Question
Let $A$ be a square matrix. Prove that its maximal eigenvalue is smaller than its maximal singular value: $\max \left|\lambda_i\right| \leq \max \sigma_i$. Hint: Use Exercise 8.7.24.
Step 1
- $\lambda_i$ are the eigenvalues of matrix $A$. - $\sigma_i$ are the singular values of matrix $A$. - The singular values of $A$ are the square roots of the eigenvalues of $A^* A$ (where $A^*$ is the conjugate transpose of $A$). Show more…
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Symmetric Matrices and Quadratic Forms
Singular Values
Let $A$ be an $n \times n$ symmetric matrix, let $M$ and $m$ denote the maximum and minimum values of the quadratic form $\mathbf{x}^{T} A \mathbf{x}$ , where $\mathbf{x}^{T} \mathbf{x}=1,$ and denote corresponding unit eigenvectors by $\mathbf{u}_{1}$ and $\mathbf{u}_{n} .$ The following calculations show that given any number $t$ between $M$ and $m,$ there is a unit vector $\mathbf{X}$ such that $t=\mathbf{x}^{T} A \mathbf{x} .$ Verify that $t=(1-\alpha) m+\alpha M$ for some number $\alpha$ between 0 and $1 .$ Then let $\mathbf{x}=\sqrt{1-\alpha} \mathbf{u}_{n}+\sqrt{\alpha} \mathbf{u}_{1},$ and show that $\mathbf{x}^{T} \mathbf{x}=1$ and $\mathbf{x}^{T} A \mathbf{x}=t$
Constrained Optimization
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