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

Solution 1:

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 $$ \det(T) I = (\text{adj}(T)^T)^T \cdot T = \text{adj}(T) \cdot T .$$

Solution 2:

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 $v\in V$.