# For each $y ∈ R^n$ show that $L_y$ is a linear transformation and compute $\dim_R(\ker(L_y))$.

Hi I'm having trouble with this homework question:

"For $y ∈ R^n$ let $L_y : R^n → R$ denote the map given by $x → L_y(x) = x · y$ where $x · y$ denotes the dot product of $x$ and $y$.

For each $y ∈ R^n$ show that $L_y$ is a linear transformation and compute $\dim_R(\ker(L_y))$."

So I've managed to show that its a linear transformation quite easily but I'm stuck on finding the dimension of the null space. This is what I've done so far:

Null space={$x : L(x)=0$}

$L(x)=x · y$ so I split it into two cases:

Case 1 is where $y=0_V$ where $0_V$ is the zero vector

Then $x$ is any chosen vector in $R^n$ so the dimension is infinite (which doesn't sound right).

Case 2 is where $y\ne0_V$

Then $x$ is the zero vector in $R^n$ so its dimension is $0$

I'm not very sure about Case 1. Any help is very much appreciated.

If $ker(L)= R^n$, then it's dimension is $n$, not $\infty$! You're simply confusing dimension with cardinality: although the vector space $R^n$ contains infinitely many elements, it's dimension is given by the number of elements in any basis of $R^n$ (this will always be $n$).

Let $y = (y_1,\dots,y_n)$.

If $y = 0$, then $ker(L_y) = \mathbb{R}^n$ which is $n$-dimensional.

If $y \neq 0$, then $y_i \neq 0$ for some $i$. Without loss of generality assume that $y_n \neq 0$. Suppose $x = (x_1,\dots,x_n) \in ker(L_y)$.

Then $$x_1y_1 + \dots + x_ny_n = 0.$$

Since $y_n \neq 0$, we can solve for $x_n$ in this equation yielding:

$$x_n = -\frac{y_1}{y_n}x_1 - \dots - \frac{y_{n-1}}{y_n}x_{n-1}.$$

So when choosing a vector $x$ in $ker(L_y)$ we have $n-1$ degrees of freedom (We choose the values of $x_1,\dots,x_{n-1}$. After that, $x_n$ is determined). That is, $dim(ker(L_y)) = n-1$.