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Dungog Shire

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Let $V$ be a subspace of $mathbb{R}^4$, spanned by $v$ and $u$. Find a linear transformation whose kernel is...

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3 0 $begingroup$ And the vectors given are $v = (1,0,3,-2)$ and $u = (0,1,4,1)$ . It asks me to find the linear transformation from $mathbb{R}^4$ to $mathbb{R}^2$ , where the kernel of that transformation is $V$ . So what I know is that: the transformation I'm trying to find, applied to every vector in the span of $(1,0,3,-2)$ and $(0,1,4,1)$ , will give the zero vector. Please let me know if that interpretation is incorrect. I've really no idea how to get started on this question. I have the equation $Av = 0$ where $A$ is the matrix of the transformation in question, and v is any vector of the subspace V, but...I don't think that gets me anywhere. Any help is greatly appreciated. linear-algebra linear-transformations ...

Pianfei

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Coordenadas: 44° 22' N 7° 42' E Pianfei      Comuna    Pianfei Localização de Pianfei na Itália Coordenadas 44° 22' N 7° 42' E Região Piemonte Província Cuneo Área  - Total 15 km² População  - Total 1,811     • Densidade 0,1 hab./km² Outros dados Comunas limítrofes Chiusa di Pesio, Margarita, Mondovì, Roccaforte Mondovì, Villanova Mondovì Código ISTAT 004165 Código postal 12080 Prefixo telefônico 0174 Sítio www.comune.pianfei.cn.it Pianfei é uma comuna italiana da região do Piemonte, província de Cuneo, com cerca de 1.811 habitantes. Estende-se por uma área de 15 km², tendo uma densidade populacional de 121 hab/km². Faz fronteira com Chiusa di Pesio, Margarita, Mondovì, Roccaforte Mondovì, Villanova Mondovì. [ 1 ] [ 2 ] [ 3 ] Demografia | Variação demográfica do município entre 1861 e 2011 [ 3 ] Fonte : Istituto Nazionale di Statistica...