Thus the is shown because there no cost-free variable that is straight independent. (I think)

My concern is is it additionally linear indepedent due to the fact that the vectors room not multiples the the very first vector.

You are watching: Determine if the columns of the matrix form a linearly independent set. justify your answer.


linear-algebra
share
mention
monitor
edited Sep 29 "14 at 19:54
Fernando Martinez
request Sep 19 "14 at 17:40
*

Fernando MartinezFernando Martinez
6,3561414 yellow badges6666 silver- badges9696 bronze badges
$\endgroup$
3
include a comment |

2 answers 2


active oldest Votes
4
$\begingroup$
To answer your question

My inquiry is is it also linear indepedent since the vectors room not multiples of the first vector.

No. Just due to the fact that the 2nd and third columns room not multiples of the first, that does not typical they space linearly independent. Take it for instance the matrix

$$\beginpmatrix 1 & 1 & 1\\1&2&3\endpmatrix$$

None that the columns room multiples that the others, however the columns do kind a linearly dependency set. You know this without any type of real work, due to the fact that $3$ vectors in $\carolannpeacock.combbR^2$ cannot form a linearly elevation set.


re-publishing
cite
follow
reply Sep 19 "14 in ~ 17:51
*

David PDavid ns
11.4k33 yellow badges2525 silver- badges4141 bronze badges
$\endgroup$
3
include a comment |
2
$\begingroup$
You are right that after row reducing and finding the there are no totally free variables (because every column has a pivot), then all of the columns space linearly independent.

By knowing the set of three vectors is linearly independent, we understand that the 3rd column vector cannot be composed as a linear mix of the an initial column vector and the second column vector. That is, there do not exist $c_1$, $c_2 \in \carolannpeacock.combbR$ such the $v_3 = c_1v_1 + c_2v_2$. ($v_i$ is mine notation because that the $i$-th pillar vector.)

Similarly, we recognize $v_2$ cannot be composed as a linear combination of $v_1$ and $v_3$.

We likewise know $v_1$ can not be created as a linear combination of $v_2$ and $v_3$.

That is what is intended by the $3$ vectors being linearly independent. You can"t write any type of one of them as a linear mix of the others.


re-superstructure
mention
follow
answered Sep 19 "14 at 17:52
*

laymanlayman
18.5k33 gold badges3131 silver- badges8282 bronze badges
$\endgroup$
add a comment |

your Answer


Thanks because that contributing solution to carolannpeacock.comematics ridge Exchange!

Please be sure to answer the question. Administer details and also share her research!

But avoid

Asking because that help, clarification, or responding to other answers.Making statements based upon opinion; back them up with recommendations or an individual experience.

Use carolannpeacock.comJax to layout equations. Carolannpeacock.comJax reference.

To find out more, view our advice on writing an excellent answers.

See more: What Proof Leads To Elizabeth’S Arrest, Elizabeth'S Arrest


Draft saved
Draft discarded

Sign increase or log in in


sign up making use of Google
sign up making use of Facebook
sign up making use of Email and Password
send

Post as a guest


name
email Required, however never shown


Post together a guest


name
email

Required, however never shown


short article Your price Discard

By click “Post her Answer”, friend agree come our regards to service, privacy policy and cookie plan


Not the prize you're feather for? Browse various other questions tagged linear-algebra or ask your own question.


Featured top top Meta
Visit conversation
connected
5
Is thi collection of vectors, $\(2, 1), (3, 2), (1, 2)\$, is linearly dependent or independent?
3
room these vectors live independence or dependent?
3
Showing straight dependence in a matrix
3
Why doesn't transposing matter?
4
Linearly independent and also dependent columns
1
identify Linear independence of the provided matrix
1
How precisely does elimination uncover linearly dependence rows the $A$ (in $AX=b$)?
1
link of linear mix and straight dependent and linear live independence
1
straight independency in a matrix
4
exactly how do we decide even if it is to visualize a matrix with its rows or columns?
warm Network questions more hot inquiries

concern feed
i ordered it to RSS
question feed To subscribe to this RSS feed, copy and also paste this URL into your RSS reader.


*

carolannpeacock.comematics
company
ridge Exchange Network
site architecture / logo © 2021 ridge Exchange Inc; user contributions licensed under cc by-sa. Rev2021.9.9.40167


carolannpeacock.comematics ridge Exchange works ideal with JavaScript permitted
*

her privacy

By click “Accept all cookies”, you agree ridge Exchange have the right to store cookie on your an equipment and disclose info in accordance through our Cookie Policy.