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

NS(A) = {~x∣∣ A~x = ~0};

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

NS(A) = {~x∣∣ A~x = ~0};

just the solution set for the homogeneous equation A~x = ~0. This is avector subspace of

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

NS(A) = {~x∣∣ A~x = ~0};

just the solution set for the homogeneous equation A~x = ~0. This is avector subspace of Rn.

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

NS(A) = {~x∣∣ A~x = ~0};

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

NS(A) = {~x∣∣ A~x = ~0};

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

NS(A) = {~x∣∣ A~x = ~0};

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

NS(A) = {~x∣∣ A~x = ~0};

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

A =[~a1 ~a2 . . . ~an

]an m × n matrix and Rn T−→ Rm is T (~x) = A~x

NS(A) = {~x∣∣ A~x = ~0} and

CS(A) = Span{~a1, ~a2, . . . , ~an}

= {~b in Rm∣∣ A~x = ~b has a solution}

= CS(A) = Rng(T )

{~x∣∣ A~x = ~b}

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

~b = T (~p)

A =[~a1 ~a2 . . . ~an

NS(A) = {~x∣∣ A~x = ~0} and

CS(A) = Span{~a1, ~a2, . . . , ~an}

= CS(A) = Rng(T )

{~x∣∣ A~x = ~b}

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

~b = T (~p)

A =[~a1 ~a2 . . . ~an

NS(A) = {~x∣∣ A~x = ~0} and

CS(A) = Span{~a1, ~a2, . . . , ~an}

= CS(A) = Rng(T )

{~x∣∣ A~x = ~b}

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

~b = T (~p)

A =[~a1 ~a2 . . . ~an

NS(A) = {~x∣∣ A~x = ~0} and

CS(A) = Span{~a1, ~a2, . . . , ~an}

= CS(A) = Rng(T )

{~x∣∣ A~x = ~b}

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

~b = T (~p)

A =[~a1 ~a2 . . . ~an

NS(A) = {~x∣∣ A~x = ~0} and

CS(A) = Span{~a1, ~a2, . . . , ~an}

= CS(A) = Rng(T )

{~x∣∣ A~x = ~b}

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

~b = T (~p)

A =[~a1 ~a2 . . . ~an

NS(A) = {~x∣∣ A~x = ~0} and

CS(A) = Span{~a1, ~a2, . . . , ~an}

= CS(A) = Rng(T )

{~x∣∣ A~x = ~b}

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

= T (~p)

A =[~a1 ~a2 . . . ~an

NS(A) = {~x∣∣ A~x = ~0} and

CS(A) = Span{~a1, ~a2, . . . , ~an}

= CS(A) = Rng(T )

{~x∣∣ A~x = ~b}

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

~b = T (~p)

A =[~a1 ~a2 . . . ~an

NS(A) = {~x∣∣ A~x = ~0} and

CS(A) = Span{~a1, ~a2, . . . , ~an}

= CS(A) = Rng(T )

{~x∣∣ A~x = ~b}

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

~b = T (~p)

A =[~a1 ~a2 . . . ~an

NS(A) = {~x∣∣ A~x = ~0} and

CS(A) = Span{~a1, ~a2, . . . , ~an}

= CS(A) = Rng(T )

{~x∣∣ A~x = ~b}

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

~b = T (~p)

Null Space and Column Space Example

Find the null space and column space of

1 −1 1 21 0 1 11 2 1 −1

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

What should we do now? How about row reducing A?

1 −1 1 21 0 1 11 2 1 −1

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

1 −1 1 21 0 1 11 2 1 −1

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

1 −1 1 21 0 1 11 2 1 −1

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

1 −1 1 21 0 1 11 2 1 −1

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

1 −1 1 21 0 1 11 2 1 −1

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

What should we do now?

How about row reducing A?

1 −1 1 21 0 1 11 2 1 −1

~y = T (~x)

~y = A~x

CS(A) = Rng(T )

What should we do now? How about row reducing A?Applied Linear Algebra Subspaces of Rn Chapter 3, Section 5 11 / 12

Vector Subspaces—Basic Fact

~0 is in V,

(~u, ~v in V =⇒ ~u + ~v in V)

If V is a vector subspace; ~v1, ~v2, . . . , ~vp in V; s1, s2, . . . , sp are scalars: thens1~v1, s2~v2, . . . , sp~vp all in V, so s1~v1 + s2~v2 + · · ·+ sp~vp is in V.

Any LC of vectors in a VSS V is a vector in V!

Basic Fact about Vector SubSpaces

Let V be a vector subspace. Suppose ~v1, ~v2, . . . , ~vp are in V.Then Span{~v1, ~v2, . . . , ~vp} lies in V.

~0 is in V,

(~u, ~v in V =⇒ ~u + ~v in V)

~0 is in V,

(~u, ~v in V =⇒ ~u + ~v in V)

~0 is in V,

V closed with respect to vector addition (~u, ~v in V =⇒ ~u + ~v in V)

~0 is in V,

(s scalar , ~v in V =⇒ s~v in V)If V is a vector subspace; ~v1, ~v2, . . . , ~vp in V; s1, s2, . . . , sp are scalars: thens1~v1, s2~v2, . . . , sp~vp all in V, so s1~v1 + s2~v2 + · · ·+ sp~vp is in V.

~0 is in V,

V closed with respect to scalar mult (s scalar , ~v in V =⇒ s~v in V)

~0 is in V,

V closed with respect to scalar mult (s scalar , ~v in V =⇒ s~v in V)If V is a vector subspace; ~v1, ~v2, . . . , ~vp in V; s1, s2, . . . , sp are scalars: then

s1~v1, s2~v2, . . . , sp~vp all in V, so s1~v1 + s2~v2 + · · ·+ sp~vp is in V.

~0 is in V,

V closed with respect to scalar mult (s scalar , ~v in V =⇒ s~v in V)If V is a vector subspace; ~v1, ~v2, . . . , ~vp in V; s1, s2, . . . , sp are scalars: thens1~v1, s2~v2, . . . , sp~vp all in V, so

s1~v1 + s2~v2 + · · ·+ sp~vp is in V.

~0 is in V,

V closed with respect to scalar mult (s scalar , ~v in V =⇒ s~v in V)If V is a vector subspace; ~v1, ~v2, . . . , ~vp in V; s1, s2, . . . , sp are scalars: thens1~v1, s2~v2, . . . , sp~vp all in V, so s1~v1 + s2~v2 + · · ·+ sp~vp is in V.

~0 is in V,

applied linear algebra math 5112/6012herronda/linear_algebra/...applied linear algebra subspaces of...

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