Proof by induction of block diagonal decomposition of a matrix

[psie]

TL;DR

I'm working two exercises, one following the other. The first asks one to prove directly the case $k=2$ of the theorem below. The other is about establishing the theorem by induction for any $k$. I'm a bit perplexed; how do you prove the theorem by induction? In particular, how do you use the induction hypothesis in a possible proof by induction?

Theorem. Let $\mathsf T$ be a linear operator on a finite dimensional vector space $\mathsf V$, and let $\mathsf W_1,\mathsf W_2,\ldots,\mathsf W_k$ be $\mathsf T$-invariant subspaces of $\mathsf V$ such that $\mathsf V=\mathsf W_1\oplus\mathsf W_2\oplus\cdots\oplus\mathsf W_k$. For each $i$, let $\beta_i$ be an ordered basis for $\mathsf W_i$, and let $\beta=\beta_1\cup\beta_2\cup\cdots\cup\beta_k$. Let $A=[\mathsf T]_\beta$ and $B_i=[\mathsf T_{\mathsf W_i}]_{\beta_i}$ for $i=1,2,\ldots, k$. Then $A=B_1\oplus B_2\oplus \cdots\oplus B_k$.

Here $A=B_1\oplus B_2\oplus \cdots\oplus B_k$ means that $A$ is block diagonal with $B_i$ along the diagonal.

A proof that I've seen goes as follows. The $k=2$ case boils down to realizing that if $v\in\beta_1\subseteq\mathsf W_1$, then $\mathsf T(v)$ is a linear combination of vectors in $\beta_1$ (since $\mathsf W_1$ is $\mathsf T$-invariant and $\beta_1$ is a basis for this subspace). A similar thing can be said about $\mathsf T(v)$ for any $v\in\beta_2$. So clearly the matrix representation with respect to $\beta_1 \cup \beta_2$ is block diagonal. I feel like one could go on with this argument for any finite number of subspaces $k$. How do you prove this by induction? In particular, how do you use the induction hypothesis in a possible proof by induction?

 

[PeroK]

Are you asking about induction in general or in this particular case?

 

[psie]

Are you asking about induction in general or in this particular case?

In this particular case. I'm confused how one would use the induction hypothesis.

 

[fresh_42]

The case $k=2$ is the induction step. The induction hypothesis is that you have done it for $k-1$ being invariant under $\beta_1\oplus \ldots\oplus \beta_{k-1}.$ Since every semidirect product in the category of vector spaces is already direct, there is not much to add. It's getting interesting in categories in which short-exact sequences do not automatically split.

 

[fresh_42]

The automatism of a split exact sequence here is equivalent to the invariance under $T.$ Means, the split exact sequences in the category of vector spaces only apply to the choice of bases, i.e. that any quotient $V/U$ can be considered as a subspace again. I was wrong to use this as an argument for the theorem since the invariance of the transformation is crucial here, not just subsapces.

If we have only one $T$-invariant subspace $U\subseteq V,$ then $T$ is a block-triangular matrix since $V/U$ can have entries in the basis for $U.$ This happens all the time in the categories of groups, rings, algebras, etc. if $U$ is a normal subgroup or an ideal but $V/U$ is not isomorphic to a normal subgroup or ideal again.

In this case, we need the $T$-invariance of all subspaces to control the off-diagonal entries.