Domain of linear operator being the direct sum of kernel and another subspace.

linear algebralinear-transformations

Let, $F: X \rightarrow Y$ be a linear operator. Let, $Q = F(X)$ be an $n$-dimensional vector space. How do I demonstrate that,

$$X = \ker(F) \oplus Z,$$
for some subspace $Z$ with dimension equal to $n$?

Best Answer

Choose a basis $y_{1}, \ldots, y_{n}$ of $F(X)$ and consider the vectors $x_{1}, \ldots, x_{n}$ in $X$ such that $$ F(x_{1}) = y_{1} $$ $$ \vdots $$ $$ F(x_{n}) = y_{n} .$$ Define $Z = \text{span}(x_{1}, \ldots, x_{n})$. It can be shown that the $x_{i}$'s are linearly independent, so $\dim Z = n$. Now let $x \in X$ be arbitrary. Then $$ F(x) = \alpha_{1}y_{1} + \ldots + \alpha_{n}y_{n} $$ for some scalars $\alpha_{1}, \ldots, \alpha_{n}$. It is clear that $$ x = \left( x - \sum_{i=1}^{n}\alpha_{i}x_{i} \right) + \sum_{i=1}^{n}\alpha_{i}x_{i},$$ and from this we deduce that $X = \ker F + Z$. It is not difficult to show that the sum is direct.

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