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GO Classes DPP | GATE CS, DA | Linear Algebra | Rank

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5 5 votes

Let $A$ be a $4\times 3$ real matrix with rank $2$. Let $B=A^T$. Which one of the following statements is true?

  1. Rank of $BA$ is less than $2$
     
  2. Rank of $BA$ is equal to $2$
     
  3. Rank of $BA$ is greater than $2$
     
  4. Rank of $BA$ can be any number between $1$ and $3$

5 Answers

1 1 vote
Since $B=A^T$, we have
$BA=A^T A$.

Now,
$\operatorname{rank}(BA)=\operatorname{rank}(A^T A)\le \min\big(\operatorname{rank}(A^T),\operatorname{rank}(A)\big)$

But
$\operatorname{rank}(A)=\operatorname{rank}(A^T)=2$

So,
$\operatorname{rank}(A^T A)\le 2$

To get the exact value, let $A^T A x=0$ for some vector $x$.
Then
$x^T A^T A x = 0$

So,
$(Ax)^T(Ax)=0$

This gives
$\|Ax\|^2=0$

Hence,
$Ax=0$

Therefore, the null space of $A^T A$ is the same as the null space of $A$.

Now $A$ has $3$ columns and rank $2$, so
$\operatorname{nullity}(A)=3-2=1$

Hence,
$\operatorname{nullity}(A^T A)=1$

Since $A^T A$ is a $3\times 3$ matrix,
$\operatorname{rank}(A^T A)=3-1=2$

Therefore,
$\operatorname{rank}(BA)=2$

So the correct option is $\boxed{(B)}$
0 0 votes
When you multiply two matrices (A*B), you are basically passing the information from B through a "filter" (A).Because of this, the final result can never have more information than what you started with. It can only stay the same or decrease.

The rank of the product is less than or equal to the smallest rank of the two matrices. Rank(AB)<=min(Rank(A),Rank(B)).
So the option :B
0 0 votes
Here we have B = A^T , A has a rank = 2

so when we compute BA so the columns of AB will be the the linear combination of a b which will get a cofficients form the columns of A so as we know that the columns of a has 2 linarly independent columns so it will also have a 2 linearly independent columns or less than that brcause we cannot create any independent column from the dependent columns so it the rank must be smallest between those matrices so it will be equal to 2

so B will be the correct answer.
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