Let A=diag(1,2) and v=(1,1), the sum of an eigenvector for λ=1 and one for λ=2. Is v an eigenvector of A?

No: Av=(1,2) is not a multiple of v

Exercise · ZW-EIGENSPACE-008 · ★★☆ Intermediate

Let \(A=\operatorname{diag}(1,2)\) and \(v=(1,1)\), the sum of an eigenvector for \(\lambda=1\) and one for \(\lambda=2\). Is \(v\) an eigenvector of \(A\)?

Choose one option.

Show answer

A. No: \(Av=(1,2)\) is not a multiple of \(v\)

Explanation

\(Av=(1,2)\) is not a multiple of \((1,1)\). Eigenspaces of different eigenvalues do not mix: only sums within the same eigenspace stay in it.

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Review the concept: Eigenspace

The eigenspace \(E_\lambda\) of a square matrix \(A\) for the eigenvalue \(\lambda\) is zero together with the eigenvectors of \(\lambda\), where \(I\) is the identity of the size of \(A\).

\[E_\lambda=\ker(A-\lambda I)\]

Let \(A=\operatorname{diag}(1,2)\) and \(v=(1,1)\), the sum of an eigenvector for \(\lambda=1\) and one for \(\lambda=2\). Is \(v\) an eigenvector of \(A\)?

Explanation

Review the concept: Eigenspace

More practice problems: eigenspace