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Force-System Resultants and Equilibrium
Published in Richard C. Dorf, The Engineering Handbook, 2018
The difference of any two solutions of the linear system Ax=b is a member of the null space of A Thus this system has at most one solution if and only if the nullity of A is zero. If the system is square (i.e., if A is n×n), then there will be a solution for every right-hand side b if and only if the collection of columns of A is linearly independent, which is the same as saying the rank of A is n. In this case the nullity must be zero. Thus, for any b, the square system Ax=b has exactly one solution if and only if rank A=n. In other words, the n×n matrix A is invertible if and only if rank A=n.
Linear Algebra and Matrices
Published in William F. Ames, George Cain, Y.L. Tong, W. Glenn Steele, Hugh W. Coleman, Richard L. Kautz, Dan M. Frangopol, Paul Norton, Mathematics for Mechanical Engineers, 2022
The difference of any two solutions of the linear system Ax = b is a member of the null space of A. Thus this system has at most one solution if and only if the nullity of A is zero. If the system is square (that is, if A is n × n), then there will be a solution for every right-hand side b if and only if the collection of columns of A is linearly independent, which is the same as saying the rank of A is n. In this case the nullity must be zero. Thus, for any b, the square system Ax = b has exactly one solution if and only if rank A = n. In other words the n × n matrix A is invertible if and only if rank A= n.
Background Principles
Published in Richard Leach, Stuart T. Smith, Basics of Precision Engineering, 2017
since A–1A = I and Ix = x. Thus, the solution x is simply the product of A–1 with b. However, this solution makes sense only as long as A–1 exists. It can be shown that the matrix A is invertible if its columns are linearly independent or equivalently the determinant of the matrix A, denoted by det A, is non-zero. For a square matrix A of order 2 with components Aij with i,j = 1,2, the determinant is simply A11A22– A12A21. For a square matrix of order 3, the determinant is given by
Lyapunov-based boundary feedback design for parabolic PDEs
Published in International Journal of Control, 2021
We are now ready to turn to the proof of the lemma. It suffices to show that the single-input, linear, time-invariant system is controllable. The Kalman rank controllability test for system (79) gives the square matrix Since for , it follows that the matrix is invertible with non-zero determinant. Since the matrix is a Vandermonde matrix and since , it follows that the matrix has a non-zero determinant. It follows from (80) that the determinant of the matrix is non-zero. Thus and system (79) is controllable. The proof is complete.
Variational approach for robust design and sensitivity analysis of mechatronic systems
Published in Journal of the Chinese Institute of Engineers, 2020
Hana Siala, Faïda Mhenni, Jean-Yves Choley, Maher Barkallah, Jamel LouatI, Mohamed Haddar
The pseudo-inverse (Buss 2004) is defined for all matrices , even if they are not a square matrix or not a full rank matrix. The pseudoinverse exists and is unique. When matrix is a full rank matrix, the pseudo-inverse can be defined as a simple algebraic formula: If has linearly independent columns, the matrix is invertible. In this case, the particular pseudo inverse is a left inverse and . can be expressed as:If has linearly independent rows, the matrix is invertible. In this case, the pseudo inverse is a right inverse and . can be expressed as:
Reconstructing real symmetric matrices from eigenvalues of finite dimensional perturbations
Published in Inverse Problems in Science and Engineering, 2020
Xuewen Wu, Pengcheng Niu, Guangsheng Wei
According to Proposition 2.1, can be expressed by (5). Let and . Then If are linearly independent, then the matrix X is invertible. By Property 2.9 of Chapter 4 in [9], the rank of matrix B is r. That means the number of nonzero real numbers is r. The proof is complete.