# Ali's question at Yahoo! Answers (ker f)

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Here is the question:

Prove that every subspace U of a finite dimensional vector space V is the kernel of a linear transformation from V to V.

Thanks.
Here is a link to the question:

Subspaces, kernel of vector spaces? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Hello Ali,

Suppose that $B_U=\{u_1,\ldots,u_r\}$ is a basis of $U$. According to the incomplete basis theorem, there exist vectors $u_{r+1},\ldots,u_n$ such that $B_V=\{u_1,\ldots,u_n\}$ is a basis of $V$. Define the linear map $$f:V\to V,\qquad\left \{ \begin{matrix}f(u_i)=0&\mbox{if}&1\le i \le r\\ f(u_i)=u_i&\mbox{if}&r+1\le i \le n\end{matrix}\right.$$ The matrix of $f$ with respect to $B_V$ is diagonal by blocks:$$A=\begin{bmatrix}{0_{r\times r}}&{0_{r\times (n-r)}}\\{0_{(n-r)r\times r}}&{I_{(n-r)\times (n-r)}}\end{bmatrix}$$ Then, $x\in V$ (with coordinates $X=(x_j)^T$ with respect to $B_V$) belongs to $\ker f$ if and only if $AX=0$ or equivalently, if and only if $$X=(\alpha_1,\ldots,\alpha_r,0,\ldots,0)^T\quad (\alpha_i\in \mathbb{K})$$ which are the coordinates of all vectors of $U$. That is, $\ker f= U$.