Mathematical Immaturity

2.11 Construction of a matrix representation in diagonal form

$\quad$ Since it is possible to obtain different matrix representations of a given linear transformation by different choices of bases, it is natural to try to choose the bases so that the resulting matrix will have a particularly simple form. The next theorem shows that we can make all the entries $0$ except possibly along the diagonal starting from the upper left-hand corner of the matrix. Along this diagonal there will be a string of ones followed by zeros, the number of ones being equal to the rank of the transformation. A matrix $(t_{ik})$ with all entries $t_{ik} = 0$ when $i \neq k$ is said to be a diagonal matrix.

$\quad$ Theorem 2.14. $\quad$ Let $V$ and $W$ be finite-dimensional linear spaces, with $\dim V = n$ and $\dim W = m.$ Assume $T \in \mathscr{L}(V, W)$ and let $r = \dim T(V)$ denote the rank of $T.$ Then, there exists a basis $(e_1, \dots, e_n)$ for $V$ and a basis $(w_1, \dots, w_m)$ for $W$ such that \begin{align*} (2.14) \qquad T(e_i) &= w_i \qquad for \quad i=1, 2, \dots, r. \\ \\ \text{and} \\ \\ (2.15) \qquad T(e_i) &= O \qquad for \quad i=r + 1, \dots, n. \end{align*} Therefore, the matrix $(t_{ik})$ of $T$ relative to these bases has all entries zero except for the $r$ diagonal entries.

$\quad$ Proof. $\quad$ Let $w_1, \dots, w_r$ be the basis elements of $T(V).$ By Theorem 1.7, we know that these elements are a subset of a basis for $W.$ If we append independent elements $w_{r+1}, \dots, w_m$ to this set, we get \begin{align*} (w_1, \dots, w_r, w_{r+1}, \dots, w_m) \end{align*} which forms a basis for $W.$ Because $T$ is a linear transformation in $\mathscr{L}(V, W),$ for each $i = 1, 2, \dots, r,$ there is at least one $e_i$ in $V$ satisfying $(2.14).$ Then, if $\dim N(T) = k$, by the nullity plus rank theorem, we have $n = k + r,$ and there are $k$ independent elements $e_{r+1}, \dots, e_{r + k}$ that satisfy $(2.15).$

$\quad$ To complete the proof, we will show that the set $(e_1, \dots, e_r, e_{r+1}, \dots, e_{r+k})$ forms a basis for $V.$ Because $e_{r + 1}, \dots, e_{r + k}$ form a basis for $N(T),$ we know that they are $k$ independent elements in $V,$ (in other words, they form a subset of a basis for $V$). As such, we only need to show that $e_1, \dots, e_r$ are independent elements of $V.$ Taking a linear combination of $e_1, \dots, e_r$ and setting it to zero, we get \begin{align*} \sum_{i=1}^r c_ie_i &= O \end{align*} Applying $T$ to the sum and using $(2.14),$ we get \begin{align*} T\left(\sum_{i=1}^r c_ie_i\right) &= \sum_{i=1}^r c_iT(e_i) \\ &= \sum_{i=1}^r c_iw_i \\ &= O \end{align*} But because $w_1, \dots, w_r$ are independent, we have $c_1 = \cdots = c_r = 0,$ which implies that $e_1, \dots, e_r$ are independent. As such, \begin{align*} (e_1, \dots, e_r, e_{r+1}, \dots, e_{r+k}) \end{align*} is an independent set of $n$ elements in $V.$ In other words, it is a basis for $V.$ Thus, we have shown that there exist bases $(w_1, \dots, w_m)$ and $(e_1, \dots, e_n)$ satisfying $(2.14)$ and $(2.15). \quad \blacksquare$