Abstract Nonsense

Direct Sum of Linear Transformations and Direct Sum of Matrices (Pt. II)

Point of post: This is a literal continuation of this post. Treat it as one contiguous object.

We may make a similar statement about the kernels of the direct sum of linear transformations. Namely:

Theorem: Let $\mathscr{V}$, $\mathscr{W}_k,\mathscr{U}_k, \phi_k, T_k$ and $T$ be as in the definition of direct sum. Then,

$\displaystyle \ker\left(T\right)=\bigoplus_{k=1}^{m}\phi_k\left(\ker \left(T_k\right)\right)$

Proof: We prove first that

$\displaystyle \ker\left(T\right)=\sum_{k=1}^{m}\phi_k\left(\ker\left(T_k\right)\right)$

To see this we first that if $\displaystyle \sum_{k=1}^{m}\phi_k(x_k)\in\displaystyle \sum_{k=1}^{m}\phi_k\left(\ker\left(T_k\right)\right)$ where of course $x_k\in\ker\left(T_k\right)$ then

\displaystyle \begin{aligned}T\left(\sum_{k=1}^{m}x_k\right) &= \sum_{k=1}^{m}T_k\left(\phi^{-1}_k\left(\phi_k\left(x_k\right)\right)\right)\\ &= \sum_{k=1}^{m}T_k(x_k)\\ &= \sum_{k=1}^{m}\bold{0}\\ &= \bold{0}\end{aligned}

So that $\displaystyle \sum_{k=1}^{m}\phi_k\left(\ker \left(T_k\right)\right)\subseteq \ker\left(T\right)$. Conversely we see that if $\displaystyle \sum_{k=1}^{m}w_k\in\ker\left(T\right)$ then

\displaystyle \begin{aligned}\bold{0} &= T\left(\sum_{k=1}^{m}w_k\right)\\ &= \sum_{k=1}^{m}T_k\left(\phi^{-1}_k\left(w_k\right)\right)\end{aligned}

but since $\displaystyle \mathscr{V}=\bigoplus_{k=1}^{m}\mathscr{W}_k$ and $T_k\left(\phi^{-1}_k\left(w_k\right)\right)\in\mathscr{W}_k,\text{ }k\in[m]$ we may conclude (from a common characterization of independence of subspaces) that $T_k\left(\phi^{-1}_k(w_k)\right)=\bold{0},\text{ }k\in[m]$ or that $\phi^{-1}(w_k)\in\ker \left(T_k\right)$ or that $w_k\in \phi\left(\ker\left(T_k\right)\right)$ from where it follows that $\displaystyle \ker\left(T\right)\subseteq\sum_{k=1}^{m}\phi_k\left(\ker\left(T_k\right)\right)$ from where it indeed follows that $\displaystyle \ker\left(T\right)=\sum_{k=1}^{m}\phi_k\left(\ker\left(T_k\right)\right)$.  Thus, to prove that $\displaystyle \ker\left(T\right)=\bigoplus_{k=1}^{m}\phi_k\left(\ker\left(T_k\right)\right)$ it suffices to notice (similarly to the proof of the last theorem) that for each $k\in[m]$ we have that

$\displaystyle \{\bold{0}\}\subseteq \phi_k\left(\ker\left(T_k\right)\right)\cap\sum_{\substack{j\in[m]\\ j\ne k}}\phi_j\left(\ker\left(T_j\right)\right)\subseteq \mathscr{W}_k\cap\sum_{\substack{j\in[m]\\ j\ne k}}\mathscr{W}_j=\{\bold{0}\}$

From these two we can prove the following theorem:

Theorem: Let $\mathscr{V}$, $\mathscr{W}_k,\mathscr{U}_k, \phi_k, T_k$ and $T$ be as in the definition of direct sum. Then, $T\in\text{GL}\left(\mathscr{V}\right)$ if and only if $T_k\in\text{Hom}\left(\mathscr{U}_k,\mathscr{W}_k\right),\text{ }k\in[m]$ is an isomorphism.

Proof: Suppose first that $T_k$ is an isomorphism for each $k\in[m]$, then we note that by the above two theorems that

$\displaystyle T\left(\mathscr{V}\right)=\bigoplus_{k=1}^{m}T_k\left(\mathscr{U}_k\right)=\bigoplus_{k=1}^{m}\mathscr{W}_k=\mathscr{V}$

and

$\displaystyle \ker\left(T\right)=\bigoplus_{k=1}^{m}\phi_k\left(\ker\left(T_k\right)\right)=\bigoplus_{k=1}^{m}\phi_k\left(\{\bold{0}\}\right)=\{\bold{0}\}$

and thus $T\in\text{GL}\left(\mathscr{V}\right)$ as required.

Conversely, suppose that $T\in\text{GL}\left(\mathscr{V}\right)$ then we see from our two previous theorems that

$\displaystyle \bigoplus_{k=1}^{m}\mathscr{W}_k=\mathscr{V}=T\left(\mathscr{V}\right)=\bigoplus_{k=1}^{m}T\left(\mathscr{U}_k\right)$

from where it clearly follows (recalling that $T\left(\mathscr{U}_k\right)\subseteq\mathscr{W}_k$)  that $T\left(\mathscr{U}_k\right)=\mathscr{W}_k$. Similarly, we have that

$\displaystyle 0=\dim_F \{\bold{0}\}=\dim_F \ker \left(T\right)=\dim_F \bigoplus_{k=1}^{m}\phi_k\left(\ker\left(T_k\right)\right)=\sum_{k=1}^{m}\dim_F \phi_k\left(\ker\left(T_k\right)\right)$

from where it follows that $\dim_F \phi_k\left(\ker\left(T_k\right)\right)=0,\text{ }k\in[m]$ so that $\phi_k\left(\ker\left(T_k\right)\right)=\{\bold{0}\}$ and so $\ker\left(T_k\right)=\{\bold{0}\}$. Thus, combining these two gives that $T_k$ is an isomorphism for each $k\in[m]$.

References:

1. Halmos, Paul R.  Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print