Question 4.3.4: Consider the linear transformation T (M) =[0 1 0 0] M − M [0......
Consider the linear transformation
T (M) =\begin{bmatrix}0&1\\0&0\end{bmatrix} M − M \begin{bmatrix}0&1\\0&0\end{bmatrix} from ℝ^{2×2} to ℝ^{2×2}.
a. Find the matrix B of T with respect to the standard basis 𝔅 of ℝ^{2×2} .
b. Find bases of the image and kernel of B.
c. Find bases of the image and kernel of T , and thus determine the rank and nullity of transformation T .
a. For the sake of variety, let us find B by means of a diagram.
We see that
B=\begin{bmatrix}0&0&1&0\\-1&0&0&1\\0&0&0&0\\0&0&-1&0\end{bmatrix}.
b. Note that columns \vec{v}_2 and \vec{v}_4 of B are redundant, with \vec{v}_2 = 0 and \vec{v}_4 = −\vec{v}_1, or \vec{v}_1 + \vec{v}_4 = 0. Thus, the nonredundant columns
\vec{v}_1=\begin{bmatrix}0\\-1\\0\\0\end{bmatrix},\vec{v}_3=\begin{bmatrix}1\\0\\0\\-1\end{bmatrix} form a basis of im(B),
and
\begin{bmatrix}0\\-1\\0\\0\end{bmatrix},\begin{bmatrix}1\\0\\0\\-1\end{bmatrix} form a basis of ker(B).
c. We apply L^{−1}_𝔅 to transform the vectors we found in part (b) back into ℝ^{2×2}, the domain and target space of transformation T :
\begin{bmatrix}0&-1\\0&0\end{bmatrix}, \begin{bmatrix}1&0\\0&-1\end{bmatrix} is a basis of im(T ),
and
\begin{bmatrix}0&1\\0&0\end{bmatrix}, I_2=\begin{bmatrix}1&0\\0&1\end{bmatrix} is a basis of ker(T ).
Thus, rank(T ) = dim(im T ) = 2 and nullity(T ) = dim(ker T ) = 2.