Review : Eigenvalues & Eigenvectors: Question 5-3.2 & Solution

Question 2

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.

For real $a$ , consider $B_a=\begin{pmatrix}2&a\\0&2\end{pmatrix}.$ A student claims that a repeated eigenvalue automatically provides two independent eigenvectors.

Tasks

Find the eigenvalue and both multiplicities for every $a$ , treating $a=0$ separately.
Classify diagonalizability over $\mathbb R$ and over $\mathbb C$ . Explain whether allowing complex eigenvectors repairs any failure.
Find an exact formula for $B_a^m$ for every nonnegative integer $m$ . Derive it rather than assuming diagonalization.
For a fixed vector $x=(x_1,x_2)^T$ , classify when $2^{-m}B_a^m x$ stays bounded as $m\to\infty$ . Compare this result with what the repeated eigenvalue alone suggests.

Original worksheet page 1: question and worked solution for 5-3-002

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Question 2 – Solution

Strategy. Root multiplicity and the number of independent eigenvectors answer different questions.

Step 1: Solve the eigenspace equation. The characteristic polynomial is $(\lambda-2)^2$ for every $a$ . The equation $(B_a-2I)x=0$ reduces to $a x_2=0$ . Thus the algebraic multiplicity is always $2$ , while $\boxed{a=0:\ E_2=\mathbb R^2,\ \dim E_2=2; \qquad a\ne 0:\ E_2=\operatorname{span}\{(1,0)^T\},\ \dim E_2=1.}$ A repeated root does not itself produce a second independent vector.

Step 2: Classify diagonalization in both fields. At $a=0$ , $B_a=2I$ is already diagonal. At $a\ne 0$ , there is only one independent eigenvector, so no eigenvector basis exists. The same equation $a x_2=0$ holds over $\mathbb C$ , with the same dimension count there. Hence $\boxed{B_a\text{ is diagonalizable exactly when }a=0}$ over either field.

Step 3: Derive the powers from a terminating product. Let $N=B_a-2I=\begin{pmatrix}0&a\\0&0\end{pmatrix}$ , so $N^2=0$ . The commuting binomial expansion gives $\boxed{B_a^m=2^m\begin{pmatrix}1&ma/2\\0&1\end{pmatrix}.}$ For $m=0$ this is $I$ . Multiplication by $B_a$ adds $a/2$ to the scaled upper-right entry, confirming the formula inductively without an eigenvector basis.

Step 4: Identify the remaining growth. The scaled iterate is $2^{-m}B_a^m x=(x_1+ma x_2/2,\,x_2)^T.$ It is bounded exactly when $\boxed{a x_2=0}$ , and then is constant. For every $a\ne 0$ and $x_2\ne 0$ , its first coordinate grows in magnitude linearly. Dividing out the common exponential factor $2^m$ does not remove this extra growth. The missing eigenvector permits behavior that the list of eigenvalues, even with multiplicity, does not describe by itself.

Original worksheet page 2: question and worked solution for 5-3-002

Review : Eigenvalues & Eigenvectors — Question 2

Question 2

Question 2 – Solution