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Let's consider a matrix A (m by n). It is true that row space R(A) and nullspace Null(A) are orthogonal to each other. Why? Because $v=A^Tz \in R(A)$, where z is any m by 1 column vector.$y\in Null(A)\implies Ay=0$
$v.y=(A^Tz).y=z^TAy=z^T(0)=0\implies v \perp y \;\;\forall v,y$
However, here is the problem: How do I prove that any vector which is orthogonal to Null space necessarily lies in row space ?

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It is easier to show that every vector which is orthogonal to the row space is necessarily in the null space.

The vectors of the row space can be written in the form $A^Tz$. If $x \in R(A)^\perp$, then it holds that for every $z \in \Bbb R^n$ we have $z^T(Ax) = (A^Tz)x = 0$. Since $z^T(Ax)$ holds for all vectors $z$, it must be that $Ax = 0$, which is to say that $x \in N(A)$.

With this along with what you have said so far, we have everything we need. Because the two spaces are orthogonal to each other, we have $R(A)^\perp \supseteq N(A)$. Because every vector orthogonal to the row space is in the null space, we have $R(A)^\perp \subseteq N(A)$. Finally, we can conclude that $R(A)^\perp = N(A)$. It follows that $N(A)^\perp = R(A)$, which is to say that a vector is contained in the row space if and only if it is orthogonal to the null space.


An alternative approach: again, as your works so far shows, we have $N(A) \subseteq R(A)^\perp$. The rank-nullity theorem tells us that $\dim(N(A)) = n - \dim(R(A))$. However, we also have $\dim(R(A)^\perp) = n - \dim(R(A))$. Because $N(A) \subseteq R(A)^\perp$ and $\dim(N(A)) = \dim(R(A)^\perp)$, we have $N(A) = R(A)^\perp$.


In fact, any proof of the fact that $N(A)^\perp \subseteq R(A)$ must make use of the fact that we are dealing with operators over a finite dimensional vector space. In the infinite dimensional setting (over Hilbert spaces), there are operators $A$ whose "row space" (image of the adjoint) fails to be closed, which means that they must satisfy $R(A) \neq N(A)^\perp$ (in fact, $R(A) \subsetneq N(A)^\perp$).

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  • $\begingroup$ Wow. Thanks a lot. Short and sweet proof. $\endgroup$ – Koro May 10 '20 at 15:18
  • $\begingroup$ You're welcome! $\endgroup$ – Ben Grossmann May 10 '20 at 15:34
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By rank-nullity theorem and independence of orthogonal vectors, if $\{v_1, v_2, \dots, v_m\}$ is an orthonormal basis for the rowspace and $\{w_1, w_2, \dots, w_n\}$ is an orthonormal basis for the nullspace, then $\{v_1, v_2, \dots, v_m, w_1, w_2, \dots, w_n\}$ is an orthonormal basis for the whole space.

Let $u$ be a vector orthogonal to the nullspace. $u = c_1v_1 + c_2v_2 + \dots + c_mv_m + b_1w_1+b_2w_2 +\dots+b_nw_n$

Since $u$ and each $v_i$ is orthogonal to the nullspace, and $\{w_1, w_2, \dots, w_n\}$ are orthonormal,

$u\cdot w_i = (c_1v_1 + c_2v_2 + \dots + c_mv_m + b_1w_1+b_2w_2 +\dots+b_nw_n)\cdot w_i=0$

$\Rightarrow b_i=0$ for $1\leq i\leq n$

Thus $u=c_1v_1 + c_2v_2 + \dots + c_mv_m$ and therefore $u$ is in the rowspace of $A$.

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  • $\begingroup$ Thanks a lot. Can you please elaborate on why ${v_1,v_2,v_3,...,w_1,...,w_n}$ is a basis for whole space? $\endgroup$ – Koro May 10 '20 at 15:22
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    $\begingroup$ @Koro Because of the rank-nullity theorem, we only need to show that they are linearly independent. Let $c_1v_1 + c_2v_2 + \dots c_nw_n = 0$ . Then $(c_1v_1 + c_2v_2 + \dots c_nw_n)\cdot w_i = 0$ $\Rightarrow c_i = 0$ for the $w_i$. Similarly do a dot product with the $v_i$ to get $c_i=0$ for all $i$ $\endgroup$ – Saad Haider May 10 '20 at 15:23
  • $\begingroup$ Wow. Thanks a lot. I have understood. Please change the last line , u should be a linear combination of $v_i$'s. $\endgroup$ – Koro May 10 '20 at 15:30
  • $\begingroup$ @Koro Changed it thanks and glad to be of help $\endgroup$ – Saad Haider May 10 '20 at 15:33

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