 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/digital-systems-linear-algebra-introduction-to-formal-proof-resit/5/d/solution

For

1 = (1, 1, 1)^{T}

, the entries of

P^{T} 1

are the column sums of

P

. Hence

P^{T} 1 = 111 = 1,

1

is an eigenvalue of

P^{T}

1

det (P - I) = det ((P - I)^{T}) = det (P^{T} - I) = 0.

Thus

P - I

1 is also an eigenvalue of P .

The same argument with

λ I

shows that a matrix and its transpose have the same characteristic polynomial. It does not say that they have the same eigenvectors: here

P 1 = (0.9, 1.2, 0.9)^{T} \neq = 1

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