This worksheet deals with matrix transformations, and in particular, kernel and image. The goal is to understand these important subspaces in a familiar context.
Let \(A\) be an \(m\times n\) matrix. We can use \(A\) to define a transformation \(T_A:\R^n\to \R^m\) given by \(T_A(\xx)=A\xx\text{,}\) where we view \(\xx\) as an \(n\times 1\) column vector.
The kernel of \(T_A\) is the set of vectors \(\xx\) such that \(T_A(\xx)=\zer\text{.}\) That is, \(\ker T_A\) is the set of solutions to the homogeneous system \(A\xx = \zer\text{.}\)
The image of \(T_A\) (also known as the range of \(T_A\)) is the set of vectors \(\yy\in \R^m\) such that \(\yy = A\xx\) for some \(\xx\in\R^n\text{.}\) In other words, \(\im(T_A)\) is the set of vectors \(\yy\) for which the non-homogeneous system \(A\xx=\yy\) is consistent.
Because \(T_A\) is a linear transformation, we can compute it as long as weβre given its values on a basis. If \(\{\vv_1, \vv_2,\ldots, \vv_n\}\) is a basis for \(\R^n\text{,}\) then for any \(\xx\in\R^n\) there exist unique scalars \(c_1,c_2,\ldots, c_n\) such that
The main challenge, computationally speaking, is that if our basis is not the standard basis, some effort will be required to write \(\xx\) in terms of the given basis.
To assist with solving this problem, a code cell is provided below. Recall that you can enter the matrix \(\begin{bmatrix}a\amp b\amp c\\d\amp e\amp f\\g\amp h\amp i\end{bmatrix}\) as Matrix([[a,b,c],[d,e,f],[g,h,i]]) or as Matrix(3,3,[a,b,c,d,e,f,g,h,i]).
The reduced row-echeleon form of A is given by A.rref(). The product of matrices A and B is simply A*B. The inverse of a matrix A can be found using A.inv() or simply A**(-1).
One note of caution: in the HTML worksheet, if you donβt import sympy as your first line of code, youβll instead use Sage syntax. Sage uses A.inverse() instead of A.inv().
Let \(M\) be the matrix whose columns are given by the values of \(T\) on the basis \(B\text{.}\) (This would be the matrix of \(T\) if \(B\) was actually the standard basis.) Let \(N\) be the matrix whose inverse you used to solve part (b). Can you find a way to combine these matrices to obtain the matrix \(A\text{?}\) If so, explain why your result makes sense.
Next we will compute the kernel and image of the transformation from the previous exercises. Recall that when solving a homogeneous system \(A\xx = \zer\text{,}\) we find the RREF of \(A\text{,}\) and any variables whose columns do not contain a leading 1 are assigned as parameters. We then express the general solution \(x\) in terms of those parameters.
The image of a matrix transformation \(T_A\) is also known as the column space of \(A\text{,}\) because the range of \(T_A\) is precisely the span of the columns of \(A\text{.}\) The RREF of \(A\) tells us which columns to keep: the columns of \(A\) that correspond to the columns in the RREF of \(A\) with a leading 1.