0 votes 0 votes If A is a non-zero column matrix of order n×1 and B is a non-zero row matrix of order 1×n then rank of AB equals ? Rank(ab) can be zero??? Linear Algebra engineering-mathematics linear-algebra matrix self-doubt + – Overflow04 asked Sep 21, 2022 Overflow04 587 views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply jugnu1337 commented Sep 21, 2022 reply Follow Share (hint) only null matrix have rank zero. 1 votes 1 votes ankitgupta.1729 commented Sep 21, 2022 reply Follow Share Say, $A = \begin{pmatrix} a_1\\ a_2\\ ...\\ a_n \end{pmatrix}$ and $B = \begin{pmatrix} b_1 & b_2 &... & b_n \end{pmatrix}$ where, $a_i, b_i \in \mathbb{R} $ for $1 \leq i \leq n$ and all $a_i,b_i$ should not be zero because these are non-zero matrices. Now, $AB = \begin{pmatrix} a_1b_1 & a_1b_2 & ... & a_1b_n \\ a_2b_1 & a_2b_2 & ... &a_2b_n \\ ... & ... & ... & ... \\ a_nb_1 & a_nb_2 & ... &a_nb_n \end{pmatrix}$ Now, you can write each column here as: $b_1\begin{pmatrix} a_1\\ a_2\\ ...\\ a_n \end{pmatrix}, b_2\begin{pmatrix} a_1\\ a_2\\ ...\\ a_n \end{pmatrix},…,b_n\begin{pmatrix} a_1\\ a_2\\ ...\\ a_n \end{pmatrix}$ So, here, all the columns are multiple of $\begin{pmatrix} a_1\\ a_2\\ ...\\ a_n \end{pmatrix}$ So, maximum number of linearly independent vectors in column space $\mathbb{R}^n$ is $1.$ Similarly, you can see all the rows are multiples of $\begin{pmatrix} b_1 & b_2 &... & b_n \end{pmatrix}.$ So, maximum number of linearly independent vectors in row space $\mathbb{R}^n$ is $1.$ Hence, rank(AB)=1. Another way might be using $rank(AB) \leq \min(rank(A),rank(B)).$ As all $a_i,$ or $b_i$ can not be zero so you can't get rank 0 here. 2 votes 2 votes Overflow04 commented Sep 21, 2022 reply Follow Share What about this example. @ankitgupta.1729 @jugnu1337 @Arjun sir @afroze 0 votes 0 votes ankitgupta.1729 commented Sep 21, 2022 reply Follow Share your question is about column matrix $\times$ row matrix, not row matrix $\times$ column matrix. 1 votes 1 votes Overflow04 commented Sep 21, 2022 reply Follow Share @ankitgupta.1729 sorry, my mistake Thank you 1 votes 1 votes Sonu12345 commented Oct 26, 2022 reply Follow Share Wrong multiplication 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes $A,B \ are\ non-zero\ matrices\ so\ matrix\ AB\ will\ be\ non-zero\ matrix$ $There\ will\ have\ at\ least\ one\ non-zero\ element.$ afroze answered Sep 21, 2022 afroze comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Rank (A*B) equals min of(rank of A, rank of B) therefore always 1.(since they are non zero Matrix) Ujjal roy answered Mar 8, 2023 Ujjal roy comment Share Follow See all 0 reply Please log in or register to add a comment.