Review : Eigenvalues & Eigenvectors — Question 7

Question 7 – Solution

Strategy. Decaying eigencomponents need not make Euclidean length decrease at every step when the eigenvectors are not orthogonal.

Step 1: Find the distinct-eigenvalue basis. The eigenvalues are $1/2$ and $1/4$, with eigenvectors $u=(1,0)^T$ and $v=(-4K,1)^T$. These are independent for every $K$, so $T_K$ is always diagonalizable over $\mathbb R$.

Step 2: Compute all powers. For $m\ge 1$, the upper-right entry in the product is $$K\sum_{j=0}^{m-1}(1/2)^{m-1-j}(1/4)^j=4K(2^{-m}-4^{-m}).$$ This finite geometric sum yields $$\boxed{T_K^m=\begin{pmatrix} 2^{-m}&4K(2^{-m}-4^{-m})\\0&4^{-m} \end{pmatrix}.}$$ The formula is $I$ at $m=0$. Every entry tends to zero for each fixed $K$, so $T_K^m x\to 0$ for every fixed starting vector $x$.

Step 3: Exhibit a transient length increase. The first image of $e_2$ is $(K,1/4)^T$, of length $\sqrt{K^2+1/16}$. It exceeds $1$ exactly when $\boxed{|K|>\sqrt{15}/4}$. An early increase and eventual decay concern different times, so there is no contradiction. The plot uses $K=2$ and shows the actual lengths at integer iteration counts; the initial increase is followed by decay.

Step 4: Test all directions, not just one. For $x=(a,b)^T$, direct expansion and completion of the square give $$\|T_Kx\|^2-\|x\|^2 =-\frac 34\left(a-\frac{2K}{3}b\right)^2 +\left(\frac{4K^2}{3}-\frac{15}{16}\right)b^2.$$ This is nonpositive for every $a,b$ exactly when the last coefficient is nonpositive: necessity follows by choosing $a=2Kb/3$ with $b\ne 0$. Hence the exact condition is $$\boxed{|K|\le\frac{3\sqrt 5}{8}.}$$ This stricter threshold cannot be recovered by checking only $e_2$ or only the eigenvalues. At equality, nonzero vectors on $a=2Kb/3$ retain their length in one step; outside that line they decrease.