2 corrected typos edited Oct 2 '14 at 9:52 amoeba 64.8k1616 gold badges218218 silver badges273273 bronze badges With your new information, that all the components of the positive-definite matrix are positive, it becomes easy. While it follows directly from the Perron-Frobenius theoremPerron-Frobenius theorem (Whichwhich is valid for square matrices with non-negative elements, symmetric or not), in the symmetric case it is much easier. Let the positive-definitdefinite matrix beebe $$S$$. theThe eigenvector corresponding to the largest eigenvector is the value of the vector $$x$$ obtaining the maximum in the following problem: $$\lambda_{\text{max}} = max_{\{x \colon \| x\|=1\}} x^T S x$$$$\lambda_{\mathrm{max}} = \mathrm{max}_{\{x \colon \| x\|=1\}} x^T S x$$(that is, the "argmax") where $$\lambda_{\text{max}}$$ is the largest eigenvalue. Suppose to get a contradiction that $$x_1$$ is negative, while the other components of $$x$$ are non-negative. We can write $$x^T S x = x_1 S_{11} x_1+2x_1 \sum_{j=2}^m s_{1j} x_j + \sum_{i=2}^m \sum_{j=2}^m x_i s_{ij} x_j$$ Note that the first and third terms are positive while the second term is negative, and we can get a strictly larger value by switching the sign of $$x_1$$, which respects the restriction on norm. That gives the contradiction you need. A similar argument can be written for any other pattern of negative/positive sign. With your new information, that all the components of the positive-definite matrix are positive, it becomes easy. While it follows directly from the Perron-Frobenius theorem (Which is valid for square matrices with non-negative elements, symmetric or not), in the symmetric case it is much easier. Let the positive-definit matrix bee $$S$$. the eigenvector corresponding to the largest eigenvector is the value of the vector $$x$$ obtaining the maximum in the following problem: $$\lambda_{\text{max}} = max_{\{x \colon \| x\|=1\}} x^T S x$$(that is, the "argmax") where $$\lambda_{\text{max}}$$ is the largest eigenvalue. Suppose to get a contradiction that $$x_1$$ is negative, while the other components of $$x$$ are non-negative. We can write $$x^T S x = x_1 S_{11} x_1+2x_1 \sum_{j=2}^m s_{1j} x_j + \sum_{i=2}^m \sum_{j=2}^m x_i s_{ij} x_j$$ Note that the first and third terms are positive while the second term is negative, and we can get a strictly larger value by switching the sign of $$x_1$$, which respects the restriction on norm. That gives the contradiction you need. A similar argument can be written for any other pattern of negative/positive sign. With your new information, that all the components of the positive-definite matrix are positive, it becomes easy. While it follows directly from the Perron-Frobenius theorem (which is valid for square matrices with non-negative elements, symmetric or not), in the symmetric case it is much easier. Let the positive-definite matrix be $$S$$. The eigenvector corresponding to the largest eigenvector is the vector $$x$$ obtaining the maximum in the following problem: $$\lambda_{\mathrm{max}} = \mathrm{max}_{\{x \colon \| x\|=1\}} x^T S x$$(that is, the "argmax") where $$\lambda_{\text{max}}$$ is the largest eigenvalue. Suppose to get a contradiction that $$x_1$$ is negative, while the other components of $$x$$ are non-negative. We can write $$x^T S x = x_1 S_{11} x_1+2x_1 \sum_{j=2}^m s_{1j} x_j + \sum_{i=2}^m \sum_{j=2}^m x_i s_{ij} x_j$$ Note that the first and third terms are positive while the second term is negative, and we can get a strictly larger value by switching the sign of $$x_1$$, which respects the restriction on norm. That gives the contradiction you need. A similar argument can be written for any other pattern of negative/positive sign. 1 answered Oct 5 '12 at 1:45 kjetil b halvorsen 35.5k99 gold badges9090 silver badges274274 bronze badges With your new information, that all the components of the positive-definite matrix are positive, it becomes easy. While it follows directly from the Perron-Frobenius theorem (Which is valid for square matrices with non-negative elements, symmetric or not), in the symmetric case it is much easier. Let the positive-definit matrix bee $$S$$. the eigenvector corresponding to the largest eigenvector is the value of the vector $$x$$ obtaining the maximum in the following problem: $$\lambda_{\text{max}} = max_{\{x \colon \| x\|=1\}} x^T S x$$(that is, the "argmax") where $$\lambda_{\text{max}}$$ is the largest eigenvalue. Suppose to get a contradiction that $$x_1$$ is negative, while the other components of $$x$$ are non-negative. We can write $$x^T S x = x_1 S_{11} x_1+2x_1 \sum_{j=2}^m s_{1j} x_j + \sum_{i=2}^m \sum_{j=2}^m x_i s_{ij} x_j$$ Note that the first and third terms are positive while the second term is negative, and we can get a strictly larger value by switching the sign of $$x_1$$, which respects the restriction on norm. That gives the contradiction you need. A similar argument can be written for any other pattern of negative/positive sign.