Science:Math Exam Resources/Courses/MATH221/April 2013/Question 07 (b)/Solution 1

What we know from part a)

Using the eigenvectors $\left[{\begin{array}{c}4\\3\end{array}}\right],\left[{\begin{array}{c}-3\\4\end{array}}\right]$, and the eigenvalues $\ 1,-1$, we know that $A\left[{\begin{array}{c}4\\3\end{array}}\right]=\left[{\begin{array}{c}4\\3\end{array}}\right]$ and $A\left[{\begin{array}{c}-3\\4\end{array}}\right]=-\left[{\begin{array}{c}-3\\4\end{array}}\right]$.

Finding the formula for the matrix $\ A$

Then, we can write the information from above in matrix notation.

${\begin{aligned}A\left[{\begin{array}{cc}4&-3\\3&4\end{array}}\right]&={\begin{bmatrix}A{\begin{bmatrix}4\\3\end{bmatrix}}\ A{\begin{bmatrix}-3\\4\end{bmatrix}}\end{bmatrix}}\\&={\begin{bmatrix}(1){\begin{bmatrix}4\\3\end{bmatrix}}\ (-1){\begin{bmatrix}-3\\4\end{bmatrix}}\end{bmatrix}}\\&=(1){\begin{bmatrix}4&0\\3&0\end{bmatrix}}+(-1){\begin{bmatrix}0&-3\\0&4\end{bmatrix}}\\&={\begin{bmatrix}4\\3\end{bmatrix}}{\begin{bmatrix}1&0\end{bmatrix}}+{\begin{bmatrix}-3\\4\end{bmatrix}}{\begin{bmatrix}0&-1\end{bmatrix}}\\&={\begin{bmatrix}4&-3\\3&4\end{bmatrix}}{\begin{bmatrix}1&0\\0&-1\end{bmatrix}}\end{aligned}}$

The first equality follows from this: when we product two 2 by 2 matrices, the first column of the resulting matrix is the product of the first matrix with the first column of the second matrix. Similarly, the second column of the resulting matrix is the product of the first matrix with the second column of the second matrix. In the last step we use, in reverse, that the product of two matrices is the sum of (columns of the first matrix) times (rows of the second matrix).

We define the matrix $B={\begin{bmatrix}4&-3\\3&4\end{bmatrix}}$ as the matrix of eigenvectors of $A$ and write the equation above as

$AB=B\left[{\begin{array}{cc}1&0\\0&-1\end{array}}\right].$

This leads us to $A=B\left[{\begin{array}{cc}1&0\\0&-1\end{array}}\right]B^{-1}$.

Calculating $\ A$

First the inverse of B is given by the usual method for 2x2 matrices (where $25=\det(B)$):

$B^{-1}={\frac {1}{25}}\left[{\begin{array}{cc}4&3\\-3&4\end{array}}\right]$

We can now compute A as claimed:

$A=\left[{\begin{array}{cc}4&-3\\3&4\end{array}}\right]\left[{\begin{array}{cc}1&0\\0&-1\end{array}}\right]\left({\frac {1}{25}}\left[{\begin{array}{cc}4&3\\-3&4\end{array}}\right]\right)={\frac {1}{25}}\left[{\begin{array}{cc}7&24\\24&-7\end{array}}\right]$.