# High dimension intuition, inner spaces

Let $\mathbb E$ be a euclidean space.

$dim (\mathbb E) = 5$.

For every $a \in \mathbb E$ define a linear transformation $T:\mathbb E \to \mathbb R$ by the standart inner product $T_a(x)=<a,x>$.

For every $a,b\in \mathbb E$ find $dim(ker(T_a) \bigcap ker(T_b))$.

Im having a hard time, obviously, visualizing the subspace $ker(T_a) \bigcap ker(T_b)$...

Since T is a linear transformation it's easy to see that: $$dim Im(T)+dim ker(T)=dim (E) \implies dim ker(T)=4$$$$dim(ker(T_a)+ker(T_b))=dimker(T_a)+dimker(T_b)-dim(ker(T_a) \bigcap ker(T_b)$$$$ker(T_a) \bigcap ker(T_b) = 8-dim(ker(T_a)+ker(T_b))$$

How do I proceed? How do I know what is $dim(ker(T_a)+ker(T_b))$?

• Hint: $\ker(T_a)$ is the set of all vectors $x$ such that $\langle a,x\rangle=0$, i.e., all vectors orthogonal to $a$. So, $\ker(T_a)\cap\ker(T_b)$ would be all vectors orthogonal to both $a$ and $b$. – amd Jan 20 '16 at 21:35
• You can write your set as a solution set of a system of linear equations, and you will see that there are different cases depending on what $a$ and $b$ is. – user302982 Jan 20 '16 at 21:37

Consider the problem first in the "more geometric" setting, working in $\mathbb{R}^3$. If $a \neq 0$ then $\ker(T_a)$ is a plane passing through the origin whose normal vector is $a$. If $a = 0$ then $\ker(T_a) = \mathbb{R}^3$. Given $a, b \in \mathbb{R}^3$ you have two possibilities:
1. If $a = b = 0$ then $\ker(T_a) \cap \ker(T_b) = \mathbb{R}^3$.
2. If $a \neq 0$ and $b$ is a multiple of $a$ then if $b \neq 0$ the planes $\ker(T_a)$ and $\ker(T_b)$ are the same planes (as they have proportional normal vector) and so $\ker(T_a) \cap \ker(T_b)$ is two-dimensional. If $b = 0$ then $\ker(T_a) \cap \ker(T_b) = \ker(T_a)$. Similarly if $b \neq 0$ and $a$ is a multiple of $b$.
3. If $a$ and $b$ are linearly independent, the planes $\ker(T_a)$ and $\ker(T_b)$ are different and so they intersect along a line and $\ker(T_a) \cap \ker(T_b)$ is one-dimensional.
Similarly, working in $\mathbb{R}^5$, if $0 \neq a \in \mathbb{R}^5$ then $\ker(T_a)$ will be a hyperplane (a four dimensional subspace) whose normal vector is $a$. To see this formally, note that $T_a$ is onto and so by the rank-nullity theorem, the dimension of $\ker(T_a)$ is four. If $a = 0$ then clearly $\ker(T_a) = \mathbb{R}^5$. Again, you have three options:
1. If $a = b = 0$ then $\ker(T_a) \cap \ker(T_b) = \mathbb{R}^5$.
2. If $a \neq 0$ and $b$ is a multiple of $a$ then if $b \neq 0$ then the hyperplanes $\ker(T_a)$ and $\ker(T_b)$ are the same (as $\left<a, x \right> = 0$ if and only if $\left< b, x \right> = 0$) and so $\ker(T_a) \cap \ker(T_b)$ is four dimensional. If $b = 0$ then again $\ker(T_a) \cap \ker(T_b) = \ker(T_a)$ is four dimensional. Similarly if $b \neq 0$ and $a$ is a multiple of $b$.
3. If $a$ and $b$ are linearly independent, then the map $T \colon \mathbb{R}^5 \rightarrow \mathbb{R}^2$ given by $T(x) = (T_a(x), T_b(x))$ will be onto (check this!) and $\ker(T_a) \cap \ker(T_b) = \ker(T)$ so by the rank-nullity theorem, $\ker(T)$ will be three dimensional and the two hyperplanes will intersect along a three-dimensional subspace.