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The question:

A $\in$ $M_{4 \times 4}$. Let $A$ and $adjA$ (transpose of cofactor matrix of A) be non null matrix, and detA = 0. Find rank of A.

My attempt:

I understand that the rank of A must be less than 4. But after that I am totally stuck. The other thing that keeps popping up in mind is the expression of $A^{-1}$ in terms of $det A$ and $adj A$ but then $detA$ has to be non zero there.

Please help.

Debu
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2 Answers2

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We have $rank(A) \le 3$, else $\det A \neq 0$. From the fact that the cofactor is not the null matrix, we know $rank\; A \ge 3$. To see this, suppose towards a contradiction that $rank(A)<3$, then the column space of $A$ is spanned by no more than two of its columns. Then the column space of any minor of $A$ is also spanned by two columns, hence the determinant of any minor is zero and the cofactor of $A$ is null.

See if you can generalize this argument to show any determinant 0 $n\times n$ matrix with non-zero cofactor has rank $n-1$.

Johann Peters
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  • Thanks a lot. I got it. Yes i am able to generalize the above to an $n \times n$ matrix. – Debu Apr 24 '24 at 18:29
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Your immediate question has been answered but a closely related useful fact is that for any $n\times n$ matrix $A$ with $n\ge 2$, $$\mathop{\mathrm{rank}}(\mathop{\mathrm{adj}}A)=\begin{cases} n&\text{if }\mathop{\mathrm{rank}}A=n\\ 1&\text{if }\mathop{\mathrm{rank}}A=n-1\\ 0&\text{if }\mathop{\mathrm{rank}}A\le n-2 \end{cases}\tag{1}$$ So the rank of the adjugate drops off very quickly. In your case the only possibility is $\mathop{\mathrm{rank}}A=n-1$.

One way of proving (1) is using Jacobi's identity which says $$(\mathop{\mathrm{adj}}A)^{(k)}=(\det A)^{k-1}D(A^{(n-k)})\tag{2}$$ where $M^{(k)}$ denotes the $k$-th compound power of $M$, and $D$ is the Poincaré dual map. Then (1) follows since the rank of $M$ is the largest $r$ for which $M^{(r)}\ne 0$.

blargoner
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