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copied!<p>I don't know BLAS or LAPACK, so I have no idea what may cause this behaviour.</p> <p>But, for matrices of the given form, calculating the inverse is quite easy. The important fact for this is</p> <pre><code>(J^T*J)^2 = (J^T*J)*(J^T*J) = J^T*(J*J^T)*J = <J|J> * (J^T*J) </code></pre> <p>where <code><u|v></code> denotes the inner product (if the components are real - the canonical <em>bilinear</em> form for complex components, but then you'd probably consider not the transpose but the conjugate transpose, and you'd be back at the inner product).</p> <p>Generalising,</p> <pre><code>(J^T*J)^n = (<J|J>)^(n-1) * (J^T*J), for n >= 1. </code></pre> <p>Let us denote the symmetric square matrix <code>(J^T*J)</code> by <code>S</code> and the scalar <code><J|J></code> by <code>q</code>. Then, for general <code>a != 0</code> of sufficiently large absolute value (<code>|a| > q</code>):</p> <pre><code>(a*I + S)^(-1) = 1/a * (I + a^(-1)*S)^(-1) = 1/a * (I + ∑ (-1)^k * a^(-k) * S^k) k>0 = 1/a * (I + (∑ (-1)^k * a^(-k) * q^(k-1)) * S) k>0 = 1/a * (I - 1/(a+q)*S) = 1/a*I - 1/(a*(a+q))*S </code></pre> <p>That formula holds (by analyticity) for all <code>a</code> except <code>a = 0</code> and <code>a = -q</code>, as can be verified by calculating</p> <pre><code>(a*I + S) * (1/a*I - 1/(a*(a+q))*S) = I + 1/a*S - 1/(a+q)*S - 1/(a*(a+q))*S^2 = I + 1/a*S - 1/(a+q)*S - q/(a*(a+q))*S = I + ((a+q) - a - q)/(a*(a+q))*S = I </code></pre> <p>using <code>S^2 = q*S</code>.</p> <p>That calculation is also much simpler and more efficient than first finding the LU decomposition.</p>

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