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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 $A \operatorname{adj}(A) = 0$?

Jendrik Stelzner
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user1942348
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2 Answers2

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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 $\textrm{adj}(A) = [0]_n$.

Sarvesh Ravichandran Iyer
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  • 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. – user1942348 Sep 10 '16 at 05:32
  • 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. – Sarvesh Ravichandran Iyer Sep 10 '16 at 05:33
  • Yes. One can understand from your writing. But I need the proof using mathematical properties.I have edited what I want to tell. – user1942348 Sep 10 '16 at 05:42
  • 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? – Sarvesh Ravichandran Iyer Sep 10 '16 at 05:45
  • Fine, I have edited it, making it little more mathematical. I don't want to improve it further because it's quite clear now. – Sarvesh Ravichandran Iyer Sep 10 '16 at 06:31
  • Fine, I have edited it, making it little more mathematical. I don't want to improve it further because it's quite clear now. – Sarvesh Ravichandran Iyer Sep 10 '16 at 06:31
  • Schur's lemma is a heavy tool to use for someone who has yet to learn even of the rank-nullity theorem, and probably has yet to touch matrix similarity. – Ben Grossmann Sep 11 '16 at 03:12
  • @Omnomnomnom I didn't know what the questioner meant when he said make it more mathematical. The triangularization algorithm is taught as a corollary of Schur's lemma. – Sarvesh Ravichandran Iyer Sep 11 '16 at 03:29
  • In the question, he specifies that he wants an answer using "elementary theory", and I'm saying your approach is not elementary. My point is distinct from the asker's incomprehensible objection that your original post was somehow insufficiently mathematical, but I imagine what he really wants is an explanation he can understand. – Ben Grossmann Sep 11 '16 at 03:33
  • Ok, my apologies on that. How would you have done this? Like you have done below? – Sarvesh Ravichandran Iyer Sep 11 '16 at 03:34
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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.

Ben Grossmann
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