Linear Algebra – Problems in Skew-Symmetric Matrices

linear algebramatricesskew-symmetric matrices

Let $A$ be a real skew-symmetric matrix. Prove that $I+A$ is non-singular, where $I$ is the identity matrix.

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As $A$ is skew symmetric, if $(A+I)x=0$, we have $0=x^T(A+I)x=x^TAx+\|x\|^2=\|x\|^2$, i.e. $x=0$. Hence $(A+I)$ is invertible.

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Linear Algebra – Determinant of a Real Skew-Symmetric Matrix as Square of an Integer

A proof by induction is given in David J. Buontempo, The determinant of a skew-symmetric matrix, The Mathematical Gazette, Vol. 66, No. 435, Mar., 1982, Note 66.15, pages 67-69. If you have access to jstor, it's here. The proof does not depend on the Pfaffian.

[Math] Problem related to positive definite , symmetric , skew symmetric matrix

(a) If $C=AB-BA$ we see that $\operatorname{tr} C = 0$. If $(I-C)^n=0$, then $I-C$ is nilpotent and all eigenvalues are zero. Hence $\operatorname{tr} (I-C) = 0$, which implies that $\operatorname{tr}C = n$, a contradiction. Hence no such matrix can exist.

(b) The arithmetic and geometric means are related by ${x_1+...+x_n \over n} \ge \sqrt[n]{x_1\cdots x_n}$, which is equivalent to $(x_1+...+x_n)^n \ge n^n x_1\cdots x_n$. Since $\operatorname{tr} A = \lambda_1+\cdots + \lambda_n$ and $\det A = \lambda_1 \cdots \lambda_n$, we obtain the desired result.

(c) All eigenvalues of a skew symmetric matrix are imaginary. Since it is real and has odd dimension, it must have one real eigenvalue. The only number that is both real and imaginary is zero. Hence $A$ is singular.

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Linear Algebra – Determinant of a Real Skew-Symmetric Matrix as Square of an Integer

[Math] Problem related to positive definite , symmetric , skew symmetric matrix

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