Let each of the following matrices M describe a deformation of the$\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$plane for each given Mfind: the Eigen values and eigenvectors of the transformation, the matrix Cwhich Diagonalizesand specifies the rotation to new axesrole="math" localid="1658833126295" $\mathbf{(}\mathit{x}\mathbf{\text{'}}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{)}$along the eigenvectors, and the matrix D which gives the deformation relative to the new axes. Describe the deformation relative to the new axes.

role="math" localid="1658833142584" $\left(\begin{array}{cc}3& 1\\ 1& 3\end{array}\right)$

The given matrix$M=\left(\begin{array}{cc}3& 1\\ 1& 3\end{array}\right)$, describing a deformation of$(x,y)$ plane.

Eigen values are the special set of scalar values that is associated with the set of linear equations most probably in the matrix equations

An Eigen vector or characteristic vector of a linear transformation is a nonzero vector that changes at most by a scalar factor when that linear transformation is applied to it.

The roots of characteristic equation$\left|A-\lambda l\right|\mathbf{=}\mathbf{0}$of matrix A are known as the Eigen values of matrix A(I the unit matrix of same order as of. The eigenvector corresponding to Eigen value${\mathit{\lambda }}_{\mathbf{I}}$is given by$\mathbf{(}\mathit{A}\mathbf{-}{\mathit{\lambda }}_{\mathbf{I}}\mathit{l}\mathbf{)}{\mathit{X}}_{\mathbf{i}}\mathbf{=}\mathit{O}$where Ois null matrix.

The characteristic equation of matrix M is:

$$\begin{gathered} \left|\begin{array}{cc}3-\lambda & 1\\ 1& 3-\lambda \end{array}\right|=0 \\ {\left(3-\lambda \right)}^2-1=0 \\ (2-\lambda )(4-\lambda )=0 \end{gathered}$$

Therefore,$\lambda =2$or 4 . These are the eigen values of the matrix M.

Now the eigenvectors for this matrix should satisfy the equation,

$$\begin{gathered} x+y=0for\lambda =2 \\ x-y=0for\lambda =4 \end{gathered}$$

Then the eigenvectors are;

For$\lambda =2$is$\left(\begin{array}{c}1\\ -1\end{array}\right)$

For$\lambda =4$is$\left(\begin{array}{c}1\\ 1\end{array}\right)$

Therefore, the two normalized eigenvectors for the matrix M are,

For$\lambda =2$is$\left(\begin{array}{c}\frac{1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}\end{array}\right)$

For$\lambda =4$is$\left(\begin{array}{c}\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right)$

Hence the matrix C is as follows:

$C=\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)$

And the diagonal matrix D can be represented by,

$D=\left(\begin{array}{cc}2& 0\\ 0& 4\end{array}\right)$

Therefore, relative to the new axes, the deformation leaves the x-coordinate unchanged while they – coordinates is multiplied by 4.