Exercise · ZW-EIGENSPACE-010 · ★★★ Advanced

Let \(\begin{pmatrix}2&0&0\\0&0&1\\0&1&0\end{pmatrix}\). Which is a basis of the eigenspace of \(\lambda=1\)?

Choose one option.

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  • B. \(\{(0,1,1)\}\)

Explanation

Solving \((A-I)v=0\) gives \(x=0\), \(y=z\): basis \(\{(0,1,1)\}\). The vector \((0,1,-1)\) corresponds to \(\lambda=-1\) and \(e_1\) to \(\lambda=2\).

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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 \(A\).

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

More practice problems: eigenspace

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