ALL-ZEBRAS
Here are some Zebras to Hunt for !!
Distinct Eigen-Values correspond to linearly independent Eigen-Vectors.(Prove it !!)
If all eigen vectors are linearly independent then the matrix is diagonalizable and the diagonal matrix has \(\lambda_1 , \lambda_2, \cdots \lambda_n\) as diagonal entries.
Let \(a_n = (\ln 3)^n \sum_{k=1}^{n} \frac{k^2}{k!(n-k)!}\) . Then the sum of the series \(a_1 + a_2 + a_3 + ... \to \infty\) , is equal to?
A is a \(n \times m\) matrix and B is a \(m \times n\) matrix . If \(|I_{n} - AB|\) is non-singular so is \(|I_{m} - BA|\)
For each positive integer \(n \ge 3\), define \(A_n\) and \(B_n\) as \(A_n = \sqrt{n^2 + 1} + \sqrt{n^2 + 3} + \cdots + \sqrt{n^2+2n-1}\) , \(B_n = \sqrt{n^2 + 2} + \sqrt{n^2 + 4} + \cdots + \sqrt{n^2 + 2n}.\) Determine all positive integers \(n\ge 3\) for which \(\lfloor A_n \rfloor = \lfloor B_n \rfloor\).
For any real number \(x\), \(\lfloor x\rfloor\) denotes the largest integer \(N\le x\).
Suppose \(f(X), g(X)\) are polynomials with coefficients that can be taken to be complex. Assume that \(\text{degree}(f) \ge \text{degree}(g) + 2\). If \(f\) has no multiple roots, prove that \(\sum_{\alpha} \frac{g(\alpha)}{f'(\alpha)} = 0\) where the sum is over the roots of \(f\).