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All solution of $AX = 0$ where $A$ is a $n \times n$ matrix and $X$ is a column vector form a subspace of $\mathbb{R}^n$. All the subspaces of $\mathbb{R}^n$ are of this type. How to prove this result? Linear Algebra: solution of homogeneous system of equation

Thank you.

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Let $S$ a subspace of $\mathbb R^n$ and choose $(e_1,\ldots,e_p)$ a basis of $S$ which we compete it by a basis $(e_1,\ldots,e_p,e_{p+1},\ldots,e_n)$ of $\mathbb R^n$.

Now let the endomorphism $f$ defined by $f(e_i)=0,\ 1\leq i\leq p$ and $f(e_i)=e_i,\ p+1\leq i\leq n$ and let $A$ the matrix of $f$ in this basis then $$AX=0\iff X\in S$$

  • Can it be solved without using the concept of basis? – Supriyo Jul 10 '13 at 11:47
  • Yes by the Arthur's hints: take the normal subspace to the subspace $S$ so we have $\mathbb R^n= S\oplus S^\perp$ and let $p$ the orthogonal projection on $S^\perp$ so $\ker p=S$. –  Jul 10 '13 at 11:53