# For $T\in \mathcal L(V)$, we have $\text{adj}(T)T=(\det T)I$.

Let $V$ be an $n$-dimensional vector space over a field of characteristic $0$. For a linear operator $T\in \mathcal L(V)$, we know that $\bigwedge^n T=(\det T)I$, where $I:V\to V$ is the identity map.

Further, from this answer we can define the adjugate of $T$ as $\bigwedge^{n-1}T^t:\bigwedge^{n-1}V^*\to \bigwedge^{n-1}V^*$, where $T^t$ is the transpose of $T$. We write $T^\sharp$ as a shorthand for $\bigwedge^{n-1}T^t$.

The Question: It is a well-known formula that if $M$ is an $n\times n$ matrix with entries from a field $F$, then $$\text{adj}(M)M=M(\text{adj}(M))=(\det M)I_n$$ where $\text{adj}(M)$ is the adjugate of $M$.

I am trying to formulate this fact in the language of linear maps rather than matrices. The problem is that it does not mean anything to take the product of $T^\sharp$ with $T$. We just need to make a connection between $T^\sharp$ , $T$, and $\bigwedge^n T^t$.

• These questions you've been asking are neat. Is there a particular text you've been working through? – Omnomnomnom Jul 21 '15 at 16:53
• @Omnomnomnom I am not following any book. Just that I have for sometime been thinking about exterior algebras. I came across exterior powers of vector spaces in a differential geometry course and realized that a lot of matrix facts can be said in a neat invariant way using this concept. I have always liked and searched for invariant and coordinate approaches whenever possible – caffeinemachine Jul 21 '15 at 18:47
• correction: coordinatefree approaches. – caffeinemachine Jul 22 '15 at 4:02

So let's assume that $V$ has a non-degenerate bilinear form $\langle\cdot,\cdot\rangle$ with a basis $e_1,\dots,e_n$ such that $\langle e_i,e_j\rangle = \delta_{ij}$, the Kronecker delta. Let $*$ denote the Hodge star operator. Note that we have the formula $$\langle x,y\rangle = *((*x)\wedge y) .$$

Let's identify any operator on $V$ with its matrix representation. Then we have the following identity: $$\langle Tx,y\rangle = \langle x,T^T y\rangle \quad (x,y \in V) .$$ Now we extend $T$ to all of $\Lambda(V)$ in the standard way by the formula $T(x\wedge y) = Tx \wedge Ty$. Then we have the identities $$\text{adj}(T)^T x = *(T(*x)) \quad (x \in V) ,$$ $$\det(T) = *(T(*1)) ,$$ noting that $*1 = e_1\wedge e_2 \wedge \cdots \wedge e_n$.

Then for all $x,y \in V$, we have $$\langle \text{adj}(T)^T x,Ty\rangle = *(T(*x) \wedge Ty) = *(T((*x)\wedge y)) = \det(T) \langle x,y\rangle .$$ Since $\langle\cdot,\cdot\rangle$ is non-degenerate, we have $$\det(T) I = (\text{adj}(T)^T)^T \cdot T = \text{adj}(T) \cdot T .$$

• It is a bit unsatisfactory that we need to put an inner product structure. Do you know if a way which does not require an inner product? – caffeinemachine Jul 22 '15 at 3:40
• I have been wondering if there is some kind of Hodge star operator that takes $\Lambda^{(k)}(V)$ to $\Lambda^{(n-k)}(V^*)$ without using an inner product. I don't yet have an answer, and I have never seen anything like this before. – Stephen Montgomery-Smith Jul 22 '15 at 4:15
• But also, once you have picked a basis, you no longer need the inner product structure, because you can define the star operation directly using the basis elements. Also, if you want to define it over fields of characteristic greater than zero, it's not obvious to me what the inner product would even mean. – Stephen Montgomery-Smith Jul 22 '15 at 4:18
• But now that I look at en.wikipedia.org/wiki/Hodge_dual, specifically at en.wikipedia.org/wiki/…, I see that the definition requires a non-degenerate bilinear form. So over the complex numbers, this would not be the standard inner product, which is sesqui-linear. – Stephen Montgomery-Smith Jul 22 '15 at 4:22
• So instead, start with a basis, and make en.wikipedia.org/wiki/Hodge_dual#Computation_of_the_Hodge_star the definition. And of course, once you have picked a basis, you also have a bilinear form for free. – Stephen Montgomery-Smith Jul 22 '15 at 4:26

I'll propose to you another (slightly different, but isomorphic) definition of the adjugate (classical adjoint). Im borrowing from section 8 of http://people.reed.edu/~jerry/332/27exterior.pdf .

Let $f:V\rightarrow V$ (with $n$ the dimension of $V$). We have a canonical isomorphism $\phi:V=\wedge^1 V\rightarrow\mathrm{Hom}(\wedge^{n-1} V,\wedge^n V)$ induced by the Wedge product. Let the adjugate $\mathrm{adj}(f):V\rightarrow V$ of $f$ be obtained from $\mathrm{Hom}(\wedge^{n-1} f,\wedge^{n} V)$ via $\phi$, i.e. $\phi\circ \mathrm{adj}(f)=\mathrm{Hom}(\wedge^{n-1} f,\wedge^{n} V)\circ\phi$. It is then easy to check that $\mathrm{adj}(f)\circ f=\det(f)\mathrm{id}_V$: simply check $\phi((\mathrm{adj}(f)\circ f)v)=\phi(\det(f)v)$ for every $v\in V$.