query: How to prove $Tr(A)$ is sum of Eigen values of $A$
positive_passages.docid: 88473
positive_passages.text: Proof that the trace of a matrix is the sum of its eigenvalues
negative_passages.docid: 6514
negative_passages.text: Prove $T$ has at most two distinct eigenvalues
negative_passages.docid: 21553
negative_passages.text: Prove that $\nabla_{\mathrm X} \mbox{tr} (\mathrm A \mathrm X^{-1} \mathrm B) = - \mathrm X^{-\top} \mathrm A^\top \mathrm B^\top \mathrm X^{-\top}$
negative_passages.docid: 94383
negative_passages.text: Prove that $A^TD-C^TB=I$
negative_passages.docid: 75536
negative_passages.text: every eigenvalue of $T$ has only one corresponding eigenvector up to a scalar multiplication
negative_passages.docid: 80015
negative_passages.text: $Tr(A^2)=Tr(A^3)=Tr(A^4)$ then find $Tr(A)$
negative_passages.docid: 106097
negative_passages.text: Spectrum of the sum of two commuting matrices
negative_passages.docid: 18939
negative_passages.text: Eigenvalues of the transpose
negative_passages.docid: 38850
negative_passages.text: How to show Von Neumann Trace inequality $ \text{Tr}(AB) \leq \sum_{i=1}^n \sigma_{A,i}\sigma_{B,i} $?
negative_passages.docid: 130253
negative_passages.text: Differentiating $\mbox{tr} (ABA^TC)$ w.r.t. $A$
negative_passages.docid: 110086
negative_passages.text: Let $A$ be an $m \times n$ matrix. Show $A^TA$ and $AA^T$ have the same eigenvalues
negative_passages.docid: 109675
negative_passages.text: Finite sum of eigenspaces (with distinct eigenvalues) is a direct sum
negative_passages.docid: 28003
negative_passages.text: How to prove $A+ A^T$ symmetric, $A-A^T$ skew-symmetric.
negative_passages.docid: 61538
negative_passages.text: Given a square matrix $A$, both $AA^T$ and $A^TA$ are symmetric
negative_passages.docid: 155884
negative_passages.text: What are the conditions on $\text{tr}(AB) \leq \text{tr(A)} \text{tr(B)}$ to be true?
negative_passages.docid: 95677
negative_passages.text: Linear Matrix Inequality of a matrix $T$ with $\| T \|_2 < \rho$
negative_passages.docid: 128453
negative_passages.text: If $A$ has two eigenvalues $\lambda _1, \lambda_2$ and $\dim (E_{\lambda_1})=n-1$, then $A$ is diagonalizable