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.

  • $\begingroup$ 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 $\Bbb V$. $\endgroup$ Oct 26 '15 at 20:07
  • $\begingroup$ Do you know the ''inner product'' of two vectors? $\endgroup$ 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 $n-1$.


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