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Let us consider $\vec{b}_{1}=\begin{bmatrix}2\\1\end{bmatrix}$, $\vec{b}_{2}=\begin{bmatrix}-1\\-1\end{bmatrix}$ vectors of $R^{2}$ and $\vec{c}_{1}=\begin{bmatrix}1\\2\\-1\end{bmatrix}$, $\vec{c}_{2}=\begin{bmatrix}0\\-1\\2\end{bmatrix}$, $\vec{c}_{3}=\begin{bmatrix}1\\0\\-1\end{bmatrix}$ vectors of $R^{3}$.
Furthermore, let us consider a homomorphism $F: R^{2} \rightarrow R^{3}$ such that:
$F(\vec{b}_{1} + \vec{b}_{2}) = \vec{c}_{2} + \vec{c}_{3}$
$F(\vec{b}_{1} - \vec{b}_{2}) = 2\vec{c}_{1} - \vec{c}_{2} + 3\vec{c}_{3}$
where $B = (\vec{b}_{1}, \vec{b}_{2})$ and $C = (\vec{c}_{1}, \vec{c}_{2}, \vec{c}_{3})$ are orderd bases of $R^{2}$ and $R^{3}$ respectively.

I need to determine the transformation matrix $A_{F}$ of $F$ with respect to the order bases $B$ and $C$.

What bugs me the most is how can I express the transformation matrix if I am given the transformations on two arbitrary vectors from the domain, concretely:
$F(\begin{bmatrix}1\\1\end{bmatrix}_{B}) = \begin{bmatrix}0\\1\\2\end{bmatrix}_{C}$
$F(\begin{bmatrix}1\\-1\end{bmatrix}_{B}) = \begin{bmatrix}2\\-1\\3\end{bmatrix}_{C}$

I tried expressing all the vectors in the standard basis for $R^{2}$ and $R^{3}$ respectively:
$F(\begin{bmatrix}1\\0\end{bmatrix}) = \begin{bmatrix}1\\-1\\1\end{bmatrix}$
$F(\begin{bmatrix}3\\2\end{bmatrix}) = \begin{bmatrix}5\\5\\-7\end{bmatrix}$

but that got me nowhere. I understand that at some point the change of basis matrices will come in, but I cannot seem to determine the transformation matrix between the standard bases in $R^{2}$ and $R^{3}$.

If anyone is curious, this is exercise 2.20.d) from the book Mathematics for Machine Learning, I am self studying and there are no solutions published, so I am having a bit of hard time, so thank you for your input.

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You can represent a given linear transformation (you use the term "homomorphism" but a homomorphism from $R^2$ to $R^2$ is a linear transformation), with given bases for both domain and range spaces, you 1) Apply the linear transformation to each basis vector in the domain, in turn 2) Write the result as a linear combination of the basis vectors in the range 3) Use the coefficients to form a column in the matrix.

You appear to know that but are not told, directly, what $F(\vec{b_1})$ and $F(\vec{b_2})$ are.

Here you are told that $F(\vec{b_1}+ \vec{b_2})= \vec{c_1}+ \vec{c_2}+ \vec{c_3}$ and that $F(\vec{b_1}- \vec{b_2})= 2\vec{c_1}- \vec{c_2}+ 3\vec{c_3}$.

The key is that, since this is a homomorphism (linear transformation), $F(\vec{b_1}+ \vec{b_2})= F(\vec{b_1})+ F(\vec{b_2})$ and $F(\vec{b_1}- \vec{b_2})= F(\vec{b_1})- F(\vec{b_2})$.

So you have $F(\vec{b_1})+ F(\vec{b_2})= \vec{c_1}+ \vec{c_2}+ \vec{c_3}$ and $F(\vec{b_1})- F(\vec{b_2})= 2\vec{c_1}- \vec{c_2}+ 3\vec{c_3}$.

Now do you see how to find $F(\vec{b_1})$ and $F(\vec{b_2})$ separately?

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  • $\begingroup$ Hello, thank you for your reply. I followed your instructions and ended up with $\begin{bmatrix}1&-1\\0&1\\-2&1\end{bmatrix}$. However, your answer left me with another question that has been secretly bugging me - why is it that sometimes applying a transformation matrix like this on a vector from the standard basis yields a vector with coordinates in the standard basis, but in this case, the transformation matrix "knows" how to convert between these two bases? $\endgroup$ Jun 2 '20 at 19:36
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    $\begingroup$ It "knows" because it was built from the two bases to convert a basis vector in one base to a basis vector in the other base. $\endgroup$
    – user247327
    Jun 3 '20 at 13:14

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