Exercise 1.9.2

Answers

$A = [\begin{matrix} 3 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$
The eigenvalues of $A^{T} A$ are $λ_{1} = 3, λ_{2} = 2$ . We have $A = [\begin{matrix} 0 & 3 \\ 2 & 0 \end{matrix}] = U Σ V^{T} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] [\begin{matrix} 3 & 0 \\ 0 & 2 \end{matrix}] [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]$ , so the rank-1 matrix approximation is: $[\begin{matrix} 1 \\ 0 \end{matrix}] [\begin{matrix} 0 & 1 \end{matrix}] = [\begin{matrix} 0 & 3 \\ 0 & 0 \end{matrix}]$
The eigenvalues of $A^{T} A$ are $λ_{1} = 3, λ_{2} = 1$ . We have $A = [\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix}] = U Σ V^{T} = \frac{1}{2} [\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix}] [\begin{matrix} 3 & 0 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix}]$ , so the rank-1 matrix approximation is: $\frac{1}{2} [\begin{matrix} 1 \\ 1 \end{matrix}] [\begin{matrix} 1 & 1 \end{matrix}] = \frac{1}{2} [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}]$

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2020-03-20 00:00