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Let $A = (a_{ij}) ∈M_n(\mathbb{R})$; $n≥3$. Let $B = (b_{ij})$ be the matrix of its co- factors, i.e. $b_{ij}$ is the cofactor of the entry $a_{ij}$ in $A$. What is the rank of $B$ when
a. the rank of $A$ is $n$?
b. the rank of $A$ is less than, or equal to, $n ≥ 2$?

I am completely stuck on it.how can i solve this.

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PLEASE AVOID TYPING IN ALL CAPPPPPPPS!!!!!! –  user1551 Dec 29 '12 at 10:20
    
For b. the rank of $A$ is always less than or equal to $n$, so what do you mean here? –  Marc van Leeuwen Dec 29 '12 at 10:31
    
Part b should be: the rank of $A$ is less than or equal to $n-1$ –  Calvin Lin Dec 29 '12 at 10:36
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2 Answers

Hint: Calculate the matrix $A B^T$. Then use the fact that rank(AB^T) $\leq \min( $Rank A, Rank B^T).

To do part b, split it into the case where the rank is $n-1$ and when the rank is $\leq n-2$ (They have different answers).

Case 1: Rank of $A$ is $n-1$ - Calculate $AB^T$ again, and then use the fact that the rank of matrix $A$ is the largest integer $r$ such that some $r-$minor of $A$ does not vanish.

Case 2: Rank of $A$ is $\leq n-2$ - Show that $B=0$.

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I don't think the argument given in case 1 suffices to determine the rank of $B$ precisely (see what I wrote about this case in my answer). Could you be more precise? –  Marc van Leeuwen Dec 29 '12 at 11:21
    
Case 1: We have $AB^T=0$, which gives that rank(A) + rank$(B^T) \leq n$ (easy to show using dim null space). By the fact, $B^T$ has a non-zero entry. Hence, rank$(B^T)=1$ –  Calvin Lin Dec 29 '12 at 11:46
    
Thanks! Rather stupid of me not to think of rank-nullity here. –  Marc van Leeuwen Dec 29 '12 at 12:28
    
Yes, this question can be done nicely, if you put together various basic facts of Linear Algebra as I did above. The 'hardest' part was to think of multiplying $AB^T$, but that is motivated from determinants and cofactor definition, which he'd see if he does the multiplication. –  Calvin Lin Dec 29 '12 at 12:45
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The rank of a matrix is maximal the size of its nonzero minors, and the cofactors are up to a sign equal to the $(n-1)\times(n-1)$ minors of $A$. So if the rank of $A$ is less than $n-1$, none of the cofactors are nonzero, and the rank of $B$ is zero. On the other hand if the rank of $A$ is $n$, then $A$ is invertible and $B$ is up to a nonzero scalar $\det A$ and transpostion equal to the inverse of $A$, so also invertible and of rank $n$. So the only interesting case is when the rank of $A$ is $n-1$: now $\det A=0$ but at least one cofactor of $A$ is nonzero, so that one has $B\neq0$ but $AB^t=0$. The latter means that the image of $B^t$ is contained in the kernel of$~A$; this kernel is of dimension$~1$ by the assumption on the rank of$~A$, so in fact the image of$~B^t$ and the kernel of$~A$ must coincide, and the rank of $B^t$ (and of$~B$) is$~1$.

Note that $\operatorname{rank}(B)=1$ usually means that $B$ is not invertible (like $A$, whose rank was suppose to be less than $n$ in this case), but for $n=1$ it means that $B$ is invertible.

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