Example 3.5.1. Linear transformaation defined by a matrix..
Let us define a linear tarnsformatrion defined by a matrix \(A\) over a rational numbers.
Clearly, here \(T\) is the matrix tranformatrion defined from \(\Q^3\) to \(\Q^2\text{.}\) We can find, the domain, codmain, parent, images, kernel, that is null space etc using the dot methods. Let us explore some of them.
This returns a basis matrix of the image along with the dimension of the image space.
We can find image of any vector using the following Sage syntax.
If we take a vector \(w\) in the range space of \(T\text{,}\) we can find a representative of the \(T^{-1}(w)\text{.}\)
Note that the same linear transformation, we can also define as follows.
Problem 3.5.2.
Define a matrix \(A=\left(\begin{array}{rrr}
-1 \amp 1 \amp 0 \\
2 \amp 0 \amp -1 \\
2 \amp -1 \amp 1 \\
0 \amp 2 \amp 1
\end{array}\right)\) and define a linear transformation \(T:\Q^3\to \Q^4\) by \(T(x)=Ax\text{.}\) Now define two vectors \(w_1=(-3, -1, 0, -31/3)\) and \(w_2=(1, -1/2, -1, 11/6)\) . Check that \(w_1,w_2\) lie in the image space of \(T\text{.}\) Let \(v_1\in T^{-1}(w_1)\) and \(v_2\in T^{-1}(w_1)\text{.}\) Show that \(v_1, v_2\) are linearly independent in \(\Q^3\text{.}\) Can you generalize this result and prove the same?
Example 3.5.3.
Let us define a linear transformation \(T:\Q^4\to\Q^4\) given by
\begin{equation*}
T(x_1,x_2,x_3,x_4)=\left(\begin{array}{r}
-x_{1} + 2 \, x_{3} - x_{4} \\
x_{1} + 3 \, x_{2} + 7 \, x_{3} - 2 \, x_{4} \\
x_{1} - 3 \, x_{2} - 21 \, x_{3} + 4 \, x_{4} \\
x_{1} + 6 \, x_{2} + 16 \, x_{3} - 5 \, x_{4}
\end{array}\right)
\end{equation*}
and explore this in Sage.
Next we find the matrix of \(T\) with respect to a bases \(\beta\) on the domain and a basis \(\gamma\) on the codomain.
Now we define a subspace \(V_1\) of \(V\) with B as a basis and a subspace \(W_1\) of \(W\) with C as a basis. After that we rectrict T on to \(V_1\) on its domain and \(W_1\) on its codomain. We call this restriction as \(T_1\text{.}\) The matrix of \(T_1\) is the matrix of \(T\) with respect to \(\beta\) on domain and \(\gamma\) on codmain, that is \([T]_\beta^\gamma\text{.}\)
Note that the matrix \([T]_\beta^\gamma\) can also be obtained by defining a column matrix whose columns are \(w_1,w_2,w_3,w_4, T(v_1), T(v_2), T(v_3), T(v_4)\text{,}\) then applying RREF and extracting the last four columns.
The image of \(T\) can also be obtained by using RREF of the matrix whose rows are vectors \(T(u1),T(u2),T(u3),T(u4)\text{.}\)
Note that the first three rows of this is sage as what we get using T.image().
Reshufle the basis elements. Let us see what happens to the matrix of a linear transformation when we reshuffle the elemnts of a basis on domain and codomain. Let us find the matrix of the abobe linear transformation with resepect the basis \(\beta_1 =\{u_1,u_3,u_2,u_4\}\) of the domain and \(\gamma\) of the codomain. Here we have interchanged 2nd and 3rd basis elements of \(\beta\text{.}\) You may also explore with other interchanges.
Clearly in this case 2nd and 3rd columns of \([T]_\beta^\gamma\) are interchanged.
Clearly in this case 2nd and 3rd rows of \([T]_\beta^\gamma\) are interchanged.
