# Column space and orthogonal complement

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$$(A), then at least one column (vector) of $$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)$$

• 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? Aug 27 '20 at 9:20
• @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. Aug 30 '20 at 23:41
• 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. Aug 31 '20 at 6:03
• @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 $C(A)^\perp.$ Aug 31 '20 at 14:23
• Please is this correct? Let me know if there is something wrong so I will correct it again. Thank you. Aug 31 '20 at 14:27

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.

• Could you give out your side of the solution? Aug 27 '20 at 15:55
• I gave a counter example, thus the statement you are trying to prove is false. What more are you asking for? Aug 28 '20 at 6:36

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.

• You are mixing up unions and direct sums. Aug 27 '20 at 9:16