Review : Eigenvalues & Eigenvectors: Question 5-3.1 & Solution

Question 1

An eigenpair satisfies $Av=\lambda v$ with $v\ne 0$ . The eigenspace $E_\lambda=\ker(A-\lambda I)$ includes zero, although zero is not an eigenvector. Algebraic multiplicity counts roots of $\det(\lambda I-A)$ ; geometric multiplicity is $\dim E_\lambda$ . Work over $\mathbb R$ unless complex scalars are explicitly requested.

Let $A=\begin{pmatrix}2&1\\1&2\end{pmatrix},\qquad u=(1,1)^T,\quad v=(1,-1)^T,$ and write a real starting vector as $x_0=\alpha u+\beta v$ .

Tasks

Find the eigenvalues, their full eigenspaces and both multiplicities. Explain why arbitrary nonzero multiples of an eigenvector are allowed but the zero vector is not.
Express $w=(3,1)^T$ in the eigenvector basis and decide whether $w$ itself is an eigenvector.
Find $A^m x_0$ for every nonnegative integer $m$ , and specialize to $w$ . Verify the result without multiplying out $m$ matrices.
For every nonzero real $x_0$ , determine the limit of $A^m x_0/\|A^m x_0\|$ . Include the cases $\alpha>0$ , $\alpha<0$ and $\alpha=0$ .

Original worksheet page 1: question and worked solution for 5-3-001

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Question 1 – Solution

Strategy. Resolve the starting vector into invariant directions before analyzing repeated multiplication.

Step 1: Compute the eigenspaces. The characteristic polynomial is $(\lambda-2)^2-1=(\lambda-3)(\lambda-1)$ . Solving the two singular systems gives $\boxed{E_3=\operatorname{span}\{u\},\qquad E_1=\operatorname{span}\{v\}.}$ Both algebraic and geometric multiplicities are $1$ . Scaling a nonzero eigenvector preserves its eigenvalue equation. Zero solves every homogeneous eigenvalue equation and supplies no direction, so it is excluded as an eigenvector.

Step 2: Decompose and test the given vector. The equations $\alpha+\beta=3$ , $\alpha-\beta=1$ give $w=2u+v$ . Both components are nonzero and their eigenvalues differ. Directly, $Aw=(7,5)^T$ is not a scalar multiple of $(3,1)^T$ , so $w$ is not an eigenvector.

Step 3: Compute all iterates. Linearity and $Au=3u$ , $Av=v$ give, by induction, $\boxed{A^m x_0=3^m\alpha u+\beta v,\qquad A^m w=(2\cdot 3^m+1,\,2\cdot 3^m-1)^T.}$ The formula holds at $m=0$ , and one multiplication advances each coefficient by its eigenvalue, proving the next case.

Step 4: Classify normalized direction limits. If $\alpha\ne 0$ , divide the iterate by $3^m|\alpha|$ before normalizing. The remaining $v$ coefficient tends to zero, so the limit is $\boxed{u/\sqrt 2}$ for $\alpha>0$ and $\boxed{-u/\sqrt 2}$ for $\alpha<0$ . If $\alpha=0$ , then $\beta\ne 0$ and the iterate is the unchanged vector $\beta v$ , giving limit $\boxed{\operatorname{sgn}(\beta)v/\sqrt 2}$ . The dominant eigenvalue determines the limit only when its component is present. The diagram shows the two directions and the decomposition $w=2u+v$ .

See the diagram in the original worksheet below.

Original worksheet page 2: question and worked solution for 5-3-001

Review : Eigenvalues & Eigenvectors — Question 1

Question 1

Question 1 – Solution