# Hyperplane Matrix Linear transformation

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.

• 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 $\Bbb V$. Oct 26 '15 at 20:07
• Do you know the ''inner product'' of two vectors? Oct 26 '15 at 20:07

A correction in the question: The codomain space should be $$\mathbb R$$ and not $$\mathbb R^n$$.
The kernel of a linear transformation $$T:V\to W$$, denoted by $$\ker(T)$$, is defined to be the set of elements in the domain space $$V$$ which are mapped to zero, that is, $$\ker (T)= \{x\in V: T(x)=0\}$$. It is a routine to show that the kernel of every linear transformation is a vector subspace of the domain space. The question here, as far as I understand, is to determine the dimension of the kernel. That too is not too difficult to see.
Note that $$\ker (T)= \{x\in \mathbb R^n: T(x)=0\}=\{x\in \mathbb R^n: u^Tx =0\}$$, which is simply the set of all vectors in $$\mathbb R^n$$ that are orthogonal to the vector $$u\in \mathbb R^n$$. Let $$\hat u$$ be the normalization of $$u$$, that is, $$\hat u=\frac{u}{\|u\|}$$. Observe that the set $$\{\hat u\}\subseteq \mathbb R^n$$ can then be extended to an orthonormal basis of $$\mathbb R^n$$, say $$\{\hat u, v_1, v_2,..., v_{n-1}\}$$. Clearly then, a vector in $$\mathbb R^n$$ would be orthogonal to $$u$$ (or $$\hat u$$) if and only if it is a linear combination of $$v_1,...,v_{n-1}$$, that is, belongs to $$span\{v_1,...v_{n-1}\}$$. It then immediately follows that $$\{v_1,...v_{n-1}\}$$ is an orthonormal basis of $$\ker T$$ and hence the dimension is $$n-1$$.