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This question already has an answer here:

Let $A \in M_n(\mathbf{C})$ and let $B = \operatorname{adj}(A)$ be the adjugate matrix.

a) Prove that if $A$ is invertible, then so is $B$.
b) Prove that if $A$ has rank $n-1$, then $B$ has rank $1$.
c) Prove that if $A$ has rank at most $n-2$, then $B = O_n$.

a) We have that $A$ is invertible, so $\det(A) \neq 0$. The rank of $A$ is $n$. Also $$ A \operatorname{adj}(A) = \operatorname{adj}(A) A = \det(A) I. $$ But what next? Please help me to solve the problem using elementary theory of matrix algebra (also please do not use the rank–nullity theorem).

EDIT:
b) Clearly $\det(A) = 0$. Then $$ A \operatorname{adj}(A) = \operatorname{adj}(A) A = \det(A) I = 0, $$ i.e. $A \operatorname{adj}(A) = 0$. But how does this imply that $\operatorname{adj}(A)$ has rank $1$?

May I request for any alternative solution of the problem? Can we conclude the result from $A \operatorname{adj}(A) = 0$?

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marked as duplicate by user1551 linear-algebra Jan 7 '17 at 22:20

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  • $\begingroup$ Any particular reason you want to avoid the rank-nullity theorem? $\endgroup$ – Omnomnomnom Sep 10 '16 at 23:50
  • $\begingroup$ @Omnomnomnom No I have not yet learned it. $\endgroup$ – user1942348 Sep 11 '16 at 2:37
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1) We know that $A\cdot \textrm{adj}(A) = \det A \cdot I_n$. That is, the inverse of $\textrm{adj}(A)$ is $\frac{1}{\det(A)} A$, hence $\textrm{adj}(A)$ is invertible.

2) Let A have rank $n-1$.

By Schur diagonalization theorem, $A=UTU^{-1}$, where $T$ is an upper triangular matrix and $U$ is a unitary matrix, by Schur's lemma.

Also, by Schur's lemma, $\textrm{rank}(T) = \textrm{rank}(A)$.

Since $T$ is a triangular matrix the rows of $T$ are all linearly independent of each other, hence one of the rows consists entirely of zeros because the rank of $T$ is $n-1$. Let this row be row $I$, where $1 \leq I \leq n$.

Note that $\textrm{adj} A = \textrm{adj}(UTU^{-1}) = \textrm{adj}(U^{-1})\textrm{adj}(T)\textrm{adj}(U)$, but then $\textrm{adj}(U)$ and $\textrm{adj}(U^{-1})$ are of full rank, hence $\textrm{rank}(\textrm{adj} A) = \textrm{rank}(\textrm{adj}(T))$.

Now, imagine that we are calculating the cofactors of elements of row $j$. Suppose that $j \neq I$. Then, omitting the row $j$ and whichever column the element is on, will leave in the cofactor, the elements of row $I$, which are all zero. Since the cofactor is a determinant containing a zero row, it must have the value zero. Thus, row $j$ of $\textrm{adj}(T)$ is zero when $j \neq I$.

Suppose that $j=I$. Then, omitting the row $I$ and whichever column, we obtain some cofactor matrix $C$. Note that $C$ is of rank $n-1$, hence at least one of the cofactor elements is non-zero. That is to say, the $I$th row of $T$ is non-zero.

Thus, $\textrm{adj}(T)$ is identically zero except for one non-zero row. Hence $\textrm{rank}(\textrm{adj}(T)) = 1 = \textrm{rank}(\textrm{adj}(A))$

3)I'll go more loosely over this, if the logic has been understood:

If A has rank $2$, then on triangularizing you will get $2$ rows containing zero. Then the cofactor matrix will be empty, because every cofactor will contain at least one row of zeros. So $\textrm{adj}(A) = [0]_n$.

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  • $\begingroup$ The 1st part is OK. Thanks. I understand (b) part. Yes it is true that rank(adjA)=1. But I need a mathematical proof of it. $\endgroup$ – user1942348 Sep 10 '16 at 5:32
  • $\begingroup$ Ok, I will edit my post. Please specify where you feel you need more mathematical justification, because I think my proof is fine for now. $\endgroup$ – астон вілла олоф мэллбэрг Sep 10 '16 at 5:33
  • $\begingroup$ Yes. One can understand from your writing. But I need the proof using mathematical properties.I have edited what I want to tell. $\endgroup$ – user1942348 Sep 10 '16 at 5:42
  • $\begingroup$ But have I not written about the triangularization of $A$, followed by the cofactor argument? What did you not understand there? I showed mathematically that it has rank $1$, didn't I? $\endgroup$ – астон вілла олоф мэллбэрг Sep 10 '16 at 5:45
  • $\begingroup$ Fine, I have edited it, making it little more mathematical. I don't want to improve it further because it's quite clear now. $\endgroup$ – астон вілла олоф мэллбэрг Sep 10 '16 at 6:31
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For a), of course we observe that $Adj(A)/\det(A)$ must be $A^{-1}$.

For b): note that $A$ has rank $n-1$. Conclude that we must have a null space with dimension at most $1$ (either apply the rank nullity theorem, or simply note that row-reducing to solve $Ax=0$ leaves only one free variable). However, since $A \; Adj(A)=0$, every column of the adjoint must be in the null space of $A$. Conclude that the rank of $Adj(A)$ is at most $1$.

Now, how do we conclude that $Adj(A)$ is not zero? For this, we need a little fact:

A matrix $A$ has rank less than $k$ if and only if every $k\times k$ submatrix has determinant zero

And with $k=n-1$, we see that not every entry of the adjoint can be zero.

For 3): directly apply the above fact.

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  • $\begingroup$ Here's a proof of that fact if this is the first time that you're seeing it. $\endgroup$ – Omnomnomnom Sep 11 '16 at 3:10
  • $\begingroup$ Please help me to understand "since $AAdj(A)=0$, every column of the adjoint must be in the null space of $A$." $\endgroup$ – user1942348 Sep 11 '16 at 11:29
  • $\begingroup$ In general, if $AB=0$, then $Ax=0$ for every column $x$ of $B$. That's just from the definition of matrix multiplication $\endgroup$ – Omnomnomnom Sep 11 '16 at 12:08
  • $\begingroup$ Thanks. I understand what you have explained. but how the above line concludes "Conclude that the rank of Adj(A) is at most 1." $\endgroup$ – user1942348 Sep 11 '16 at 16:58
  • $\begingroup$ Note that every column of $Adj(A)$ is a multiple of the same vector (namely our basis for the null space). $\endgroup$ – Omnomnomnom Sep 11 '16 at 16:59

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