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The stiffness matrix $K$ in a finite element analysis is a symmetric positive definite matrix. When we introduce essential boundary conditions, we remove rows and columns associated with prescribed displacements, for example. And we obtain a linear system of equations $\bar{K}\bar{X}=\bar{F}$ in which $\bar{K}$ is the reduced stiffness matrix and is obtained by removing rows and columns of stiffness matrix $K$ associated with essential boundary conditions. My project is about damage in cyclic loading (low-cycle, $10^3$). This is a non-linear analysis because of plasticity and damage involved in. So, the run-time of the code that I am writing will be too large. I want to use Conjugate Gradient method for solving the systems of linear equations during the analysis. But the method requires that the reduced stiffness matrix $\bar{K}$ is symmetric and positive definite. The symmetry of $\bar{K}$ is trivial, but I do not know whether it is always positive definite or not.

P. Mir
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    First of all, consider the simple case of linear elasticity. The original stiffness matrix $K$ typically has zero eigenvalue and, hence, it is not positive definite. By removing the rows and columns corresponding to Dirichlet boundary conditions you obtain a matrix $\overline{K}$ that is invertible and positive definite.

    Second, in the more complex case of plasticity, especially if you use Lagrange multipliers, the tangent matrix is not necessary positive definite. Sometimes it is not even symmetric. You should provide more details of your model.

     – knl Jan 10 '18 at 07:55
  • @knl Thank you very much! So, my question is basically wrong. Because I thought that the original stiffness matrix is positive definite, but you say it is not. Honestly, we (engineers) don not study mathematics in a suitable manner. For example, we are taught that the stiffness matrix is positive definite without proof and mathematical preliminaries and always we are said that "proofs are not needed". – P. Mir Jan 10 '18 at 09:25
  • @P.Mir What kind of statment is that? I'm an engineer, I didn't study mathematics and I know that the stiffness matrix is not positive definite, hence the need for boundary conditions..... Don't rely on what was taught to you, look into it yourself. – P. G. Jan 12 '18 at 11:23
  • @P.Mir, here's a post that shows the positive definiteness of $\mathbf{K}$ for a Poisson problem. For linear elastic (static) problems, a nice proof of the positive definiteness of $\mathbf{K}$ is given in the FEM book by TJR Hughes. – hpc Jan 12 '18 at 18:34

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