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Unlocking Eigenvalues: The Geometry of Linear Transformations

We explore the concept of eigenvalues and eigenvectors by examining how a specific 2x2 matrix stretches space along specific axes. We step through the characteristic equation and the algebraic process of finding these fundamental directions.

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What are Eigenvectors?
Av=λvAv = \lambda v
Think of a matrix as a transformation that pushes and pulls space. Usually, vectors get knocked off their span, but some special vectors just get stretched or squished. These are our eigenvectors.
Step-by-step solver
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(a) Characteristic Equation

Compute the determinant of A minus lambda times the identity to find the characteristic polynomial.

det⁡(2−λ112−λ)=(2−λ)2−1=λ2−4λ+3\det\begin{pmatrix}2-\lambda & 1\\ 1 & 2-\lambda\end{pmatrix} = (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3
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(b) Find Eigenvalues

Factor the characteristic polynomial to find the roots, which are the eigenvalues.

λ2−4λ+3=(λ−3)(λ−1)=0  ⟹  λ1=3,λ2=1\lambda^2 - 4\lambda + 3 = (\lambda-3)(\lambda-1) = 0 \implies \lambda_1=3, \lambda_2=1
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(c) Eigenvector for λ=3

Solve the system (A-3I)v=0 to find the eigenvector corresponding to lambda=3.

(−111−1)(xy)=(00)  ⟹  x=y,v1=(11)\begin{pmatrix}-1 & 1\\ 1 & -1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix} \implies x=y, v_1=\begin{pmatrix}1\\1\end{pmatrix}
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(d) Eigenvector for λ=1

Solve the system (A-I)v=0 to find the eigenvector corresponding to lambda=1.

(1111)(xy)=(00)  ⟹  x=−y,v2=(1−1)\begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix} \implies x=-y, v_2=\begin{pmatrix}1\\-1\end{pmatrix}

Original question

Find the eigenvalues and a corresponding eigenvector for each, of the matrix [[2,1],[1,2]].

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