Abstract Nonsense

Crushing one theorem at a time

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

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December 22, 2010 - Posted by | Algebra, Halmos, Linear Algebra, Uncategorized | ,

1 Comment »

  1. […] of post: This is a literal continuation of this post. Treat the two posts as one contiguous […]

    Pingback by Direct Sum of Linear Transformations and Direct Sum of Matrices (Pt. III) « Abstract Nonsense | December 22, 2010 | Reply


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