How do you show that if the adjoint of a matrix $A$, adj$A$, is invertible then $A$ is invertible?
I know how to prove the opposite, but I can't figure out how to prove it.
Eden Segal
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2$\begingroup$ What do you know about the product of a matrix and its adjoint? $\endgroup$– Gerry MyersonCommented Feb 16, 2013 at 11:47
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1$\begingroup$ Eden --- say something --- anything --- let us know you're still with us. You've had an answer up for a couple of days now, and a lengthy string of comments; what do you reckon? $\endgroup$– Gerry MyersonCommented Feb 19, 2013 at 2:23
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2 Answers
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If you mean adjoint: $\det A^* = \det {\overline{A^T}} = \overline{\det A^T} = \overline {\det A}$.
If you mean adjugate: $A \operatorname{adj}(A) = \operatorname{adj}(A) A= (\det A) I$.
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$\begingroup$ This does not agree with the way I use the word adjoint. $\endgroup$ Commented Feb 16, 2013 at 11:52
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$\begingroup$ This does agree with my use of both the words "adjoint" and "adjugate", but of course: the adjoint has such a simple expression only if we're using an orthonormal basis... $\endgroup$ Commented Feb 16, 2013 at 13:42
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$\begingroup$ @Don, the definition of adjoint with which I am familiar makes no reference to a basis. $\endgroup$ Commented Feb 17, 2013 at 6:05
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$\begingroup$ @Gerry If $V, W$ are linear spaces equipped with bilinear forms $γ_V \colon V × V → K$ and $γ_W \colon W × W → K$ to some base field, you can define a map $g \colon W → V$ to be adjoint to a linear map $f \colon V → W$ if for all $v ∈ V, w ∈ W$: $γ_W(f(v),w) = γ_V(v,g(w))$. $\endgroup$– k.stmCommented Feb 17, 2013 at 9:41
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$\begingroup$ @GerryMyerson, neither does the one I'm familiar with. I just remarked that the adjoint is the transpose of the conjugate when we're using an orthonormal basis. $\endgroup$ Commented Feb 17, 2013 at 11:23
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There seems to be some dispute as to the definition of the adjoint of a matrix. Given a square matrix with entries $a_{ij}$, the minor of entry $a_{ij}$ is denoted $M_{ij}$ and is defined to be the determinant of the submatrix remaining after the $i$th row and $j$th column are deleted from $A$; the cofactor of entry $a_{ij}$ is defined by $C_{ij}=(-1)^{i+j}M_{ij}$; the adjoint of $A$ is the matrix whose $ij$-entry is $C_{ji}$.
These definitions are taken from Anton, Elementary Linear Algebra, 8th edition. The same definition is given in Noble and Daniel, Applied Linear Algebra, 3rd edition, and in Wedderburn, Lectures on Matrices.
eden --- is this the definition you are using for the adjoint?
answered Feb 17, 2013 at 23:32
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$\begingroup$ What you've described there, @Gerry, is what I used to call "classical adjoint" and it is nowadays also known as "adjugate", most probably to distinguish between these two very different things with identical names. Google it. $\endgroup$ Commented Feb 18, 2013 at 2:28
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$\begingroup$ @Don, I see what you mean. But Maple still uses Adjoint(A) in its LinearAlgebra package to calculate the matrix I define above. In any event, only eden knows which adjoint the question is about --- and eden is maintaining radio silence. $\endgroup$ Commented Feb 18, 2013 at 3:06