Question

Given the rref(A)= U a 4x5 matrix and you are given column vectors of A: a1, a2 where rank(A) is 3 and a1, a2, and a4 are linearly independent. Is there a way to determine the remaining column vectors without using elementary row operations from U to a1, and a2?

Answer 1

This is a purely conceptual question. I don't see fit to include the actual example matrices. So I was thinking performing the Row operations were very tedious and that in theory you should be able to write the remaining column vectors in some form of linear combination.

Answer 2

I actually think that is the point of the question..

Answer 3

@amistre64 : tell your friends, and their friends..

Answer 4

i dont have any friends :(

Answer 5

it seems like the remaining vectors would form a nul space

Answer 6

Openstudy doesn't have a feature that lets you cash out your fans for friends? I mean you have 1,739 fans....you'd be rich with friends

Answer 7

:)

Answer 8

i forget what rank is in matrixes

Answer 9

number of linearly independent columns

Answer 10

rank(A)+dim(N(A))= n

Answer 11

yeah, that sounds familiar. and the remaining columns define the nul space. in this case its either a plane of a line depending on if the remaining vectors are dependant of each other or not

Answer 12

im not so sure theres a simple way to go about it other than row operations

Answer 13

i only have column vectors a1, a2 and rref(A)=U. Since there are 3 independent vectors and I only have two of them is it possible to determine the other column vectors of A?

Answer 14

oh okay

Answer 15

tedious, tedious

Answer 16

matrixes are tedious ...

Answer 17

Unfortunatly, I spent more time trying to find a way around the tediousness than actually just doing the operations.

Answer 18

so:\[U=\begin{pmatrix}
a_1&a_2&a_4&a_3&a_5\\
1&0&0&c_1&c_2\\
0&1&0&c_3&c_4\\
0&0&1&c_5&c_6\\
0&0&0&0&0
\end{pmatrix}\]

\[\begin{matrix}
a_1=&0+&-c_1 a_3& - c_2 a_5\\
a_2=&0+&-c_3 a_3& - c_4 a_5\\
a_3=&0+&1~a_3& +0~ a_5\\
a_4=&0+&-c_5 a_3& - c_6 a_5\\
a_5=&0+&+0~a_3& +1~a_5
\end{matrix}\]

Answer 19

so, depending on what your 3 independant vectors are, and how it tediums to the rref, you can define the nulspace in terms of multiples of the dim vectors

Answer 20

tediums?

Answer 21

that could either be tedious or ties but not sure

Answer 22

you say you are given 2 vectors, do you have to find the 3rd linearly independant one?

Answer 23

A=original matrix, U=rref(A), and I have a1, a2 which are the column vectors of A. I need to find a3, a4, a5 for A

Answer 24

give me the 2 vectors you got :)

Answer 25

a1= (2,1,3,1)^T and a2=(-1,2,-3,1)^T

Answer 26

U= [u1, u2, (2,3,0,0), u4, (-1,-2,5,0)] where u1,u2,u4 are lin. independent.

Answer 27

obviously u1, u2,u4 contains pivotal 1s

Answer 28

I'm honestly even having difficulty doing the row operations to find A

Answer 29

\[A=\begin{pmatrix}
a_1&a_2&a_4&a_3&a_5\\
2&-1&k_1&n_1&n_2\\
1&2&k_2&n_3&n_4\\
3&-3&k_3&n_5&n_6\\
1&1&k_4&n_7&n_8
\end{pmatrix}\]
\[U=\begin{pmatrix}
a_1&a_2&a_4&a_3&a_5\\
1&0&0&c_1&c_2\\
0&1&0&c_3&c_4\\
0&0&1&c_5&c_6\\
0&0&0&0&0
\end{pmatrix}\]



\[\begin{pmatrix}
1&0&\frac{2k_1+k_2}{5}&\frac{2n_1+n_3}{5}&\frac{2n_2+n_4}{5}\\

0&1&\frac{-k_1+2k_2}{5}&\frac{-n_1+2n_3}{5}&\frac{-n_2+2n_4}{5}\\

0&0&1&\frac{-3n_1+6n_3-15n_1+10n_5}{-3k_1+6k_2-15k_1+10k_3}&\frac{-3n_2+6n_4-15n_2+10n_6}{-3k_1+6k_2-15k_1+10k_3}\\

0&0&1&\frac{-3n_1+6n_3+5n_1-10n_7}{-3k_1+6k_2+5k_1-10k_4}&\frac{-3n_2+6n_4)+5n_2-10n_8}{-3k_1+6k_2+5k_1-10k_4}
\end{pmatrix}\]
\[2k_1+k_2=0=2k_2-k_1\]
\[2k_1+k_2=4k_2-2k_1~;~k_2=0~:~k_1=0\]

Answer 30

with any luck, i didnt mismath it along the way :)

Answer 31

oh god, after all that work. I'd hate to just go: i don't understand what you just did.

Answer 32

im thinking tho

Answer 33

\[A=\begin{pmatrix}
2&-1&k_1\\
1&2&k_2\\
3&-3&k_3\\
1&1&k_4
\end{pmatrix}\]


\[\begin{pmatrix}
1&0&2k_4-k_2\\
0&1&k_2-k_4\\
0&0&\frac{k_1+3k_2-5k_4}{3}\\
0&0&\frac{6k_2-9k_4+k_3}{6} \\
\end{pmatrix}\]

\[k_1+3k_2-5k_4 = 3\]


\[\begin{pmatrix}
1&0&2k_4-k_2\\
0&1&k_2-k_4\\
0&0&1\\
0&0&\frac{6k_2-9k_4+k_3}{6} \\
\end{pmatrix}\]

so, we have 4 equations and 4 unknowns

\[k_1+3k_2-5k_4 = 3\\ 2k_4-k_2=0\\k_2-k_4=0\\6k_2-9k_4+k_3=0\]
in order to turn this into\[\begin{pmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
0&0&0 \\
\end{pmatrix}\]

Answer 34

im just doing row operations to reduce it to row echelon form

Answer 35

from this i would say that k2=k4=0, k1=3, k3=0

Answer 36

wow kudos! My opinion on this problem: stupid problem. If they were to give you (hypothetically) A(x)=b where x is some solution that equals some b. Couldn't you solve this problem way quicker?

Answer 37

yes, with a guass jordon setup

Answer 38

If I could thank you for your time - other than some virtual medal, I would.

Answer 39

lol, im pretty much doing it for the practice, and the challenge :)

Answer 40

and since a3 and a5 are dependant i would assume they form some sort of linear combination of a1, a2 and a4

Answer 41

kudos for the exceptional attitude! hah

Answer 42

... i broke the wolf trying to dbl chk myself :)

Answer 43

ha

Answer 44

Not to take up anymore time but I am thinking if you had some solution for the system x and the vector b that =Ax. You can essentially write the matrix A as (a1, a2, a4)x=b since a3, and a5 are dependent, is that right?

Answer 45

yes, but that would not find a3 and a5

Answer 46

oh, right

Answer 47

wait so how would guass jordan setup go then? you'd have to write out the unknown vectors and just perform the operations and solve for the unknown after reducing?

Answer 48

do you know if there is a specific solution? or is it such that there are many solutions and you need only find one?

Answer 49

yes, that was one idea i had

another idea is a grimm schmidy run

Answer 50

grim schmidy? I know for sure I've never seen that.

Answer 51

its a way to develop an orthogonal basis given 3 non orthoganal vectors

Answer 52

oh, that must be further on in the course.

Answer 53

also, a cross product will produce another vector that is linearly independant

Answer 54

I've done calc 3 so it rings some bells

Answer 55

completed/studied**

Answer 56

another idea is to transpose the vectors into row vectors, add a vector <w,x,y,z>, or k subs if you like

row reduce such that you get a last row of zeros ....

Answer 57

Well, I must go chase Velocirapters out of my garden before they eat all the money on my money tree! Keep on keeping on, @amistre64 . I will be back with more tedious questions in due time.

Answer 58

Cheers

Answer 59

i dont think there is a unique results for this.

assume that we know U, and lets pick some random stuff for the last 2 vectors\[
U=\begin{pmatrix}
1&0&0&1&1\\
0&1&0&1&0\\
0&0&1&1&1\\
0&0&0&0&0
\end{pmatrix}\]

we can then work "backwards" into A that contains the given vectors

\[R_2+R_4=R_{new4}
\begin{pmatrix}
1&0&0&1&1\\
0&1&0&1&0\\
0&0&1&1&1\\
0&1&0&1&0
\end{pmatrix}\]

\[-3R_2+R_3=R_{new3}
\begin{pmatrix}
1&0&0&1&1\\
0&1&0&1&0\\
0&-3&1&-2&1\\
0&1&0&1&0
\end{pmatrix}\]

\[R_1+R_4=R_{new4}
\begin{pmatrix}
1&0&0&1&1\\
0&1&0&1&0\\
0&-3&1&-2&1\\
1&1&0&2&1
\end{pmatrix}\]

\[3R_1+R_3=R_{new3}
\begin{pmatrix}
1&0&0&1&1\\
0&1&0&1&0\\
3&-3&1&1&4\\
1&1&0&2&1
\end{pmatrix}\]

\[-\frac12R_2+R_1=R_{new1}
\begin{pmatrix}
1&-1/2&0&1/2&1\\
0&1&0&1&0\\
3&-3&1&1&4\\
1&1&0&2&1
\end{pmatrix}\]

\[2R_1=R_{new1}
\begin{pmatrix}
2&-1&0&1&2\\
0&1&0&1&0\\
3&-3&1&1&4\\
1&1&0&2&1
\end{pmatrix}\]

\[R_4+R_2=R_{new2}
\begin{pmatrix}
2&-1&0&1&2\\
1&2&0&3&1\\
3&-3&1&1&4\\
1&1&0&2&1
\end{pmatrix}\]

therefore one possible setup for A is
\[A=
\begin{pmatrix}
a_1&a_2&a_3&a_4&a_5\\
2&-1&1&0&2\\
1&2&0&0&1\\
3&-3&1&1&4\\
1&1&2&0&1
\end{pmatrix}\]

the swap in columns then creates a slightly altered U form, but it still works

http://www.wolframalpha.com/input/?i=rref%7B%7B2%2C-1%2C1%2C0%2C2%7D%2C%7B1%2C2%2C3%2C0%2C1%7D%2C%7B3%2C-3%2C1%2C1%2C4%7D%2C%7B1%2C1%2C2%2C0%2C1%7D%7D