theoretical question involving determinant and adjoint

**Jskid** · February 9th 2011, 11:59 AM

Show that if A is an nxn matrix then $det(adj A)=[det(A)]^{n-1}$
I'm not good with adjoints...first thing that comes to mind is $A(adj A)=(adj A)A=det(A)I_n$

**Drexel28** · February 9th 2011, 04:16 PM

Originally Posted by Jskid

Show that if A is an nxn matrix then $det(adj A)=[det(A)]^{n-1}$
I'm not good with adjoints...first thing that comes to mind is $A(adj A)=(adj A)A=det(A)I_n$

Clearly you mean adjugate, and not adjoint. Use precisely the property you claimed. Namely, $\displaystyle \text{adj}(A)A=\det(A)I_n$ . Taking the determinants of both sides gives $\det(\text{adj}(A))\det(A)=\det(\text{adj}(A)A)=\d et(\det(A)I_n)=\det^n(A)\det(I_n)=\det(A)^n$ .

**Jskid** · February 9th 2011, 07:40 PM

Originally Posted by Drexel28

Clearly you mean adjugate, and not adjoint.

My text book calls it adjoint. What's the difference?

**Drexel28** · February 9th 2011, 07:46 PM

Originally Posted by Jskid

My text book calls it adjoint. What's the difference?

What you are calling the adjoint is what I call the adjugate. The adjoint of an endomorphism $A:V\to V$ (if it exists) on an inner product space $\left(V,\langle\cdot,\cdot\rangle\right)$ is the unique endomorphism $A^\ast:V\to V$ such that $\left\langle A(x),y\right\rangle=\left\langle x,A^\ast(y)\right\rangle$ for every $x,y\in V$ . It has the property that if you fix an orthonormal ordered basis (assuming $\dim V<\infty$ ) $\mathcal{B}$ then $\left[A^\ast\right]_B=A\right]_B^\ast$ (where here $[\cdot]_B$ is the matrix representation with respect to that ordered basis, and $[A]_B^\ast=\left(\overline{[A]_B}\right)^{\top}$ ).

**dwsmith** · February 9th 2011, 08:17 PM

Originally Posted by Jskid

My text book calls it adjoint. What's the difference?

My linear book uses the term adjoint to so now worries.

**mathguy25** · April 9th 2013, 07:05 PM

My linear book describes adjoint as the linear operator T* such that (T*(v), w) = (v, T(w)). * relates to linear operators in much the same way as * relates to matrices.

Of course, this isn't the first time mathematicians use the same word to describe multiple mathematical concepts.

**topsquark** · April 10th 2013, 06:31 AM

Originally Posted by mathguy25

My linear book describes adjoint as the linear operator T* such that (T*(v), w) = (v, T(w)). * relates to linear operators in much the same way as * relates to matrices.

Of course, this isn't the first time mathematicians use the same word to describe multiple mathematical concepts.

Ummm...It's occasionally useful to add a post on older material, but you are aware you just helped out on a post that is more than 2 years old?

-Dan

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