12 votes 12 votes Consider a set of $n$ linearly independent vectors $\left\{\vec{w}_1, \ldots, \vec{w}_n\right\} \in \mathbb{R}^n$. A vector $\vec{u} \in \mathbb{R}^n$ will:Option $1.$ Always be a linear combination of $\left\{\vec{w}_1, \ldots, \vec{w}_n\right\}$Option $2.$ Never be a linear combination of $\left\{\vec{w}_1, \ldots, \vec{w}_n\right\}$Enter the correct option as the numeric number. That is if option $2$ is correct then enter $\text{“2”}.$ Linear Algebra goclasses2025_csda_wq3 numerical-answers goclasses linear-algebra vector-space 1-mark + – GO Classes asked Mar 14, 2023 • edited Mar 13 by shadymademe GO Classes 852 views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply himanshud2611 commented Mar 15, 2023 reply Follow Share Maybe I missed this part in lecture, how a vector can be expressed as LC of LI vectors? 0 votes 0 votes geekybread commented Mar 15, 2023 reply Follow Share Just consider {w1,w2,..,wn} as convenient vectors and think about it. 1 votes 1 votes ManasaM commented Mar 16, 2023 reply Follow Share @Sachin Mittal 1 sir ,if we have the n+1 vectors of Rn then we can definetly say them as linearly dependent vectors but if we have n vectors but some times it could be linearly combination right but not always why option 2 is incorrect?? Please give me hint 0 votes 0 votes JoyBoy commented Mar 16, 2023 reply Follow Share silly mistake, not reading question properly.Here we are talking about vector which is not present in Linearly independent set.i,.e u does not belong to {w1,w2….} set. Hence it will always be combination of other vectors. 0 votes 0 votes KG commented Mar 16, 2023 reply Follow Share @ManasaM In the question it is given that the n vectors are LI. 0 votes 0 votes ITACHI_003 commented Mar 16, 2023 reply Follow Share @JoyBoy I have also done the same mistake . I have considered that u is one of the independent vector I considered it to be part of basis 1 votes 1 votes 0z4rk commented Mar 16, 2023 reply Follow Share Same Silly mistake , considered that u in one of the LI vectors 0 votes 0 votes seba16 commented Mar 16, 2023 reply Follow Share I have not read the question properly and thought vector u is also belong to the set of n linearly independent vectors {→w1,…,→wn}. Silly Mistake 2 votes 2 votes Please log in or register to add a comment.
13 votes 13 votes In a vector space $R^{n}$, any set of n linearly independent vectors forms the basis vectors. i.e A set of n linearly independent vectors belonging to the vector space $R^{n}$ can span the entire space of $R^{n}$. So a vector $\overrightarrow{u} $ $\epsilon$ $R^{n}$ can always be expressed as the linear combination of n linearly independent vectors {$\overrightarrow{w_{1}} $,…,$\overrightarrow{w_{n}} $} $\epsilon$ $R^{n}$. Option1 is correct Vineeth Rambhiya answered Mar 15, 2023 Vineeth Rambhiya comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes {w1,w2,w3,…,wn}∈Rn and the set is linearly independent and the set consists of exactly n vectors so this set is the minimum and enough to fill the space of Rn.If you include one vector u∈Rn ,then u always can be written as the linear combination of these n vectors .so option 1 is correct. SubhamAdhikary answered Mar 25, 2023 SubhamAdhikary comment Share Follow See all 0 reply Please log in or register to add a comment.
