Science:Math Exam Resources/Courses/MATH152/April 2012/Question 03 (c)/Solution 1

For any matrix A and basis vector $\mathbf {e} _{i}$ computing

$A\mathbf {e} _{i}$

obtains column i of A. If A is our linear transformation matrix then computing $A\mathbf {e} _{1}$, $A\mathbf {e} _{2}$, and $A\mathbf {e} _{3}$ will give us the three columns of A and hence we'll know the matrix. From part (b), we already determined the transformation on the vector $\mathbf {e} _{1}$,

$A\mathbf {e} _{1}={\begin{bmatrix}1\\-1\\-1\end{bmatrix}}$

and so this is the first column of A. To get the other relations we can use our work from part (a) where we obtained the basis vectors in terms of the eigenvectors. For $\mathbf {e} _{2}$ we have

$\mathbf {e} _{2}=\mathbf {v} _{1}-\mathbf {v} _{3}$

and so

$A\mathbf {e} _{2}=A\mathbf {v} _{1}-A\mathbf {v} _{3}=\mathbf {v} _{1}-3\mathbf {v} _{3}={\begin{bmatrix}-2\\1\\-2\end{bmatrix}}$

where we have used the eigenvalue/eigenvector relationships. This is the second column of A. Finally for $\mathbf {e} _{3}$ we have from part (a),

$\mathbf {e} _{3}=-\mathbf {v} _{1}+\mathbf {v} _{2}+\mathbf {v} _{3}$

and so

$A\mathbf {e} _{2}=-A\mathbf {v} _{1}+A\mathbf {v} _{2}+A\mathbf {v} _{3}=-\mathbf {v} _{1}+2\mathbf {v} _{2}+3\mathbf {v} _{3}={\begin{bmatrix}2\\1\\4\end{bmatrix}}$

gives us the third column of A. Therefore we have that the linear transformation matrix A is

$A={\begin{bmatrix}1&-2&2\\-1&1&1\\-1&-2&4\end{bmatrix}}.$