GATE 2022 EE Q.40: solution and answer

GATE 2022 · EEQ.40 · 2 marks · MCQ

$e^{A}$ denotes the exponential of a square matrix A. Suppose $λ$ is an eigenvalue and $v$ is the corresponding eigen-vector of matrix A.

Consider the following two statements:

Statement 1: $e^{λ}$ is an eigenvalue of $e^{A}$ .

Statement 2: $v$ is an eigen-vector of $e^{A}$ .

Which one of the following options is correct?

(A)
Statement 1 is true and statement 2 is false.
(B)
Statement 1 is false and statement 2 is true.
(C)
Both the statements are correct.
(D)
Both the statements are false.

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Answer and solution

ANSWER

Both statements are true: $e^{A} v = e^{λ} v$ , so $v$ is an eigenvector of $e^{A}$ with eigenvalue $e^{λ}$ . Answer (C).

Official answer: CSource: Official answer key, GATE 2022 EE (IIT Kharagpur)

Start from the eigenvalue relation: $A v = λ v$ with $v \neq = 0$ .
Apply $A$ repeatedly (induction): $A^{2} v = A (λ v) = λ (A v) = λ^{2} v$ , and in general $A^{k} v = λ^{k} v$ for every $k \geq 0$ (with $A^{0} = I$ ).
Use the power-series definition of the matrix exponential, which converges for every square matrix: $e^{A} = k = 0 \sum \infty \frac{A ^{k}}{k !} = I + A + \frac{A ^{2}}{2 !} + \dots$
Multiply the series by $v$ term by term (allowed because the series converges and multiplying by a fixed vector is linear): $e^{A} v = k = 0 \sum \infty \frac{A ^{k} v}{k !} = k = 0 \sum \infty \frac{λ ^{k}}{k !} v = e^{λ} v$
Since $v \neq = 0$ and $e^{A} v = e^{λ} v$ , the same vector $v$ is an eigenvector of $e^{A}$ (Statement 2 true), and the eigenvalue it belongs to is $e^{λ}$ (Statement 1 true).
This is a special case of the general rule: for any polynomial or convergent power series $f$ , $f (A)$ has eigenvalue $f (λ)$ with the same eigenvector $v$ . Both statements are true, so the answer is (C).

KEY IDEA

From $A v = λ v$ you get $A^{k} v = λ^{k} v$ for every $k$ . Put that into the series $e^{A} = \sum A^{k} / k!$ and it gives $e^{A} v = e^{λ} v$ . More generally, $f (A)$ keeps the eigenvectors of $A$ and maps each eigenvalue $λ$ to $f (λ)$ .

WATCH OUT FOR

Thinking the function changes the eigenvectors along with the eigenvalues, and so picking (A). Applying a polynomial or power series to $A$ changes only the eigenvalue ( $λ \to f (λ)$ ). The eigenvector $v$ stays the same. A second slip is to treat $e^{A}$ as the entry-wise exponential of $A$ . It is the matrix power series, not $e^{a_{ij}}$ taken entry by entry.

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GATE 2022 EE Q.40: solution and answer

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