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Let $K$ be a field and $f$ the endomorphism of $\mathcal M_n(K)$ that sends a matrix to its transpose. I want to determine the determinant of $f$. I know that since $f^2=id$ then $det(f)=1\ or \ -1 $ and I know by choosing the canonical basis of $\mathcal M_n(K)$ that the determinant is actually $-1$ but I was asked to solve this using the decomposition of $$\mathcal M_n(K)=\mathcal A_n(K)\oplus \mathcal S_n(K)$$ as the sum of antisymmetric matrices and the symmetric matrices. I know that for any matrix $M$ we have $$\displaystyle M=\dfrac{M+t_M}{2}+\dfrac{M-t_M}{2}$$ but then what ? Thank you for your help!

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$\mathcal A(n)$ and $\mathcal S(n)$ are stable by transposition. Let's denote by $f_A$ and $f_S$ the restrictions of $f$ to $\mathcal A(n)$ and $\mathcal S(n)$ respectively.Transposition is the direct sum $f_A\oplus f_S=(-\operatorname{Id}_{\mathcal A(n)})\oplus \operatorname{Id}_{\mathcal S(n)}$, hence $$\det f=\det(-\operatorname{Id}_{\mathcal A(n)})\cdot\det(\operatorname{Id}_{\mathcal S(n)})=(-1)^{\dim\mathcal A(n)}=(-1)^{\tfrac{n(n-1)}2}.$$

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  • $\begingroup$ Any matrix $M$ is written uniquely as $M=A+S$ so $f(M)=f(A)+f(S)=-A+S=id(A)+id(S)$. After that i don't see how do you get to the product of the determinants especially that the function $det$ is not linear !! I know that this is true for composition of endomorphisms : $det(v\circ u)=det(v).det(u)$ but here i don't see what you did. $\endgroup$ – palio Apr 29 '16 at 20:06
  • $\begingroup$ It is also true for block diagonal determinants. $\endgroup$ – Bernard Apr 29 '16 at 20:13
  • $\begingroup$ basically you are saying that if $E=E_1\oplus E_2$ and $f$ is an endomorphism of $E$ such that we can write $f$ as $f=f_1\oplus f_2$ then $det(f)=det(f_1).det(f_2)$ ? is this what you are saying ? if yes where can i find something to read on this. Thanks ! $\endgroup$ – palio Apr 29 '16 at 20:19
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    $\begingroup$ The condition is of course that $f_i$ maps $E_i$ to itself, so that, taking as a basis the union of a basis of $E_1$ and a basis of $E_2$, the matrix of $f$ is a block diagonal matrix. You can look at Wikipedia but in the present case, it's obvious; the matrix of $f$ is a diagonal matrix, starting with $-1$s, the $1$s. $\endgroup$ – Bernard Apr 29 '16 at 20:36
  • $\begingroup$ ah I see now what you mean thank you!! $\endgroup$ – palio Apr 29 '16 at 20:42
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Hint: The determinant is equal to $\prod_{i=1}^k \lambda_i^{m_i}$, where $m_i$ is the algebraic multiplicity of the eigenvalue $\lambda_i$. Because the transpose operator decomposes the space into the direct sum of eigenspaces, the geometric and algebraic multiplicities coincide.

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