# [SOLVED] Linear transformation

• Apr 10th 2009, 08:14 AM
Spec
[SOLVED] Linear transformation
Let F be the linear transformation that switches $\displaystyle \vec e_1 + \vec 2e_2$ for $\displaystyle \vec 2e_1 + \vec e_2$ and vice versa. Find the transformation matrix A for this linear transformation.

Then find the vectors $\displaystyle \vec f_1, \vec f_2$ so that $\displaystyle F(\vec f_1)=\vec f_1$ and $\displaystyle F(\vec f_2)=-\vec f_2$. Choose this as a basis, and then find F's transformation matrix in this basis.
• Apr 10th 2009, 09:18 AM
NonCommAlg
Quote:

Originally Posted by Spec

Let F be the linear transformation that switches $\displaystyle \vec e_1 + \vec 2e_2$ for $\displaystyle \vec 2e_1 + \vec e_2$ and vice versa. Find the transformation matrix A for this linear transformation.

so we have $\displaystyle F(e_1) + 2F(e_2)=2e_1 + e_2$ and $\displaystyle 2F(e_1) + F(e_2)=e_1 + 2e_2,$ which gives you $\displaystyle F(e_1)=e_2, \ F(e_2)=e_1.$ hence $\displaystyle A=\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}.$

Quote:

Then find the vectors $\displaystyle \vec f_1, \vec f_2$ so that $\displaystyle F(\vec f_1)=\vec f_1$ and $\displaystyle F(\vec f_2)=-\vec f_2$. Choose this as a basis, and then find F's transformation matrix in this basis.
let $\displaystyle f_1=ae_1+be_2, \ f_2=ce_1 + de_2.$ we want to have $\displaystyle A \begin{pmatrix} a \\ b \end{pmatrix} =\begin{pmatrix} a \\ b \end{pmatrix}$ and $\displaystyle A \begin{pmatrix} c \\ d \end{pmatrix} =\begin{pmatrix} -c \\ -d \end{pmatrix}.$ thus $\displaystyle a=b, \ c = -d.$ so we may choose $\displaystyle f_1=e_1 + e_2, \ f_2 = e_1 - e_2.$

the matrix of F in this new basis is obviously: $\displaystyle B=\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}.$

remember: if $\displaystyle F(e_j)=\sum_{i=1}^n c_{ij}e_i,$ then $\displaystyle \begin{pmatrix}c_{1j} \\ . \\ . \\ . \\ c_{nj} \end{pmatrix}$ would be the j-th column of the matrix of F with respect to the basis $\displaystyle \{e_1, \cdots , e_n \}.$
• Apr 10th 2009, 09:58 AM
Spec
Quote:

Originally Posted by NonCommAlg
the matrix of F in this new basis is obviously: ...

I can find it using the formula: $\displaystyle A_f = T^{-1}A_eT$, but I don't think that the solution you get from this is obvious. Is there an easier way to find it?
• Apr 10th 2009, 10:26 AM
NonCommAlg
Quote:

Originally Posted by Spec

I can find it using the formula: $\displaystyle A_f = T^{-1}A_eT$, but I don't think that the solution you get from this is obvious. Is there an easier way to find it?

yes, of course you may use the change of basis formula but my solution is direct and much easier: since $\displaystyle F(f_1)=f_1=1 \times f_1 + 0 \times f_2$ and $\displaystyle F(f_2)=-f_2=0 \times f_1 + (-1) \times f_2,$

the matrix of F in this basis, which i called it B, is just the matrix of coefficients. (see the remember note that i gave you at the end of my previous post)