Engineering mathematics

Question

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???

Comments

(hint)  only null matrix have rank zero.

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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.

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What about this example. @ankitgupta.1729 

@jugnu1337 @Arjun sir @afroze

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your question is about column matrix $\times$ row matrix, not  row matrix $\times$ column matrix.

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@ankitgupta.1729

sorry, my mistake

 

Thank you

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Wrong multiplication

Answer 1

$A,B \ are\ non-zero\ matrices\ so\ matrix\ AB\ will\ be\ non-zero\ matrix$

$There\ will\ have\ at\ least\ one\ non-zero\ element.$

Answer 2

Rank (A*B) equals min of(rank of A, rank of B) therefore always 1.(since they are non zero Matrix)