Question-36

eigenvalue

The eigenpairs of a \(\displaystyle 2\times 2\) matrix are given below:

\[ \begin{equation*} \left( 8,\begin{bmatrix} 1\\ 1 \end{bmatrix}\right) ,\left( 4,\begin{bmatrix} 1\\ -1 \end{bmatrix}\right) \end{equation*} \]

Find the matrix.

\(\displaystyle \begin{bmatrix} 6 & 2\\ 2 & 6 \end{bmatrix}\)
\(\displaystyle \begin{bmatrix} 8 & 0\\ 0 & 4 \end{bmatrix}\)
\(\displaystyle \begin{bmatrix} 4 & 1\\ 1 & 8 \end{bmatrix}\)
\(\displaystyle \begin{bmatrix} 1 & 0\\ 7 & 11 \end{bmatrix}\)

Answer

\(\displaystyle \begin{bmatrix} 6 & 2\\ 2 & 6 \end{bmatrix}\)
\(\displaystyle \begin{bmatrix} 8 & 0\\ 0 & 4 \end{bmatrix}\)
\(\displaystyle \begin{bmatrix} 4 & 1\\ 1 & 8 \end{bmatrix}\)
\(\displaystyle \begin{bmatrix} 1 & 0\\ 7 & 11 \end{bmatrix}\)

Solution

Note that the two eigenvectors are orthogonal. In general, if \(\displaystyle ( \lambda _{1} ,v_{1})\) and \(\displaystyle ( \lambda _{2} ,v_{2})\) are orthogonal eigenpairs, we can express \(\displaystyle A\) as: \[ \begin{equation*} A=\frac{\lambda _{1}}{v_{1}^{T} v_{1}} v_{1} v_{1}^{T} +\frac{\lambda _{2}}{v_{2}^{T} v_{2}} v_{2} v_{2}^{T} \end{equation*} \] Now, it is just a question of plugging in the values.