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C28. Let T : C4\(\rightarrow\) M22 be given by
\[\begin{equation} T= \Bigg(\Bigg[ \begin{matrix} a \\ b \\ c \\ d \end{matrix} \Bigg]\Bigg) = \Bigg(\Bigg[ \begin{matrix} a+b&a+b+c \\ a+b+c&a+d \end{matrix} \Bigg]\Bigg) \end{equation}\]Find a basis of R(T). Is T surjective?
Answer:
\[\begin{equation} V = \Bigg(\Bigg[ \begin{matrix} a+b&a+b+c \\ a+b+c&a+d \end{matrix} \Bigg]\Bigg) = a \Bigg[ \begin{matrix} 1&1 \\ 1&1 \end{matrix} \Bigg] + b \Bigg[ \begin{matrix} 1&1 \\ 1&0 \end{matrix} \Bigg] + c \Bigg[ \begin{matrix} 0&1 \\ 1&0 \end{matrix} \Bigg] + d\Bigg[ \begin{matrix} 0&0 \\ 0&1 \end{matrix} \Bigg] \end{equation}\]So
\[\begin{equation} R(T) = \Bigg\langle \Bigg[ \begin{matrix} 1&1 \\ 1&1 \end{matrix} \Bigg] , \Bigg[ \begin{matrix} 1&1 \\ 1&0 \end{matrix} \Bigg] , \Bigg[ \begin{matrix} 0&1 \\ 1&0 \end{matrix} \Bigg] , \Bigg[ \begin{matrix} 0&0 \\ 0&1 \end{matrix} \Bigg] \Bigg\rangle = \Bigg\langle \Bigg[ \begin{matrix} 1&1 \\ 1&0 \end{matrix} \Bigg] , \Bigg[ \begin{matrix} 0&1 \\ 1&0 \end{matrix} \Bigg] , \Bigg[ \begin{matrix} 0&0 \\ 0&1 \end{matrix} \Bigg] \Bigg\rangle \end{equation}\]To see whether this is the basis of V,
\[\begin{equation} \alpha_1 \Bigg[ \begin{matrix} 1&1 \\ 1&0 \end{matrix} \Bigg] +\alpha_2 \Bigg[ \begin{matrix} 0&1 \\ 1&0 \end{matrix} \Bigg] +\alpha_3 \Bigg[ \begin{matrix} 0&0 \\ 0&1 \end{matrix} \Bigg] = \Bigg[ \begin{matrix} 0&0 \\ 0&0 \end{matrix} \Bigg] \label{eq:1}\tag{1} \end{equation}\] \[\begin{equation} \alpha_1 =0 \\ \alpha_1 + \alpha_2 = 0\\ \alpha_3 = 0 \end{equation}\]then we know \(\alpha_1 =\alpha_2 =\alpha_3 = 0\)
The equation (1) only has trival slution, so the above three matrices in R(T) are linear independent and span V, it is the range of T.
Because dim(R(T)) = 3 \(\neq\) dim(\(M_{22}\)), T is not surjective.