If we know that $A$ is a symmetric $2$ by $2$ matrix, $\displaystyle \max_{\|x\|=1}\:x^{T}Ax=a$, and $\displaystyle\min_{\|x\|=1}\:x^{T}Ax=b$ for some given $a, b$, can we compute the eigenvalues of $A$?
Compute eigenvalues
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3You already did. They are $a$ and $b$. This is due to the relation $\lambda_{min}x^Tx \leq x^TAx \leq \lambda_{max}x^Tx$. – 2011-12-01
1 Answers
Since $A$ is a symmetric matrix, the eigenvalues are real and the eigenvectors can be chosen to be orthonormal. Let $x_1$, $x_2$ be the two orthonormal eigenvectors and $\lambda_1$, $\lambda_2$ be the two eigenvalues respectively. Any vector in $\mathbb{R}^2$ can be hence written as $x = c_1 x_1 + c_2 x_2$. Hence, we get that $ \begin{align} x^TAx & = (c_1 x_1 + c_2 x_2)^T A (c_1 x_1 + c_2 x_2)\\ & = c_1^2 x_1^T A x_1 + c_1c_2 x_1^T A x_2 + c_1c_2 x_2^T A x_1 + c_2^2 x_2^T A x_2\\ & = c_1^2 \lambda_1 + c_2^2 \lambda_2 \end{align} $ Hence, $\displaystyle \max_{||x||=1} x^TAx$ is equivalent to $\displaystyle \max_{c_1^2 + c_2^2 = 1} c_1^2 \lambda_1 + c_2^2 \lambda_2$. Hence, $\displaystyle \max_{||x||=1} x^TAx = \lambda_{max}.$
By the same argument, $\displaystyle \min_{||x||=1} x^TAx$ is equivalent to $\displaystyle \min_{c_1^2 + c_2^2 = 1} c_1^2 \lambda_1 + c_2^2 \lambda_2$. Hence, $\displaystyle \min_{||x||=1} x^TAx = \lambda_{min}.$
The same argument works in any $\mathbb{R}^n$ as well.
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0if \lambda_1 > \lambda_2, then $max_{c_1^2+c_2^2=1}c_1^2\lambda_1+c_2^2\lambda_2=\lambda_1, min_{c_1^2+c_2^2=1}c_1^2\lambda_1+c_2^2\lambda_2=\lambda_2$. – 2011-12-04