# Null Space

• May 26th 2011, 09:52 PM
kinkong
Null Space
Please can you guys help me to solve the following questions

Q1.if Z is an (n-1)-dimensional subspace of an n-dimensional vector space X, show that Z is the null space of suitable linear functional f on X, which is uniquely determined to within a scalar multiple.
• May 26th 2011, 10:24 PM
Drexel28
Quote:

Originally Posted by kinkong
Please can you guys help me to solve the following questions

Q1.if Z is an (n-1)-dimensional subspace of an n-dimensional vector space X, show that Z is the null space of suitable linear functional f on X, which is uniquely determined to within a scalar multiple.

Use the exact same logic as my other post. Take a basis $\displaystyle \{x_1,\cdots,x_{n-1}\}$ for $\displaystyle Z$ and extend it to a basis $\displaystyle \{x_1,\cdots,x_n\}$ for $\displaystyle X$ and define $\displaystyle \varphi:X\to F$ by $\displaystyle \varphi(x_k)=\delta_{k,n}$. To prove it's unique up to a scaling factor note that any linear functional $\displaystyle \psi$ must have the same form except $\displaystyle \psi(x_n)\ne 0$ may be different and so evidently $\displaystyle \varphi=\frac{1}{\psi(x_n)}\psi$
• May 27th 2011, 10:26 AM
Deveno
note that the linear functional φ in this case is the dual vector to the basis element xn.

note also that since n = dim(X) = null(φ) + rank(φ) = null(φ) + 1, dim(null(φ)) = n-1.

it is a theorem that every hyper-space (or hyper-plane) arises in this fashion.