I am trying to solve a question where I am told the transformation matrix $A$ representing a linear transformation $T$ by $T(x)=Ax$.
I am asked to find a basis $(f_1, f_2, f_3)$ and I am not sure about how to solve this and start my response.
Thanks for the help.
P.S. If it is of any help the exercise is part of the chapter on diagonalisation.
A vector regarding a basis can be represented as column vector of coordinates regarding the basis: $$ x = x_1 f_1 + x_2 f_2 + x_3 f_3 = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \end{pmatrix} $$ Applying $x$ to a linear transformation $T$ gives \begin{align} T(x) &= T(x_1 f_1 + x_2 f_2 + x_3 f_3) \\ &= x_1 T(f_1) + x_2 T(f_2) + x_3 T(f_3) \\ &= x_1 a_1 + x_2 a_2 + x_3 a_3 \end{align} However the image vectors $a_j = T(f_j)$ can be expressed by the basis as well: $$ a_j = a_{1j} f_1 + a_{2j} f_2 + a_{3j} f_3 = \begin{pmatrix} a_{1j} \\ a_{2j} \\ a_{3j} \end{pmatrix} $$ So we get \begin{align} T(x) &= x_1 \begin{pmatrix} a_{11} \\ a_{21} \\ a_{31} \end{pmatrix} + x_2 \begin{pmatrix} a_{12} \\ a_{22} \\ a_{32} \end{pmatrix} + x_3 \begin{pmatrix} a_{13} \\ a_{23} \\ a_{33} \end{pmatrix} \\ &= \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix} \\ &= A x \end{align} Coordinate vectors and matrices of linear transformation depend on the choice of the basis.
It turns out that for certain linear transformations one can choose a basis which results in a diagonal matrix: $$ A = \begin{pmatrix} a_{11} & 0 & 0 \\ 0 & a_{22} & 0 \\ 0 & 0 & a_{33} \end{pmatrix} $$
And your task is probably to find such a basis.
Hint: It means $$ T(f_1) = \begin{pmatrix} a_{11} \\ 0 \\ 0 \end{pmatrix} = a_{11} f_1 \iff \\ A f_1 = a_{11} f_1 $$ and so on.