I know that according to the Fundamental theorem of linear algebra the row space and the null space are orthogonal, but I don't really understand why. Could someone give an intuitive explanation of why this is with maybe some examples from $\mathbb{R}^2$ or $\mathbb{R}^3$ with the standard Euclidean inner product?
Orthogonality of row space and null space
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linear-algebra
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0It's just how matrix multiplication works --- when you multiply a matrix $A$ times a vector $v$, the $j$th entry in the product is the inner product of the $j$th row of $A$ with $v$. So $v$ is in the nullspace if and only if it's orthogonal to the generators of the row space. – 2012-10-14
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The row space is the set of $A^Tx$ for every vector $x$, the null space is the set of vectors $y$ such that $Ay=0$. The scalar product between a vector in the row space and a vector in the null space is $\langle y,A^Tx\rangle=y^T(A^Tx)=x^T(Ay)=x^T0=0$. The second equality follows from the fact that $y^TA^Tx$ has size $1\times 1$, hence is equal to its transpose $x^TAy$.
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0Sure. And $\langle u,v\rangle=\langle v,u\rangle$ because this is the $1\times1$ case... – 2012-10-15