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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?

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  • $\begingroup$ It'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. $\endgroup$ – Gerry Myerson Oct 14 '12 at 8:46
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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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  • $\begingroup$ And the first equality follows from the fact that $\langle u,v\rangle = u^Tv$ $\endgroup$ – Robert S. Barnes Oct 15 '12 at 8:55
  • $\begingroup$ Couldn't you have written directly without resorting to the $1\times 1$: $\langle y,A^Tx\rangle=(A^Tx)^Ty=(x^TA)y=x^T(Ay)=x^T0=0$ since $\langle u,v\rangle = \langle v,u\rangle$? $\endgroup$ – Robert S. Barnes Oct 15 '12 at 9:00
  • $\begingroup$ Sure. And $\langle u,v\rangle=\langle v,u\rangle$ because this is the $1\times1$ case... $\endgroup$ – Did Oct 15 '12 at 9:54

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