# Prove that there exists a matrix $B$ s.t ker$B=$Im$A$, Im$B=$ker$A$

Let $A$ be a square matrix.

a) Show that there always exists a square matrix B such that Ker $B =$ Im $A$ and Ker$A =$ Im$B$.

Here is my approach:

Let $A:\ V \rightarrow P$. Denote the basis of Ker$A$ by $u_{1}, \dots , u_{r}$ where $r\leq n$, $n$ is the size of $A$. Let $B$ be a matrix with $u_{1}, u_{2}, \dots , u_{r}$ as the first $r$ columns and the rest columns from $r$ to $n$ are some linear combinations of $u_{i}$. Thus the first r columns of $B$ span a subspace of $V$, this is by definition the image of $B, \ \rightarrow$ Im$B=$Ker$A$. From here its blank. Im not even sure if my first approach is correct.

I will be grateful for any hint. I want to prove this with matrices since next question is to prove that IF A is normal then B is normal aswell.

## 1 Answer

It is probably better to think in terms of linear maps instead of matrices. So you have a linear map $a$ on a vector space $V$. Let $v_{1}, \dots , v_{n}$ be a basis of $V$ such that $v_{1}, \dots, v_{k}$ are a basis of $\ker(a)$, so that $f(v_{k+1}), \dots , f(v_{n})$ is a basis of the image of $a$. Extend the $f(v_{i})$ to a basis $w_{1}, \dots, w_{k}, f(v_{k+1}), \dots , f(v_{n})$ of $V$, and define a linear map $b$ which is zero on the $f(v_{i})$ and maps each $w_{j}$ to $v_{j}$.

• The proof is even is way more general: Let $U$ and $W$ be any subspaces of $V$, that satisfy the dimension formula, i.e. their dimensions add up to $\dim V$. Then there exists an endomorphism with kernel and image being $U$ and $W$ respectively. – MooS Dec 8 '15 at 21:05
• Yeah I follow. But one has to prove that this map exists aswell or is it evident? – Olba12 Dec 8 '15 at 21:08
• The point is that one can define a linear map by assigning to it arbitrary values on a basis. – Andreas Caranti Dec 8 '15 at 21:19