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Let a: V → V be a projection, where V a 3-dimensional vector space over C, and suppose that the nullity v(a) > 0, Tr(a) = 2. Can you deduce the minimal polynomial ma(x) and if so, what is it?

Let a: V → V be a projection, where V a 3-dimensional vector space over C, and suppose that the nullity v(a) > 0, Tr(a) = 2. Can you deduce the minimal polynomial ma(x) and if so, what is it?

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Let a : V → V be a projection, where V a 3-dimensional vector space over C, and suppose that the nullity v(a) > 0, Tr(a) = 2. Can you deduce the minimal polynomial ma(x) and if so, what is it? Select one: O None of the others apply O No, there is not enough information to deduce this. O Yes. Since a has a nontrivial kernel, it cannot be invertible. Hence det (a) = 0, and we are also told that the trace is 2. Hence pa = x(x² - 2x + c) for some coefficient c by results in Lectures. Since a is a projection, pa(a) = 0 reduces to (c-1)α = 0. But a 0 due to its non-zero trace, so c = 1 and Pa(x) = x(x - 1)². But a(a − 1) = 0, soma = x(x − 1) as it must divide pa and have the same roots. O Yes, by results in Lectures there is a direct sum decomposition V = UW on which a acts as 0 or the identity. Hence, choosing bases for these subspaces, we have diagonalised the matrix A representing a in this basis with diagonal entries 0,1. It follows that ma(x) = m₁(x) = (x − 1)². O We can put the matrix A for a in a suitable basis in Jordan form. This must consist of a single 3-block and squaring this to see if it a projection, one finds a unique such matrix with trace 2. We then compute ma(x) =m₁(x)= x(x + 1).
Transcribed Image Text:Let a : V → V be a projection, where V a 3-dimensional vector space over C, and suppose that the nullity v(a) > 0, Tr(a) = 2. Can you deduce the minimal polynomial ma(x) and if so, what is it? Select one: O None of the others apply O No, there is not enough information to deduce this. O Yes. Since a has a nontrivial kernel, it cannot be invertible. Hence det (a) = 0, and we are also told that the trace is 2. Hence pa = x(x² - 2x + c) for some coefficient c by results in Lectures. Since a is a projection, pa(a) = 0 reduces to (c-1)α = 0. But a 0 due to its non-zero trace, so c = 1 and Pa(x) = x(x - 1)². But a(a − 1) = 0, soma = x(x − 1) as it must divide pa and have the same roots. O Yes, by results in Lectures there is a direct sum decomposition V = UW on which a acts as 0 or the identity. Hence, choosing bases for these subspaces, we have diagonalised the matrix A representing a in this basis with diagonal entries 0,1. It follows that ma(x) = m₁(x) = (x − 1)². O We can put the matrix A for a in a suitable basis in Jordan form. This must consist of a single 3-block and squaring this to see if it a projection, one finds a unique such matrix with trace 2. We then compute ma(x) =m₁(x)= x(x + 1).
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