LADR 3.B.5

This is taken from Sheldon Axler's magisterial book, aptly named “Linear Algebra Done Right”. The exercise (3.B.5 on p. 67) asks for an example of a linear map $T: \mathbb{R}^4 \rightarrow \mathbb{R}^4$ such that $\mathrm{range}(T)=N(T)$ where $N(T)$ is the null space of $T$.

We know that a matrix $A$, \begin{equation} A = \begin{bmatrix} A_1 & A_2 & A_3 & A_4 \end{bmatrix} \end{equation} with $A_1$, $A_2$ two independent vectors and $A_3=aA_1+bA_2$, $A_4=cA_1+dA_2$ has a range equal to $\mathrm{Span}\{A_1,A_2\}$. $N(A)$ contains vectors $x$ with the property, \begin{equation} x_1 A_1 + x_2 A_2 + x_3 A_3 + x_4 A_4 = 0 \end{equation} or, \begin{equation} (x_1+ax_3+cx_4) A_1 + (x_2+bx_3+dx_4) A_2 = 0 \label{null_cond} \end{equation} Since $A_1$, $A_2$ are independent vectors ($\ref{null_cond}$) holds only if, \begin{align} x_1 & = - a x_3 - c x_4 \\ x_2 & = - b x_3 - d x_4 \end{align} Therefore any $x \in N(A)$ can be written as, \begin{equation} x = x_3 \mqty( -a \\ -b \\ 1 \\ 0) + x_4 \mqty(-c \\ -d \\ 0 \\ 1) \end{equation} If, \begin{equation} A_1 = \mqty( -a \\ -b \\ 1 \\ 0) \end{equation} and, \begin{equation} A_2 = \mqty(-c \\ -d \\ 0 \\ 1) \end{equation} then $N(A)=\mathrm{Span}\{A_1,A_2\}=\mathrm{range}(A)$. Note that $A_1$, $A_2$ are independent since, \begin{equation} x \mqty( -a \\ -b \\ 1 \\ 0) + y \mqty(-c \\ -d \\ 0 \\ 1) =0 \end{equation} holds only for $x=y=0$. Therefore we can write, \begin{equation} A=\mqty[ -a & -c & -a^2-bc & -(a+d)c \\ -b & -d & -(a+d)b & -d^2-bc \\ 1 & 0 & a & c \\ 0 & 1 & b & d ] = \mqty[ -{\tilde A} & -{\tilde A}^2 \\ I & {\tilde A} ] \label{form} \end{equation} where, \begin{equation} {\tilde A}= \mqty[a & c \\ b & d] \end{equation}

The form of ($\ref{form}$) can be extended to any $2n \times 2n$ matrix so that the first $n$ columns are independent vectors and the last $n$ columns are linear combinations of these vectors. All matrices of this form have $\mathrm{range}(A)=N(A)$.

Note that this is not possible for $T:\mathbb{R}^{2n+1}\rightarrow \mathbb{R}^{2n+1}$ since $\mathrm{dim}\;\mathrm{range}(A)+\mathrm{dim}N(A)=2n+1$ and therefore $\mathrm{dim}\;\mathrm{range}(A) \neq \mathrm{dim}N(A)$ so $\mathrm{range}(A)\neq N(A)$.

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