1
$\begingroup$

I have to prove that the rank of the matrix $B=2$ and given that $B$ is a $5\times 5$ non zero matrix.

My attempt : Let the columns of $B$ be $X_1,X_2,...,X_5$.Then $AX_1=0$,$AX_2=0$,...,$AX_5=0$.Now since the rank of the matrix $A$ is given as $3$ .So there are $2$ free variables and the possible dimensions of $X_1,..,X_5$ are $2$.

Hence combining we can find the possible rank of $B$ is 2 .

Where am I going wrong in the proof?Also I have not been taught linear transformation so I cannot use it here.

$\endgroup$
8
  • 2
    $\begingroup$ Why don't you take $B=0$? $\endgroup$ Commented Dec 2, 2020 at 18:18
  • 1
    $\begingroup$ If you want to restrict what cannot be used, it would likely be more helpful to explain what can be used. IE Are we allowed to use matrix theory about dimensions? $\endgroup$
    – Calvin Lin
    Commented Dec 2, 2020 at 18:18
  • $\begingroup$ Yes @Calvin we can use that but am I going wrong? $\endgroup$
    – Antimony
    Commented Dec 2, 2020 at 18:20
  • $\begingroup$ If $AB = 0$, what can you say about the dimensions of their kernels? $\endgroup$
    – Calvin Lin
    Commented Dec 2, 2020 at 18:21
  • $\begingroup$ Yes, but looking at the matrix $B$ in this way where am I going wrong? $\endgroup$
    – Antimony
    Commented Dec 2, 2020 at 18:24

1 Answer 1

3
$\begingroup$

$B$ can be any $5 \times k$ matrix whose columns are in the null space of $A$ which is 2 dimensional. So $B$ has rank at most $2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .