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Let $$u =\begin{bmatrix}u_1\\.\\.\\.\\u_n\end{bmatrix}$$ be a nonzero vector in $\mathbb R^n$, and let $T:\mathbb R^n \to \mathbb R^n$ be the linear transformation given by $T(x) = u^\top x$. Show that the kernel of $T$ is an $(n−1)$-dimensional vector space by finding a basis. (We call this space a hyperplane.)

I understand the linear transformation that result from $T(x)$, but I don't know what the question means by "kernel of $T$" and how to find the basis of the vector space. Help would be appreciated.

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  • $\begingroup$ The kernel of a linear transformation $T: \Bbb V \to \Bbb W$ is the set $\ker T := \{v \in \Bbb V : T(v) = 0_{\Bbb W}\}$. It is closed under addition and scalar multiplication, so it is a vector subspace of $\Bbb V$. $\endgroup$ – Travis Oct 26 '15 at 20:07
  • $\begingroup$ Do you know the ''inner product'' of two vectors? $\endgroup$ – Emilio Novati Oct 26 '15 at 20:07
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A correction in the question: The codomain space should be $\mathbb R$ and not $\mathbb R^n$.

The kernel of $T$ is the set of vectors in $\mathbb R^n$ which are orthogonal to $u$, that is, $ker (T)= \{x: x\in \mathbb R^n, \langle {x,u}\rangle =0\}$. It is then a trivial observation that the dimension of this subspace is $n-1$.

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