Question

Linear Algebra,


So I have to prove this:


"Let a be a 1*n matrix and B an n*p matrix. Show that the matrix product aB can be written as a linear combination of the rows of B, where the coefficients are the entries of a"



Can you basically explain to me the problem? I don't want the solution just explain to me ? Thanks

Answer 1

@agent0smith

Answer 2

@amistre64

Answer 3

@Callisto

Answer 4

@dan815

Answer 5

@dumbcow

Answer 6

@ganeshie8

Answer 7

@hartnn

Answer 8

@hba

Answer 9

@phi

Answer 10

basically I don't understand what I need to "show"

Answer 11

Let a be a 1*n matrix

A is a 1 row, n column matrix ...

and B an n*p matrix.

B has n rows and p columns


Show that the matrix product aB can be written as a linear combination of the rows of B

a linear combination is a "sum" of terms; can you better
define what aB operation is?

where the coefficients are the entries of a

Answer 12

im thinking AB is an inner procduct; so you will most likely have to employ the defintion of an inner product

Answer 13

hmm does that have anything to do with
\[\sum\limits_{}^{} {something} \]

Answer 14

this type of thing ?

Answer 15

a linear combination of the vector <3,2,1> is 3i + 2j + k

Answer 16

but here I don't play with vectors or do I ?

Answer 17

I think I play with matrices

Answer 18

or whatnot

Answer 19

matrixes are built out of vectors

Answer 20

so for example a 5*5 matrice has 25 vectors ?

Answer 21

each column is a vector from a given span

Answer 22

a 5x5 has 5 column vectors

Answer 23

each column vector of a 5x5 matrix represents a spacial vector with 5 parts:\[\begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{pmatrix}\]

Answer 24

and 5 row vectors ?

Answer 25

or doesn't go that way ?

Answer 26

row vectors are something different ... but yes, you could look at it like that

Answer 27

spose we have a 1xn matrix; each column vector has 1 part; and there are n column vectors that span the space

Answer 28

\[A = \{a_1,a_2,a_3,a_4,...,a_n\}\]
each column vector in B has n parts
\[B = \{b_1,b_2,b_3,b_4,...,b_p\}\]
\[b_1=\begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\\.\\.\\.\\x_n\end{pmatrix}\]

Answer 29

ive got no good way to try to define the AB operation off the fly

Answer 30

I have already a definition of this thing here

Answer 31

\[\sum\limits_{k = 1}^n {aikbkj} \]

Answer 32

i and k are "smaller" letters down a and b

Answer 33

but yet that definition I know it

Answer 34

i = 1 by default since A only has 1 row

Answer 35

ok

Answer 36

\[\sum\limits_{k = 1}^n {a_{1k}~b_{kj}}~:~j=\{1,2,3,4,...,p\}\]

Answer 37

im not sure about how i stated j tho

Answer 38

hmm what do you mean by this part ? : j={1,2,3,4,...,p}

Answer 39

ok ok I got what you mean

Answer 40

thats the j of B

Answer 41

yeah, since there are p column vectors in total, the inner product will contain 1 column with n parts; but each jth part is a calcuation of Arow1, and B column j

Answer 42

do you base this on a being 1xp and also being the "first" in the inner product operation ? aB

Answer 43

\[\begin{pmatrix}a_1~a_2~a_3\end{pmatrix}\begin{pmatrix}b_{11}~b_{12}\\b_{21}~b_{22}\\b_{31}~b_{32}\\\end{pmatrix}=\begin{pmatrix}a_1b_{11}+a_2b_{12}+a_3b_{13}\\a_1b_{21}+a_2b_{22}+a_3b_{23}\\a_1b_{31}+a_2b_{32}+a_3b_{33}\\\end{pmatrix}\]

Answer 44

ok

Answer 45

I see

Answer 46

in trying to code that out, i did not produce the "second" column vector ....

Answer 47

on which of the 3 matrices ?

Answer 48

ah the third, its np

Answer 49

\[\begin{pmatrix}a_1~a_2~a_3\end{pmatrix}\begin{pmatrix}b_{11}~b_{12}\\b_{21}~b_{22}\\b_{31}~b_{32}\\\end{pmatrix}=\begin{pmatrix}(a_1b_{11}+a_2b_{21}+a_3b_{31})~~(a_1b_{12}+a_2b_{22}+a_3b_{32}) \end{pmatrix}\]

Answer 50

ok

Answer 51

so apparently thats aB

Answer 52

inner product

Answer 53

AB should have 1 row, with p columns

each column is a linear combination of the proper parts :)

Answer 54

"proper parts" ?

Answer 55

yeah, how to prove it tho ... im so far removed from any thrms and such to know what would make that more useful of simpler

Answer 56

yes, an inner product is specifically defined; and the proper parts need to be used appropriately to fit the definition

Answer 57

I can prove it just fine, the problem is

Answer 58

that I don't quite understand what I have to prove :D

Answer 59

I mean I understand kind of , but not entirely

Answer 60

prove that AB is properly defined: A has n columns, B has n rows ..... they match and can be worked

is the result a linear combination? that will have to be textbook defined to me

Answer 61

so the aB has is a linear combination of the rows of B

Answer 62

but how can it even be a linear combination of any sort since its a matrix * matrix = matrix (not linear combination)

Answer 63

:D

Answer 64

so basically I will find a 1xp AB matrix

Answer 65

that's x + 2y + 3z ..... +tp (thats the linear combination ....? )

Answer 66

if memory serves; a linear combination is defines as:

For a given span: v1, v2, v3, ..., vk

a linear combination is: av1 + bv2 + cv3 + ... + n vk

Answer 67

and THOSE coefficients has to be the elements of a = { blah , blah , blah }

Answer 68

so aB = a

Answer 69

let A = {a1, a2, a3, ... , an}
let B = {b1, b2, b3, ... , bp}

let C = {c1,c2,c3,...,cp}

show:

c1 = a1b11 +a2b12 +a3b13 + ... + anb1n
c2 = a1b21 +a2b22 +a3b23 + ... + anb2n
c3 = a1b31 +a2b32 +a3b33 + ... + anb3n
....
cp = a1bp1 +a2bp2 +a3bp3 + ... + anbpn

can be written as a linear combination of the rows of B, where the coefficients are the entries of a

Answer 70

class is starting .... gotta get