query: How is this a matrix representation of $a+ib$?
positive_passages.docid: 31999
positive_passages.text: Relation of this antisymmetric matrix $r = \left(\begin{smallmatrix}0 &1\\-1&0\end{smallmatrix}\right)$ to $i$
negative_passages.docid: 90475
negative_passages.text: Coordinate-free proof of $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$?
negative_passages.docid: 149219
negative_passages.text: If $A$ is a non-square matrix with orthonormal columns, what is $A^+$?
negative_passages.docid: 118304
negative_passages.text: If $A^2 = I$ (Identity Matrix) then $A = \pm I$
negative_passages.docid: 55056
negative_passages.text: To show: $\det\left[\begin{smallmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{smallmatrix}\right]=(ab+bc+ca)^3$
negative_passages.docid: 62030
negative_passages.text: $A^2=A^*A$. Why is matrix $A$ Hermitian?
negative_passages.docid: 111769
negative_passages.text: Why is inverse of orthogonal matrix is its transpose?
negative_passages.docid: 114271
negative_passages.text: Why is a matrix invertible if it can be written as the product of elementary matrices?
negative_passages.docid: 108589
negative_passages.text: The inverse of the matrix $\{1/(i+j-1)\}$
negative_passages.docid: 140517
negative_passages.text: Sum of the entries in the matrix $A^3$
negative_passages.docid: 140829
negative_passages.text: What does $b_i\mid b_{i+1}$ mean in this context?
negative_passages.docid: 84626
negative_passages.text: Why in these matrices are $AB=BA$ not equal? What is the logic behind them?
negative_passages.docid: 109736
negative_passages.text: For a 2x2 matrix A satisfying $A^k=I$, compute $e^A$
negative_passages.docid: 119301
negative_passages.text: Is the following matrix invertible?
negative_passages.docid: 46128
negative_passages.text: Representation of Matrix with Rank 1
negative_passages.docid: 120092
negative_passages.text: Express the complex numbers given in Problem in the form $a+ib$
negative_passages.docid: 55062
negative_passages.text: $2\times 2$ matrix $C=AB-BA$