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Part BCSIR NET December 2023rank-one-eigenvalue

Rank one eigenvalue

Let A=(aij)A = (a_{ij}) be the n × n real matrix with aij=a_{ij} = ij for all 1 ≤ i, j ≤ n. If n ≥ 3, which one of the following is an eigenvalue of A?

  1. A.1
  2. B.n
  3. C.n(n+1)/2
  4. D.n(n+1)(2n+1)/6✓ Correct

The idea, in plain words

Every entry is one number times another, which means every row is a multiple of the same row. A grid like that is as flat as it gets: it has a single meaningful direction, so there is only one eigenvalue worth having, and it equals the sum of the squares.

Solution

A = vvᵀ with v = (1, 2, …, n) has rank 1; its only non-zero eigenvalue is vTv=∑i2=n(n+1)(2n+1)/6v^{T}v = \sum{}i^{2} = n(n+1)(2n+1)/6.

Why each option is right or wrong

Checked against 4 symbolic computations
  • A.Wrong

    False.

    A = vvᵀ with v = (1, 2, …, n) has rank 1

    so its eigenvalues are 0 (with multiplicity n − 1) and the single non-zero value vᵀv.

    At n = 3, SymPy gives the eigenvalues of A as 14 and 0 — 1 is not among them.

  • B.Wrong

    False. n is the size of the matrix, not a spectral quantity here.

    At n = 3 the non-zero eigenvalue is 14, not 3.

  • C.WrongExecution slip

    False, and it is the engineered near-miss: n(n+1)/2 is ∑i\sum{}i, whereas the eigenvalue is ∑i2\sum{}i^{2}.

    At n = 3, SymPy gives ∑i=6\sum{}i = 6 but the actual eigenvalue is 14.

    Writing A = vvᵀ makes the non-zero eigenvalue vTv=∑i2v^{T}v = \sum{}i^{2}

    so a moment's inattention between 'sum' and 'sum of squares' lands exactly here.

  • D.Correct

    True.

    Since aij=i⋅j,A=a_{ij} = i\cdot{}j, A = vvᵀ with v = (1, 2, …, n).

    A rank-one matrix vvᵀ has one non-zero eigenvalue, vᵀv, with eigenvector v.

    Here vTv=∑ii2=n(n+1)(2n+1)/6v^{T}v = \sum_{i} i^{2} = n(n+1)(2n+1)/6.

    SymPy confirms all three at n = 3: the eigenvalues are {14, 0, 0}, ∑k2=14\sum{}k^{2} = 14, and the formula evaluates to 14.

    The trigger: an entry pattern aij=f(i)g(j)a_{ij} = f(i)g(j) means rank one — write it as an outer product and read the eigenvalue off.

1 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: Same characteristic polynomial ⇒ similar

More on this topic

The chapter behind this: Characteristic vs minimal polynomial — what each tells you — free to read

From Eigenvalues and Canonical Forms › Eigenvalues, characteristic & minimal polynomials, Cayley–Hamilton

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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