# Linear transformation and it's properties Let M be a p-1-dimensional subspace

Linear transformation and its
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venenolundicoofw

First, the statement is false if because in this case, t is the class of an element t′ of M and $M=kerB⇒B{t}^{\prime }=0\ne 1$.
Now, if $t\ne 0$, then ${t}^{\prime }\text{⧸}\in M$ and the line $L\phantom{\rule{0.222em}{0ex}}=⟨{t}^{\prime }⟩$ satisfies $M\oplus L={\mathbb{R}}^{p}$. Thus if $\pi$ is the projection on L along M (i.e., if $x={x}_{M}+{x}_{L}$, then $\pi \left(x\right)={x}_{L}\right)$ and if $\psi :L\to \mathbb{R}$ is defined by $\psi \left(\lambda {t}^{\prime }\right)=\lambda ,\mathrm{\forall }\lambda \in \mathbb{R}$, we see that $B=\psi \circ \pi$ satisfies .
Edit: 1. B is linear: indeed, if so
$B\left(ax+y\right)=\psi \left(\pi \left(ax+y\right)\right)=\psi \left(\left(a\lambda +\mu \right){t}^{\prime }\right)=a\lambda +\mu$
and $aB\left(x\right)+B\left(y\right)=a\psi \left(\pi \left(x\right)\right)+\psi \left(\pi \left(y\right)\right)=a\psi \left(\lambda {t}^{\prime }\right)+\psi \left(\mu {t}^{\prime }\right)=a\lambda +\mu$
By definition of . Now, if is a subspace of ${\mathbb{R}}^{p}$ containing strictly a $\left(p-1\right)$-subspace, hence $kerB$ would be all ${\mathbb{R}}^{p}$. This is false because $B{t}^{\prime }=1$. Thus $kerB=M$.