Help me to solve this exercise.
- Thread starter rumble182
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Dr.Peterson
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Please show us where you are having trouble, so we can know what help to give you. What matrix did you make? Why are you not sure of it?
Or, at least, show us an example you have been given, of a matrix associated with a mapping, so we can see what you have learned.
Or, at least, show us an example you have been given, of a matrix associated with a mapping, so we can see what you have learned.
I know how to solve steps b, c, d but I cannot understand how change linear map to canonical bases on (a).
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it's a lot easier than you seem to be making it.
[MATH]A \begin{pmatrix}x\\y\\z\\t\end{pmatrix} = \begin{pmatrix}y\\x\\z\\t\end{pmatrix}[/MATH]
Solve for [MATH]A[/MATH]
HallsofIvy
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Do you understand what the "canonical basis" is? The canonical basis for R4 consists of the vectors
[math]\begin{pmatrix}1 \\ 0 \\ 0 \\ 0 \end{pmatrix}[/math], [math]\begin{pmatrix}0 \\ 1 \\ 0 \\ 0 \end{pmatrix}[/math], [math]\begin{pmatrix}0 \\ 0 \\ 1 \\ 0 \end{pmatrix}[/math] and [math]\begin{pmatrix}0 \\ 0 \\ 0 \\ 1 \end{pmatrix}[/math].
Equivalently to what Romsek suggested, since A is from R4 to R4, it can be written as a 4 by 4 matrix, [math]A= \begin{pmatrix}a & b & c & d \\ e & f & g & h \\i & j & k & i \\ m & n & o & p\end{pmatrix}[/math].
Now, since A maps (x, y, z. t) to (y, x, z. t). it maps (1, 0, 0, 0) to (0, 1, 0, 0):
[math]\begin{pmatrix}a & b & c & d \\ e & f & g & h \\i & j & k & i \\ m & n & o & p\end{pmatrix}\begin{pmatrix}1 \\ 0 \\ 0 \\ 0 \end{pmatrix}= \begin{pmatrix} a \\ e \\ i \\ m\end{pmatrix}= \begin{pmatrix}0 \\ 1 \\ 0 \\ 0 \end{pmatrix}[/math] so that a= 0, e= 1, i= 0, and m= 0.
And
[math]\begin{pmatrix}a & b & c & d \\ e & f & g & h \\i & j & k & i \\ m & n & o & p\end{pmatrix}\begin{pmatrix}0 \\ 1 \\ 0 \\ 0 \end{pmatrix}= \begin{pmatrix} b \\ f\\ j \\ n\end{pmatrix}= \begin{pmatrix}1 \\ 0 \\ 0 \\ 0 \end{pmatrix}[/math] so that b= 1, f= 0, j= 0, and n= 0.
[math]\begin{pmatrix}1 \\ 0 \\ 0 \\ 0 \end{pmatrix}[/math], [math]\begin{pmatrix}0 \\ 1 \\ 0 \\ 0 \end{pmatrix}[/math], [math]\begin{pmatrix}0 \\ 0 \\ 1 \\ 0 \end{pmatrix}[/math] and [math]\begin{pmatrix}0 \\ 0 \\ 0 \\ 1 \end{pmatrix}[/math].
Equivalently to what Romsek suggested, since A is from R4 to R4, it can be written as a 4 by 4 matrix, [math]A= \begin{pmatrix}a & b & c & d \\ e & f & g & h \\i & j & k & i \\ m & n & o & p\end{pmatrix}[/math].
Now, since A maps (x, y, z. t) to (y, x, z. t). it maps (1, 0, 0, 0) to (0, 1, 0, 0):
[math]\begin{pmatrix}a & b & c & d \\ e & f & g & h \\i & j & k & i \\ m & n & o & p\end{pmatrix}\begin{pmatrix}1 \\ 0 \\ 0 \\ 0 \end{pmatrix}= \begin{pmatrix} a \\ e \\ i \\ m\end{pmatrix}= \begin{pmatrix}0 \\ 1 \\ 0 \\ 0 \end{pmatrix}[/math] so that a= 0, e= 1, i= 0, and m= 0.
And
[math]\begin{pmatrix}a & b & c & d \\ e & f & g & h \\i & j & k & i \\ m & n & o & p\end{pmatrix}\begin{pmatrix}0 \\ 1 \\ 0 \\ 0 \end{pmatrix}= \begin{pmatrix} b \\ f\\ j \\ n\end{pmatrix}= \begin{pmatrix}1 \\ 0 \\ 0 \\ 0 \end{pmatrix}[/math] so that b= 1, f= 0, j= 0, and n= 0.
