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I have a question and I don't really know how to solve it. I tried it and probably my solution is $\textit{wrong}$ to say the least. Here we go!

Assume we have $A_{m \times n}.$ Prove or disprove the statement: Every vector in $\mathbf{R^n}$ is either $C(A)$ or $C(A)^\perp$ or both. Where $C(A)$ is span of column space of $A$ and $C(A)^\perp$ is the orthogonal complement of $A$.

My take: Every matrix say $A$, has the property rank$(A) \leq n$.

So if rank$(A)<n$, then at least one column (vector) of $A$ is linearly dependent and rank$(A) <$ rank $(\mathbf{R^n})$ which implies that $A$ does not span $\mathbf{R^n}$ so that we have $$\text{rank } \mathbf{(R^n)} = \text{ rank} (A) + \text{ rank } (A^\perp) $$ Similarly, if rank$(A)=n$, then all columns (vectors) of $A$ are linearly independent and rank$(A) =$ rank $(\mathbf{R^n})$ which implies that $A$ spans $\mathbf{R^n}$ so that we have $$\text{rank } \mathbf{(R^n)} = \text{ rank} (A) $$

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  • $\begingroup$ What does $\operatorname{rank}(\mathbf R^n)$ mean? The rank is a an invariant of linear maps or matrices, not of spaces. Also, shouldn't you be looking at $\mathbf R^m$ instead? $\endgroup$
    – Christoph
    Aug 27 '20 at 9:20
  • $\begingroup$ @Christoph. Since the statement is false, it is good enough to disprove it using an example as: for $n=1$ and $p=2.$ Then $X = \begin{bmatrix} 1 & 0 \end{bmatrix}$. So that $C(X) = \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right\}\in R^{p} , C(X)^\perp = \left\{ \begin{bmatrix} 0 \\ 1 \end{bmatrix}\right\}\in R^{p}, \left(\text{rank} (X) = 1\right) + \left(\text{Nullity}(X) = 1 \right)= p=2. $ Clearly, $C(X) \not\in R^n$ or $C(X)^\perp \not \in R^n$ or both. $\endgroup$
    – holala
    Aug 30 '20 at 23:41
  • $\begingroup$ What is $p$? Why does your $X$ have two columns in $\mathbf R^1$ but your column space is spanned by one column in $\mathbf R^2$? Why are you considering “$C(X) \in \mathbf R^n$” at all: a subspace of $\mathbf R^p$ will never be an element of $\mathbf R^n$. The question is about whether every $x\in\mathbf R^m$ is an element of $C(A)$ or $C(A)^\perp$ or both. $\endgroup$
    – Christoph
    Aug 31 '20 at 6:03
  • $\begingroup$ @Christoph Sorry for mix-ups. For any $A_{m \times n}$, we have that $C(A) \in \mathbf{R}^m$ and $R(A) \in \mathbf{R}^n$ so we consider the case that $A_{1 \times 2} = [1, 0] \in \mathbf{R}^1$. Then, $C(A) = [1] \in \mathbf{R}^1$ and $C(A)^\perp = [0] \in \mathbf{R}^1$ and the rank-nullity Thm is satisfied $C(A) + C(A)^\perp = n = 2.$ Thus, it is obvious that no vector in $\mathbf{R}^2$ is either in $C(X)$ or $C(A)^\perp.$ $\endgroup$
    – holala
    Aug 31 '20 at 14:23
  • $\begingroup$ Please is this correct? Let me know if there is something wrong so I will correct it again. Thank you. $\endgroup$
    – holala
    Aug 31 '20 at 14:27
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This is wrong. Consider the $2\times 1$ matrix $\left(\begin{smallmatrix} 1 \\ 0\end{smallmatrix}\right)$, then the column space is $U=\operatorname{span}\left( \left(\begin{smallmatrix} 1 \\ 0\end{smallmatrix}\right)\right)$ and its orthogonal complement is $U^\perp=\operatorname{span}\left( \left(\begin{smallmatrix} 0 \\ 1\end{smallmatrix}\right)\right)$.

The vector $\left(\begin{smallmatrix} 1 \\ 1\end{smallmatrix}\right)$ is not an element of $U$ and neither of $U^\perp$.

The other answers seem to mix up unions and direct sums.

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  • $\begingroup$ Could you give out your side of the solution? $\endgroup$
    – holala
    Aug 27 '20 at 15:55
  • $\begingroup$ I gave a counter example, thus the statement you are trying to prove is false. What more are you asking for? $\endgroup$
    – Christoph
    Aug 28 '20 at 6:36
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Any subspace and its orthogonal complement partition the given space $R^n$. So any vector in $R^n$ either belongs to some subspace U in $R^n$ or belongs to $U^{\perp}$ (zero vector belong to both). To me, it seems we do not need to analyze the property of $C(A)$.

Then I take a closer look, $C(A)$ here is a subspace in $R^m$, so if there is no typo, I would claim the statement is incorrect.

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  • $\begingroup$ You are mixing up unions and direct sums. $\endgroup$
    – Christoph
    Aug 27 '20 at 9:16

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