GATE 2023 EC Q.15: solution and answer

GATE 2023 · ECQ.15 · 1 mark · MCQ

Let the sets of eigenvalues and eigenvectors of a matrix $B$ be ${λ_{k} ∣ 1 \leq k \leq n}$ and ${v_{k} ∣ 1 \leq k \leq n}$ , respectively. For any invertible matrix $P$ , the sets of eigenvalues and eigenvectors of the matrix $A$ , where $B = P^{- 1} A P$ , respectively, are

(A)
${λ_{k} det (A) ∣ 1 \leq k \leq n}$ and ${P v_{k} ∣ 1 \leq k \leq n}$
(B)
${λ_{k} ∣ 1 \leq k \leq n}$ and ${v_{k} ∣ 1 \leq k \leq n}$
(C)
${λ_{k} ∣ 1 \leq k \leq n}$ and ${P v_{k} ∣ 1 \leq k \leq n}$
(D)
${λ_{k} ∣ 1 \leq k \leq n}$ and ${P^{- 1} v_{k} ∣ 1 \leq k \leq n}$

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Answer and solution

ANSWER

(C): the eigenvalues of $A$ are ${λ_{k} ∣ 1 \leq k \leq n}$ and its eigenvectors are ${P v_{k} ∣ 1 \leq k \leq n}$ .

Official answer: CSource: Official answer key, GATE 2023 EC (IIT Kanpur)

Write $A$ in terms of $B$ . Since $P$ is invertible, $B = P^{- 1} A P \Rightarrow A = P B P^{- 1}$ .
A similarity transformation does not change eigenvalues. Using $λ I = P (λ I) P^{- 1}$ : $det (A - λ I) = det (P (B - λ I) P^{- 1}) = det (P) det (B - λ I) det (P^{- 1}) = det (B - λ I),$ because $det (P) det (P^{- 1}) = 1$ . So $A$ and $B$ have the same characteristic polynomial, and the eigenvalues of $A$ are exactly the $λ_{k}$ . The extra factor $det (A)$ in (A) is wrong: it would only give the right values in the special case $det (A) = 1$ .
For the eigenvectors, start from $B v_{k} = λ_{k} v_{k}$ and substitute $B = P^{- 1} A P$ : $P^{- 1} A P v_{k} = λ_{k} v_{k} .$
Premultiply both sides by $P$ : $A (P v_{k}) = λ_{k} (P v_{k}) .$
$P v_{k} \neq = 0$ , because $P$ is invertible (its null space is only $0$ ) and an eigenvector $v_{k}$ is non-zero. So $P v_{k}$ is an eigenvector of $A$ for the eigenvalue $λ_{k}$ . In general $v_{k}$ itself (B) and $P^{- 1} v_{k}$ (D) are not: $P^{- 1} v_{k}$ would be an eigenvector of $P^{- 1} B P$ , not of $A = P B P^{- 1}$ .
So the eigenvalues of $A$ are ${λ_{k}}$ and its eigenvectors are ${P v_{k}}$ . This is option (C).

KEY IDEA

Similar matrices have the same characteristic polynomial, so they have the same eigenvalues. Their eigenvectors are different, though: they are linked by the same matrix that links the two matrices. Write $A = P B P^{- 1}$ . If $B v = λ v$ , then $A (P v) = λ (P v)$ , so each eigenvector of $A$ is $P$ times an eigenvector of $B$ .

WATCH OUT FOR

The usual slip is getting the direction of $P$ wrong. Students see $B = P^{- 1} A P$ and pick $P^{- 1} v_{k}$ (D), when solving for $A$ gives $A = P B P^{- 1}$ and so the eigenvectors are $P v_{k}$ . The other slip is assuming similar matrices share eigenvectors as well as eigenvalues (B). To check the direction, substitute the answer back: $A (P v_{k}) = P B P^{- 1} P v_{k} = P B v_{k} = λ_{k} P v_{k}$ .

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GATE 2023 EC Q.15: solution and answer

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