I do not understand how to interpret this statement/syntax on linear transformations

wduk

New member
Joined
Dec 11, 2016
Messages
46
So this book has:


If f is a linear transformation, and f(1,0) = (a,c) and f(0,1) = (b,d)

Then the matrix of f is:

a b
c d

What i do not understand is what it means by f(1,0) = (a,c) and f(0,1) = (b,d). So position x becomes a,c but say x was 5 - it tells me little on the logic of how it becomes a and c. The syntax is a bit unusual as it's not entirely clear how it maps.
Can some one clarify this in a more clear way so i understand what it is trying to tell me please.

Currently my guess is, if i had a point (3,2), the matrix becomes:

3 0
0 2

But then b and c are always 0 so that seems a bit redundant in the rule. Hope some one can clear my confusion on this or perhaps has a cleaner way to write the rule so i just change it in a more understandable way.

Thanks
 
So this book has:

If f is a linear transformation, and f(1,0) = (a,c) and f(0,1) = (b,d)

Then the matrix of f is:

a b
c d

What i do not understand is what it means by f(1,0) = (a,c) and f(0,1) = (b,d). So position x becomes a,c but say x was 5 - it tells me little on the logic of how it becomes a and c. The syntax is a bit unusual as it's not entirely clear how it maps.
Can some one clarify this in a more clear way so i understand what it is trying to tell me please.

Currently my guess is, if i had a point (3,2), the matrix becomes:

3 0
0 2

But then b and c are always 0 so that seems a bit redundant in the rule. Hope some one can clear my confusion on this or perhaps has a cleaner way to write the rule so i just change it in a more understandable way.

Thanks

I suspect that you are expecting some different syntax, so you don't see what it is saying quite clearly:

When it says, f(1,0) = (a,c), that means that the transformation f maps the vector (1,0) to the vector (a,c).

In fact, you can see this if you check it out by multiplying the matrix by the column vector (1,0):

\(\displaystyle \begin{pmatrix}a & b\\ c & d\end{pmatrix}\begin{pmatrix}1\\ 0\end{pmatrix}
=\begin{pmatrix}a\\ c\end{pmatrix}\)

If this does not make it clear, perhaps you can tell us what you mean by "position x".
 
Top