Eigenvalues and Eigenvectors Questions and Answers

Answer: A

All the statements are true.

If \(B = P^{-1}AP\), then for any polynomial \(g\), \(g(B) = P^{-1}g(A)P\), so \(g(A) = 0\) exactly when \(g(B) = 0\): the minimal polynomials agree. The characteristic polynomials agree because

\[ \begin{aligned} \det(B - \lambda I) &= \det(P^{-1}(A - \lambda I)P) \\ &= \det(A - \lambda I) \end{aligned} \]

The trace and the determinant are coefficients of the characteristic polynomial (up to sign), so they agree as well. Every option that leaves out one of (i) to (iv) is incomplete.

Answer: A

The determinant of \(I_{4} - A\) is 0.

The eigenvalues of a real orthogonal matrix have modulus 1. The non-real ones come in conjugate pairs \(e^{\pm i\theta}\), each pair contributing 1 to the determinant, and the real ones are \(\pm 1\).

Since \(\det A = -1\), the eigenvalue \(-1\) occurs an odd number of times. In dimension 4 the real eigenvalues are even in number (4 minus twice the number of pairs), so if \(-1\) occurs an odd number of times, \(+1\) must occur an odd number of times as well, at least once. An eigenvalue 1 means \(I - A\) is singular:

\[ \det(I_{4} - A) = 0 \]

For example, A = diag(1, 1, 1, −1) gives \(\det(I - A) = 0 \cdot 0 \cdot 0 \cdot 2 = 0\).

Answer: C

The generic eigenvalues are a, -a, 0.

For a \(3 \times 3\) matrix, the trace is the sum of the eigenvalues and the determinant is their product:

\[ \begin{aligned} \lambda_{1} + \lambda_{2} + \lambda_{3} &= 0 \\ \lambda_{1}\lambda_{2}\lambda_{3} &= 0 \end{aligned} \]

Zero determinant forces one eigenvalue to be 0; then the other two must sum to zero, so they are \(a\) and \(-a\). This is the most general form with one free constant.

0, 0, 0 is a special case (\(a = 0\)). 0, 0, \(a\) has trace \(a\), not 0, and \(a, -2a, a\) has determinant \(-2a^{3}\), not 0.

Answer: A

The eigenvalues are 0, 0, 6.

A is lower triangular, so its eigenvalues are its diagonal entries 1, 2 and 4. If \(\lambda\) is an eigenvalue of A, then \(p(\lambda)\) is an eigenvalue of \(p(A)\) for any polynomial p. With \(p(t) = t^{2} - 3t + 2 = (t - 1)(t - 2)\):

\[ \begin{aligned} p(1) &= 0, \quad p(2) = 0 \\ p(4) &= 16 - 12 + 2 = 6 \end{aligned} \]

Option D (0, 3, 12) evaluates something else, and \(-1, 1, 0\) and \(-2, 4, 18\) do not come from substituting 1, 2, 4 into p.

Answer: C

The true statement is that the transpose of A is nilpotent.

If \(A^{k} = 0\), then \((A^{T})^{k} = (A^{k})^{T} = 0\), so \(A^{T}\) is nilpotent too.

The other statements fail. Every nilpotent matrix has only the eigenvalue 0, so its characteristic polynomial \(t^{n}\) splits and A is triangulable (it is similar to a strictly upper triangular matrix); for the same reason it has exactly one eigenvalue, not two. The minimal polynomial is \(t^{k}\) where k is the index of nilpotency, which can be smaller than n: for the 3 × 3 matrix with a single 1 in the corner, \(A^{2} = 0\), so the minimal polynomial is \(t^{2}\) while the characteristic polynomial is \(t^{3}\).

Answer: D

A has exactly one non-zero eigenvalue, namely n.

Every row of A is the same, so A has rank 1 and its null space has dimension \(n - 1\): the eigenvalue 0 has multiplicity \(n - 1\). The trace is n, which equals the sum of the eigenvalues, so the remaining eigenvalue is n, with eigenvector \((1, 1, \ldots, 1)\).

So for \(n \ge 2\) the eigenvalues are 0 (repeated \(n - 1\) times) and n. That rules out 'all eigenvalues positive', '0 has multiplicity 1' (it has multiplicity \(n - 1\)), and 'at least three distinct eigenvalues' (there are only two).

Answer: D

The eigenvalues are 1, −1.

As a real vector space, \(V\) has the basis \(\{1, i\}\). Complex conjugation leaves real numbers unchanged and reverses the sign of imaginary ones:

\[ T(1) = 1, \qquad T(i) = -i \]

So in this basis \(T\) is the diagonal matrix with entries 1 and −1, and those are its eigenvalues. Equivalently, \(T^{2} = I\), so any eigenvalue satisfies \(\lambda^{2} = 1\), and since \(T\) is neither \(I\) nor \(-I\) both signs occur.

The pair 1, 1 would make \(T\) the identity, 0, 0 would make it nilpotent (impossible since \(T^2 = I\)), and 2, −2 would need \(T^{2} = 4I\).

Answer: B

The recurrences are \(a_{n+1} = 2a_{n} + b_{n}\); \(b_{n+1} = 3a_{n}\).

The characteristic polynomial of \(T\) has roots −1 and 3, so it is \(\lambda^{2} - 2\lambda - 3\). By the Cayley-Hamilton theorem \(T^{2} = 2T + 3\,\mathrm{II}\). Multiply \(T^{n+1} = a_{n}T + b_{n}\mathrm{II}\) by \(T\):

\[ \begin{aligned} T^{n+2} &= a_{n}T^{2} + b_{n}T \\ &= a_{n}(2T + 3\,\mathrm{II}) + b_{n}T \\ &= (2a_{n} + b_{n})T + 3a_{n}\,\mathrm{II} \end{aligned} \]

Comparing with \(T^{n+2} = a_{n+1}T + b_{n+1}\mathrm{II}\) gives the answer. A quick check: \(T^{1} = T\) means \(a_{0} = 1, b_{0} = 0\), so \(T^{2}\) should have \(a_{1} = 2, b_{1} = 3\), as Cayley-Hamilton says. The other recurrences do not reproduce \(T^{2} = 2T + 3\,\mathrm{II}\).

Answer: A

An eigen value of the matrix is 2.

For the matrix with rows (3, -1) and (4, -2), the trace is 1 and the determinant is 3(-2) - (-1)(4) = -2, so the characteristic equation is

\[ \lambda^{2} - \lambda - 2 = 0 \]

\[ (\lambda - 2)(\lambda + 1) = 0 \]

The eigen values are 2 and -1. Of the options, only 2 is one of them; 0, 1 and 3 do not satisfy the equation.

Answer: A

Both statements are correct and the reason explains the assertion. The determinant of a matrix equals the product of its eigenvalues, so the determinant is zero exactly when some eigenvalue is zero. A matrix is non-singular when its determinant is not zero, which means none of its eigenvalues is zero.

So (R) is the correct explanation of (A), option A.

Testing each of the remaining options against the same fact shows that every one of them fails on the point the question is asking about, which confirms the keyed answer.