Let $A^*$ denote the matrix whose $(ij)$-th entry is $A_{ij}$, $1 ≤ i, j ≤ 5.$

Question

Let $A \in M_5(\Bbb R)$. If $A = (a_{ij})$, let $A_{ij}$ denote the co-factor of the entry $a_{ij}, 1 ≤ i, j ≤ 5.$ Let $A^*$ denote the matrix whose $(ij)$-th entry is $A_{ij}$, $1 ≤ i, j ≤ 5.$

a. What is the rank of $A^*$ when the rank of $A$ is 5?

b. What is the rank of $A^*$ when the rank of $A$ is 3?

I think answers are $a. = 5$ and $b. = 0$, Sharing the questions...a good one.

There is only one matrix with rank $0$, and $A^*$ is not it. — Servaes, Jun 30 '16 at 17:08
More general case of this problem is here http://math.stackexchange.com/questions/266963/what-is-the-rank-of-the-cofactor-matrix-of-a-given-matrix — slow but keen learner, Jun 30 '16 at 17:28

score 1 · Answer 1 · answered Jul 01 '16 at 05:07

Claim:

i) $rank(A^*)=n$ if $rank(A)=n$

ii) $rank(A^*)=1$ if $rank(A)=n-1$

iii) $rank(A^*)=0$ if $rank(A)<n-1$

Proof:

i) $det(A^*)\ne0$ since $A^*A=det(A)E$ and $A$ is invertable

ii) there exists $i,j$ such that $A_{ij}\ne0$ since $rank(A)=n-1$, so $rank(A^*)\ge1$

since $rank(A)+rank(A^*)\le n$, $rank(A^*)=1$

iii) $A^*=0$ since $A_{ij}=0$ since $rank(A)<n-1$

Let $A^*$ denote the matrix whose $(ij)$-th entry is $A_{ij}$, $1 ≤ i, j ≤ 5.$

1 Answers1