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Exercise 3.B.31

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Proof. Let $v_{1}, v_{2}, v_{3}, v_{4}, v_{5}$ be a basis of $ℝ^{⋬}$ and $w_{1}, w_{2}$ of $ℝ^{⊭}$ . Define $T_{1}, T_{2} \in L (ℝ^{⋬}, ℝ^{⊭})$ such that

\begin{aligned} T_{1} (a_{1} v_{1} + \dots + a_{5} v_{5}) & = a_{1} w_{1} + a_{2} w_{2} \\ T_{2} (a_{1} v_{1} + \dots + a_{5} v_{5}) & = a_{2} w_{1} + a_{1} w_{2} \end{aligned}

Clearly $null T_{1} = null T_{2} = span (v_{3}, v_{4}, v_{5})$ , but $T_{1} v_{1} = w_{1}$ and $T_{2} v_{1} = w_{2}$ . Because $w_{1}$ is not a scalar multiple of $w_{2}$ (they are linearly independent), we have that $T_{1}$ cannot be a scalar multiple $T_{2}$ . □

celiopassos

2017-10-06 00:00

Exercise 3.B.31

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