Find bases $(e_1,e_2,e_3)$ and $w_1,w_2$ in $R^2$ relative to which the matrix of $T$ = $ \left(\begin{array}{ccc}0 & 1 & 1 \\0 & 1 & -1\end{array}\right)$ is in diagonal form.
Ans: $e_1 = j, e_2 = k, e_3 = i$, $w_1 = (1,1)$, $w_2 = (1,-1)$
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Best Answer
To write $T$ in a diagonal form one has to find a basis $(e_1, e_2, e_3)$ in ${\mathbb R}^3$ and a basis $(w_1, w_2)$ in ${\mathbb R}^2$ such that $$ Te_1=w_1, \quad Te_2=w_2, \quad \text{and}\quad Te_3=0. \tag1$$ (This is possible because $T$ has rank $2$ which is the dimension of the second space.) Let $(i,j,k)$ be the standard basis in ${\mathbb R}^3$ and $(i,j)$ the standard basis in ${\mathbb R}^2$. Observe that $$ T i=0,\quad Tj=i+j, \quad \text{and}\quad Tk=i-j. $$ Since $w_1=i+j$ and $w_2=i-j$ are linearly independent they form a basis of ${\mathbb R}^2$. Hence if one set $e_1=j, e_2=k, e_3=0$, then we have (1).
Note that the above linear form is not unique, conditions in (1) can be replaced by $$ Tf_1=\lambda u_1, \quad Tf_2=\mu u_2, \quad \text{and}\quad Tf_3=0 $$ where $\lambda, \mu$ are nonzero numbers and $(f_1, f_2, f_3)$ is a basis of ${\mathbb R}^3$ and $(u_1, u_2)$ is a basis of ${\mathbb R}^2$.
Explanation
Let $V$ and $W$ be vector spaces over ${\mathbb R}$, $\dim(V)=m$ and $\dim(W)=n$, and let $T:V\to W$ be a linear transformation (which means that for arbitrary $v_1, v_2 \in V$ and $\alpha_1, \alpha_2 \in {\mathbb R}$ one has $T(\alpha_1 v_1+\alpha_2 v_2)=\alpha_1 Tv_1+\alpha_2 Tv_2$). If $(e_1, \ldots, e_m)$ is a basis of $V$ and $(f_1, \ldots, f_n)$ is a basis of $W$, then $T$ can be represented by an $n\times m$ matrix as follows. Let $j\in \{ 1, \ldots, m\}$ and consider vector $Te_j$. It is a vector in $W$ and therefore can be uniquely represented as $$ Te_j=t_{1j}f_1+\cdots+t_{nj}f_j $$ because $(f_1, \ldots, f_n)$ is a basis of $W$. Numbers $t_{1j}, \ldots t_{nj}$ form the $j$-th column of the matrix which represents $T$ with respect to the bases $(e_1, \ldots, e_m)$ and $(f_1, \ldots, f_n)$. Hence, the whole matrix is $$ \left[ \begin{array}{cccc} t_{11} & t_{12} & \cdots & t_{1m}\\ t_{21} & t_{22} & \cdots & t_{2m}\\ \vdots & \vdots & \ddots & \vdots\\ t_{n1} & t_{n2} & \cdots & t_{nm} \end{array} \right]. $$ If one represents $v=\lambda_1 e_1+\cdots+\lambda_m e_m$ as a column vector $$ \left[ \begin{array}{c} \lambda_1\\ \lambda_2\\ \vdots\\ \lambda_m\end{array}\right], $$ then $$ \left[ \begin{array}{cccc} t_{11} & t_{12} & \cdots & t_{1m}\\ t_{21} & t_{22} & \cdots & t_{2m}\\ \vdots & \vdots & \ddots & \vdots\\ t_{n1} & t_{n2} & \cdots & t_{nm} \end{array} \right]\left[ \begin{array}{c} \lambda_1\\ \lambda_2\\ \vdots\\ \lambda_m\end{array}\right]= \left[ \begin{array}{c} t_{11}\lambda_1+t_{12}\lambda_2+\cdots+t_{1m}\lambda_m\\ t_{21}\lambda_1+t_{22}\lambda_2+\cdots+t_{2m}\lambda_m\\ \vdots\\ t_{n1}\lambda_1+t_{n2}\lambda_2+\cdots+t_{nm}\lambda_m\end{array}\right]=\left[ \begin{array}{c} \mu_1\\ \mu_2\\ \vdots\\ \mu_n\end{array}\right], $$ where $\mu_1, \cdots, \mu_n$ are the coefficients in the expansion of vector $Tv$ in basis $(f_1, \ldots, f_n)$, i.e., $$ Tv=\mu_1 f_1+\cdots+\mu_n f_n.$$
In the case of the vector space ${\mathbb R}^k$ one has a natural basis, called the standard basis: $$ \left[ \begin{array}{c} 1 \\ 0\\ \vdots \\ 0 \end{array} \right],\quad \left[ \begin{array}{c} 0 \\ 1\\ \vdots \\ 0 \end{array} \right],\quad \ldots, \quad \left[ \begin{array}{c} 0 \\ 0\\ \vdots \\ 1 \end{array} \right]. $$ If an $n\times m$ matrix is given, then one considers it as a matrix which represents a linear transformation from ${\mathbb R}^m$ to ${\mathbb R}^n$ with respect to the standard bases in ${\mathbb R}^m$, resp. in ${\mathbb R}^n$.
In the case above one has $m=3$ and $n=2$. The standard bases are usualy denoted as $$ i=\left[ \begin{array}{c}1\\ 0\\ 0\end{array}\right], \quad j=\left[ \begin{array}{c}0\\ 1\\ 0\end{array}\right], \quad k=\left[ \begin{array}{c}0\\ 0\\ 1\end{array}\right]\quad \text{in}\quad {\mathbb R}^3$$ and $$ i=\left[ \begin{array}{c}1\\ 0\end{array}\right], \quad j=\left[ \begin{array}{c}0\\ 1\end{array}\right]\quad \text{in}\quad {\mathbb R}^2.$$ We have a matrix $$ \left[ \begin{array}{rrr} 0 & 1 & 1\\0 & 1 & -1\end{array}\right] $$ which represents a linear transformation $T:{\mathbb R}^3 \to {\mathbb R}^2$ with respect to the standard bases $(i,j,k)$ in ${\mathbb R}^3$ and $(i,j)$ in ${\mathbb R}^2$. The question is to find such bases $(e_1, e_2, e_3)$ in ${\mathbb R}^3$ and $(w_1, w_2)$ in ${\mathbb R}^2$ that $T$ will be represented by a diagonal matrix with respect to these bases. It means that the bases have to be chosen in a way that one has $$ Te_1=\lambda w_1,\quad Te_2=\mu w_2,\quad Te_3=0 $$ for some numbers $\lambda$ and $\mu$. Since $T$ is surjective $\lambda$ and $\mu$ are non-zero. Also one can assume that $\lambda=\mu =1$ (if there was an additional condition that the bases have to consist of vectors with norm $1$, then we couldn't assume that those numbers are $1$ but we would calculate them - however in our case there is no condition about norms, so we can assume that those numbers are $1$). Hence we would like to have $$ Te_1=w_1,\quad Te_2=w_2,\quad Te_3=0. \tag2$$ Now we observe that $$ T i=0,\quad Tj=i+j, \quad \text{and}\quad Tk=i-j. $$ Hence if we choose new bases as follows:
$e_1=j,\quad e_2=k,\quad e_3=0\quad$ the new basis in ${\mathbb R}^3$ and
$w_1=i+j, \quad w_2=i-j\quad$ the new basis in ${\mathbb R}^2$,
then (2) is fulfilled. Hence with respect to these new bases $T$ is represented by the matrix $$ \left[ \begin{array}{rrr} 1 & 0 & 0\\0 & 1 & 0\end{array}\right]. $$ It means, if we expand $v\in {\mathbb R}^3$ with respect to the basis $(e_1,e_2,e_3)$ as $$ v=\alpha_1 e_1+\alpha_2 e_2+\alpha_3 e_2, $$ then $Tv$, which is a vector in ${\mathbb R}^2$, will have an expansion $$ Tv=\alpha_1 w_1+\alpha_2 w_2 $$ with respect to the basis $(w_1,w_2)$ in ${\mathbb R}^2$: $$ \left[ \begin{array}{rrr} 1 & 0 & 0\\0 & 1 & 0\end{array}\right] \left[ \begin{array}{c} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{array}\right]=\left[ \begin{array}{c} \alpha_1 \\ \alpha_2 \end{array}\right]. $$ $\left[ \begin{array}{c} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{array}\right]$ is the column of $v$ with respect to the basis $(e_1, e_2, e_3)$ and $\left[ \begin{array}{c} \alpha_1 \\ \alpha_2 \end{array}\right]$ is the column of $Tv$ with respect to the basis $(w_1,w_2)$.