0 votes 0 votes It is given that m < n lets consider m = 4 and n = 5. If rank(A) = n, this means no of linearly independent columns in A are 5, but we cannot have 5 linearly independent vectors in R4. Where is the problem ? KrishnaR72 asked Aug 2, 2023 • edited Aug 3, 2023 by KrishnaR72 KrishnaR72 453 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply parth023 commented Aug 3, 2023 reply Follow Share if A is m*n matrix then rank(A) $\leq$ min(m,n) 1 votes 1 votes SolVer commented Oct 10, 2023 reply Follow Share Exactly rank(A)<= min(m,n) 0 votes 0 votes Please log in or register to add a comment.
6 votes 6 votes Lets consider a small matrix of size m * n such that m<n where m=2 and n=3 Rank of A is given as n which means we have n pivots,which is impossible .Lets understand by the diagram Here we can have atmost 2 pivots,Rank of the matrix can be 1 or 2 or 0(only for NULL matrix) so the statement u made is wrong as rank cannot be n ...that’s why the conclusion u r deriving is true that we cannot have more than n linearly independent vectors in R$^{n}$ space Psy Duck answered Aug 3, 2023 Psy Duck comment Share Follow See all 0 reply Please log in or register to add a comment.
4 votes 4 votes Given m<n for given matrix A In $R^{m}$, Atmost m linearly independent vectors are possible . Number of linear independent vectors must be $\leq$ m Spiderman 🕷️ answered Aug 3, 2023 Spiderman 🕷️ comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes In mXn matrix if m<n then rank will never be n , Because we can form maximum pivot elements m . And , in matrix of space R4 there will at most 4 linearly independent vectors can be formed, because that 4 vector will form whole space in R4 Aniket1710 answered Aug 3, 2023 Aniket1710 comment Share Follow See all 0 reply Please log in or register to add a comment.