silemus.

12Eigenvector for eigenvalue 3

A=[2112]\displaystyle A = \begin{bmatrix}2 & 1\\1 & 2\end{bmatrix} has eigenvalue 3\displaystyle 3. Find an eigenvector for it, scaled so its first entry is 1\displaystyle 1.

Solve (A − 3I)x = 0: subtracting 3 from the diagonal gives [[-1,1],[1,-1]], and both rows say y = x. Scaling so the first entry is 1 gives the eigenvector (1, 1). Check: A(1,1)ᵗ = (3,3)ᵗ = 3(1,1)ᵗ ✓.

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