span f columns of AT ) = span( rows of A )

Lecture 15: Quick review from previous lecture Let A be an m ⇥ n matrix.

• The kernel of A is

ker A = {x 2 Rⁿ : Ax = 0}

• The image of the matrix A is the set containing of all images of A, that is, img A = {Ax : x 2 Rⁿ}

• The coimage of A is the image of its transpose, A^T. It is denoted coimg A:

coimg A = img A^T = {A^Ty : y 2 R^m} ⇢ Rⁿ

• The cokernel of A is the kernel of its transpose, A^T. It is denoted coker A:

coker A = ker A^T = {w 2 R^m : A^Tw = 0} ⇢ R^m

—————————————————————————————————

Today we will

• continue discussing Sec. 2.5 the kernel and image, coker, and coimg.

• discuss Sect. 3.1 Inner Products

- Lecture will be recorded -

—————————————————————————————————

• Exam (2/14, Wed.) is closed book and everyone needs to open camera.

• During the exam, you can see Exam 1 problems through 1) Canvas:

Assignments ! Exam 1

2) instructor’s share screen via Zoom (contains first couple of problems due to the limit of screen).

MATH 4242-Week 6-1 1 Spring 2021

= span(^{columns of A} f

.

= span f ^{columns of} AT) ⁼ span(^{rows of A})

24I

However, there is an alternate way of building a basis for coimg A.

This method will let us see a profound connection between“ img A” and “coimg A”.

Observation.

• Performing for Gaussian elimination: (a) adding a multiple of one row to an- other row; and (b) permuting the order of rows, we have

A |{z}!

row operations

U (row echelon form)

• Both row operations (a) and (b) above obviously do not change the row span (the row space of A).

Consequently, we have

Conclusion 1: The row echelon matrix U has exactly the same row space as the original matrix A.

Conclusion 2: Therefore, we can construct a basis for the (row space) coimg A

by bringing A to row echelon form using Gaussian elimination, and take the nonzero rows as the basis vectors.

Example 5. The same matrix as in Example 3:

A ! U = 0 BB B@

1 2 2 3 2

0 0 1 1 1

0 0 0 4 2

0 0 0 0 0

1 CC CA

Then a basis of coimg (A) (img (A^T)) is

MATH 4242-Week 6-1 2 Spring 2021

I

mmfxn

.

co.mg A ^{= span / w . .- - .,} Wnt ⁼ _span ^{/ u. --- um} }

.

e-

one

, 1. ^{ii. hi :D}

Fact 6: If the rank of A is r, the basis we construct for coimg A will have r vectors. Thus,

dim(img A) = dim(coimg A) = r

§ coker A

To build a basis for coker A, solve the n-by-m homogeneous system A^Ty = 0, and set each free variable to 1, and the others to zero.

*In other words, apply the exact same procedure as for finding a basis for ker A.

Q: What is the dimension of coker A?

It is the number of free variables in A^Ty = 0. Since A^T has m columns and rank r, there are m r free variables, hence

Fact 7: If A is an m ⇥ n matrix with rank(A) = r, then dim(coker A) = m r

Summary:

We can summarize what we’ve learned about the four fundamental subspaces in the following theorem, called the Fundamental Theorem of Linear Algebra:

**Again, a very useful (and surprising) aspect of this theorem is that the column space and row space of A have the same dimension, equal to the rank r of A.

MATH 4242-Week 6-1 3 Spring 2021

-

ing ^{A = span 1 columns of A} /

wing ^{A = span} f ^{rows of A})

o

Summary

Let A be an m ⇥ n matrix with rank(A) = r.

A |{z}!

row operations

U (row echelon form)

dim Vector Space Basis

n-r ker(A) Solve Ax = 0, each free variable gives a basis vector r img (A) columns of A where the pivots occur

r coimg (A) (1) columns of A^T where the pivots occur or (2) nonzero rows of U containing pivots

m-r coker (A) Solve A^Tx = 0, each free variable gives a basis vector Example 4: Consider the matrix

A = 0

@ 1 1 2 3

2 1 3 4

1 0 1 1

1 A

Find a basis for ker A, img A, coimg A, coker A, respectively.

MATH 4242-Week 6-1 4 Spring 2021

①

"

⑦^-210 O

A-

( ^{÷ I (}

① Aborting ( ( ^!) , ( f) ) ^{. landings)=2}

@ strong

f ^{( t}

^{( ÷ )} f.

③ AbasntwkaA^{: dm ( Kors ) = 4 - 2=2}

Find ^{X so that A x = O.} Free variables ^X,,X¢

(2) ^{: - X} z ^- X z ^- 2×4 ^{= 0} .

Xz ^{= -} _X} ^- 2×4 .

t ^{) :} X_{, t Xz t} 2 Xs ^t 3×4 ^{= 0}

X^{, = - X}_, ^- Xa

Ker A

=/ ( ¥44 ⁾

X, =/

, X4=0

( f

" '

f

(4)

AbasistwwkerA-kerlstt-d.mu/orA=

.

= m ^- V

AT ^{x = O =}

- f ^exercise.

. AT ^→

( )

^{¥ ;)}

A _basis ^for _Coker A ^is

f ^{( Y}

/

nAm#IR ^{- IR"}

KerA ing A

wing ^{A coke A}

3 Inner Products and Norms 3.1 Inner Products

§ Inner products in the Euclidean space Rⁿ

Definition: If x = (x₁, . . . , x_n)^T and y = (y₁, . . . , y_n)^T are any two vectors in Rⁿ, then we define their inner product, denoted hx, yi, by:

hx, yi = x¹y₁ + . . . + x_ny_n = Xn

i=1

x_iy_i

Note that

hx, yi = y^Tx ( = x^Ty)

As in R², if x = (x, y)^T is a vector, then the “Pythagorean Theorem” tells us that its length is given by p

x² + y², and is denoted by kxk = p

x² + y².

Definition: We will use this to define the length of vectors in Rⁿ and denote the length of a vector x by

kxk = p

hx, xi = q

x²₁ + . . . + x²_n We call kxk the norm of x.

If x 6= 0, then kxk > 0. In addition, we also have kxk = 0 , x = 0.

MATH 4242-Week 6-1 5 Spring 2021

cy_n^{,-- Yn)} ! ⁿ)

- (^x, ^---xn'

f !^!)

y

iffy ^cx.si

- = ×

¥

-

Example 1. If x = (1, 0, 1)^T and y = ( 2, 1, 2)^T, then (1) find kxk, kyk, hx, x + 3yi and also normalize x and y.

(2) Is hx, yi = hy, xi?

Q: Based on these observation on hx, yi on Rⁿ above, what properties do you think they should hold for a “general” inner product.

MATH 4242-Week 6-1 6 Spring 2021

4) _11×11 ⁼ Fto ^{= M2.}

Hyll ⁼ EEE ⁼ 59 ^{= 3}.

normalize ^X , if ^,

¥_, ^{-- ri}_'ll:L . ¥^{" --}

III )

Cx, xtsy ^{> =} ( ( f) ^, (1/+3/11)

If

-

( ^X , t 3C _{X. Y}> ⁼ 11×112+3( ( 11,1713

= 2-13 ^. O ^'- I

e) Yes ,

§ Abstract definition of general inner products

Definition: Let V be a vector space. An inner product on V is a functions that assigns every pairing two vectors x and y in V to obtain a real number, denoted

hx, yi,

such that for all u, v, w 2 V and scalars c, d 2 R, the following hold:

(1) Bilinearity:

hcu + dv, wi = chu, wi + dhv, wi, hu, cv + dwi = chu, vi + dhu, wi, (2) Symmetry: hv, wi = hw, vi,

(3) Positivity: hv, vi > 0 whenever v 6= 0. Moreover, hv, vi = 0 if and only if v = 0.

Definition: A vector space V equipped with a specific inner product is called an inner product space.

The associate norm of a vector v 2 V is defined as kvk = p

hv, vi.

In other words, an inner product space V that is a vector space equipped with an additional way of pairing two vectors x and y in V to obtain a real number, denoted hx, yi.

Example 2. We have known that the inner product on Rⁿ defined earlier by hx, yi =

Xn i=1

x_iy_i satisfies these three axioms.

MATH 4242-Week 6-1 7 Spring 2021

=

( ^{By 13 ) , Hull 20).}

Vector space ^{t "inner} product ^{" inner product}

space

0

Example 3. Show that for all vectors x and y in an inner product space V , kx + yk² +kx yk² = 2(kxk² + kyk²)

§ The same vector space V can have many di↵erent inner products.

For example, while we originally equipped Rⁿ with the standard inner product hx, yi = P_n

i=1 x_iy_i, we can also define “other” inner products on Rⁿ as well. See discussions below.

Example 4.(Another inner products on Rⁿ) If c1, . . . , cn are positive numbers, we can define

hx, yi = c1x1y1 + . . . + cnxnyn = Xn

i=1

cixiyi. (1) This is a legitimate inner product (check this as an exercise). It is called a weighted inner product, with weights c₁, . . . , c_n.

Observe that while we can write the ordinary inner product on Rⁿ as x^Ty, we can write the above weighted inner product as

MATH 4242-Week 6-1 8 Spring 2021

IIX ⁺y 112 ⁼ (_{Xty, Xty} > I^' ( x _, xty) tCy,x

Cx, X > + Cx

, y^{> +Cy, X> Ky, y})

(27 ^2-

= 11×112 ^t 2 Cx

, y) t 119112. -

"x ^-yi ⁼

To be continued !