How the Householder method is better than Given’s method?

Using Householder’s transformation, reduce the following matrices in tridiagonal form and find all the eigen values and eigen vectors. Write the strum sequence.

   \(\left[ \begin{array}{cc} 1&2&2 \\ 2&1&2 \\2&2&1 \end{array} \right]\)

Using Householder’s transformation, reduce the following matrices in tridiagonal form and find all the eigen values and eigen vectors. Write the strum sequence.

     \(\left[ \begin{array}{cc} 1&\sqrt{2}&\sqrt{2} &2\\ \sqrt{2}&-\sqrt{2}&-1&\sqrt{2}\\\sqrt{2} &-1&\sqrt{2}&\sqrt{2}\\ 2&\sqrt{2}&\sqrt{2}&-3 \end{array} \right]\)

If the orthogonal transform is of the form \(p=1-2WW^T\) where \(W\) is a column vector such that \(W^TW=1\). Prove that \(P\) is symmetric and orthogonal.

Write the strum sequence and find the eigen values of following matrix:

    \(\left[ \begin{array}{cc} 1&-1&0&0&0&0 \\ 1&1&-1&0&0&0 \\ 0&1&1&-1&0&0 \\ 0&0&1&1&-1&0 \\0&0&0&1&1&-1 \\ 0&0&0&0&1&1\end{array} \right]\)

Use LR and QR both the methods to find the eigen values of the following matrix:

    \(A =\left[ \begin{array}{cc} 3&2 \\ 1&4 \end{array} \right]\)

Let \(R\) be a ring in which \(a^2=a, \forall a \in R\) (such a ring is called a Boolean ring). Show that \(2a=0, \forall a\in R\) , and \(R\) is commutative. Further show that the only Boolean ring that is an integral domain is \(\mathbb{Z_2}\).

Let \(R\) be a ring and \(a,b \in R\)such that \(ab=ba\) . Prove that for any positive integer  \((a+b)^n =a^n +\left( \begin{array}{c} n \\ 1 \end{array} \right)a^{n-1}b+ \cdots +\left( \begin{array}{c} n \\ n-1 \end{array} \right) ab^{n-1}+b^n\) .

Let \(d\) be any integer. Prove that \(\mathbb{Z}[\sqrt{d}] =\{a+b\sqrt{d} \; |a.b \in \mathbb{Z} \}\) is an integral domain and \(\mathbb{Z}[\sqrt{d}] =\{a+b\sqrt{d} \; |a.b \in \mathbb{Z} \}\) is a field.

Show that any ring with 6 elements is commutative.

If a non-zero subring \(S\) of a ring \(R\) has an identity \(1^\prime\) but \(R\) has either no identity or the identity of \(R\) is different from \(1^\prime\) , then show that \(1^\prime\) is a zero-divisor in \(R\) .