https://yutsumura.com/determinant-trace-and-eigenvalues-of-a-matrix/ Sat, 12 Aug 2017 05:01:47 +0000 hourly 1 https://wordpress.org/?v=5.3.4 https://yutsumura.com/determinant-trace-and-eigenvalues-of-a-matrix/#comment-1981 Sun, 30 Jul 2017 01:18:53 +0000 https://yutsumura.com/?p=81#comment-1981 […] $det(A)=0$. Note that the product of all eigenvalues of $A$ is $det(A)$. (See the post “Determinant/Trace and Eigenvalues of a Matrix” for a proof.) Thus, $0$ is an eigenvalue of […]

]]> https://yutsumura.com/determinant-trace-and-eigenvalues-of-a-matrix/#comment-1618 Fri, 23 Jun 2017 17:51:52 +0000 https://yutsumura.com/?p=81#comment-1618 […] that the product of all eigenvalues of $A$ is the determinant of $A$. Thus, we have [frac{-1+sqrt{3}i}{2} cdot frac{-1-sqrt{3}i}{2}cdot lambda =det(A)=1.] […]

]]> https://yutsumura.com/determinant-trace-and-eigenvalues-of-a-matrix/#comment-1484 Mon, 05 Jun 2017 06:50:07 +0000 https://yutsumura.com/?p=81#comment-1484 […] Let us give a more theoretical explanation. If an $ntimes n$ matrix $A$ is diagonalizable, then there exists an invertible matrix $P$ such that [P^{-1}AP=begin{bmatrix} lambda_1 & 0 & cdots & 0 \ 0 & lambda_2 & cdots & 0 \ vdots & vdots & ddots & vdots \ 0 & 0 & cdots & lambda_n end{bmatrix},] where $lambda_1, dots, lambda_n$ are eigenvalues of $A$. Then we consider the determinants of the matrices of both sides. The determinant of the left hand side is begin{align*} det(P^{-1}AP)=det(P)^{-1}det(A)det(P)=det(A). end{align*} On the other hand, the determinant of the right hand side is the product [lambda_1lambda_2cdots lambda_n] since the right matrix is diagonal. Hence we obtain [det(A)=lambda_1lambda_2cdots lambda_n.] (Note that it is always true that the determinant of a matrix is the product of its eigenvalues regardless diagonalizability. See the post “Determinant/trace and eigenvalues of a matrix“.) […]

]]> https://yutsumura.com/determinant-trace-and-eigenvalues-of-a-matrix/#comment-1322 Thu, 18 May 2017 01:16:48 +0000 https://yutsumura.com/?p=81#comment-1322 […] the product of all eigenvalues of $A$ is the determinant of $A$. (For a proof, see the post “Determinant/trace and eigenvalues of a matrix“.) Thus we have [alpha beta gamma=det(A)=1.] Thus, at least one of $alpha, beta, […]

]]> https://yutsumura.com/determinant-trace-and-eigenvalues-of-a-matrix/#comment-1186 Wed, 26 Apr 2017 04:27:51 +0000 https://yutsumura.com/?p=81#comment-1186 […] 5=tr(A^2)=lambda_1^2+lambda_2^2. end{align*} Here we used two facts. The first one is that the trace of a matrix is the sum of all eigenvalues of the matrix. The second one is that $lambda^2$ is an eigenvalue of $A^2$ if $lambda$ is an eigenvalue of $A$, […]

]]> https://yutsumura.com/determinant-trace-and-eigenvalues-of-a-matrix/#comment-1133 Mon, 17 Apr 2017 01:12:24 +0000 https://yutsumura.com/?p=81#comment-1133 […] Computing the determinant directly by hand is tedious. So use the fact that the determinant of a matrix $A$ is the product of all eigenvalues of $A$. […]

]]> https://yutsumura.com/determinant-trace-and-eigenvalues-of-a-matrix/#comment-10 Mon, 01 Aug 2016 14:48:29 +0000 https://yutsumura.com/?p=81#comment-10 […] For (b), use (a) and consider the trace of $B$ and its relation to eigenvalues. For this relation, see the problem Determinant/trace and eigenvalues of a matrix. […]

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