A direct sum decomposition into kernels and images.

Algebra Level 3

True or False: For every linear operator O : V V O: V \to V , there exist linear operators O i O_i such that there is a direct sum decomposition of V V as follows: V = i [ ker O i im O i ] \displaystyle V = \bigoplus_{i} [\ker O_i \oplus \text{im} O_i] .

True False

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1 solution

Hobart Pao
Jun 26, 2018

Every linear operator can be written as a linear combination of projections. Every projection operator O i : V V O_i: V \to V admits a decomposition of V = ker O i im O i V = \ker O_i \oplus \text{im} O_i . The result follows. \square

I have my doubts. How would this work for the matrix [ 0 1 0 0 ] \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} , for example?

Otto Bretscher - 2 years, 6 months ago

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Hmm perhaps I then made a mistake. I was thinking that in an operator, each vector in the domain (same as the range) either has to get killed or sent to something nonzero. But as your comment suggests, if v \vec{v} isn't sent to zero, it's not necessarily sent to something in the subspace it spans. So then my initial idea/proof, if it were correct, needs more.

Can you however show that the matrix you gave is indeed not a linear combination of projections?

Hobart Pao - 2 years, 6 months ago

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My intuition says that you might be right, actually, that the statement is not true, due to the nature of projections either killing a basis vector or sending it back to itself. But I'm not sure how to prove that the statement is false.

Hobart Pao - 2 years, 6 months ago

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@Hobart Pao We know that the statement is false because we have a counterexample ;)

The matrix I gave can indeed be written as linear combination of projection matrices, but that does not mean much, and it does not give you the kind of decomposition (as a direct sum) you seek.

Furthermore, your formula V = i [ ker O i im O i ] \displaystyle V = \bigoplus_{i} [\ker O_i \oplus \text{im} O_i] does not make much sense since we have V = ker O i im O i \displaystyle V = \ker O_i \oplus \text{im} O_i for each one of the projection operators O i O_i .

Otto Bretscher - 2 years, 6 months ago

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@Otto Bretscher I think i see what you mean. I'll withdraw it.

Hobart Pao - 2 years, 6 months ago

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@Hobart Pao I've put in a report.

Hmm maybe I meant V i V V_i \subset V for each of the individual projections, and then to sum all those together, but that still doesn't sound necessarily true.

Hobart Pao - 2 years, 6 months ago

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@Hobart Pao Yes, you would need an invariant subspace V i V_i for each O i O_i , but, clearly, you don't always have those.

Otto Bretscher - 2 years, 6 months ago

Can you email me about this, if you are interested in discussing the problem further? [email protected]. I sporadically check Brilliant. Thanks.

Hobart Pao - 2 years, 6 months ago

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