Question

Question: In Exercise 7, find a unit vector x at which Ax has maximum length.

Short answer

The unit vector is, \(\left( {\begin{array}{*{20}{c}}{\frac{2}{{\sqrt 5 }}}\\{\frac{1}{{\sqrt 5 }}}\end{array}} \right)\).

Step-by-step solution

Write the result from Exercise 7

For the matrix \(A = \left( {\begin{array}{*{20}{c}}2&{ - 1}\\2&2\end{array}} \right)\), the eigenvalues of matrix \({A^T}A\) are 9 and 4.

Find the vector x for which length of Ax is maximum

As the maximum eigenvalue is 9, so the maximum value of \(\left\| {A{\bf{x}}} \right\|\) is 9.

The eigenvectors of the matrix for \(\lambda = 9\) are:

\({A^T}A - 9I = \left( {\begin{array}{*{20}{c}}{ - 1}&2\\2&{ - 4}\end{array}} \right)\)

Write the row-reduced Augmented matrix as:

\(\begin{array}{c}M = \left( {\begin{array}{*{20}{c}}{ - 1}&2&0\\2&{ - 4}&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 1}&2&0\\0&0&0\end{array}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_2} \to {R_2} + 2{R_1}} \right)\end{array}\)

So, the corresponding eigenvector is \(\left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right)\).

Normalize the eigenvector \(\left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right)\) is:

\(\begin{array}{c}{{\bf{v}}_1} = \frac{1}{{\sqrt {{2^2} + {1^2}} }}\left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{\frac{2}{{\sqrt 5 }}}\\{\frac{1}{{\sqrt 5 }}}\end{array}} \right)\end{array}\)

Thus, the vector for which length of Ax is maximum is \(\left( {\begin{array}{*{20}{c}}{\frac{2}{{\sqrt 5 }}}\\{\frac{1}{{\sqrt 5 }}}\end{array}} \right)\).