<p><em>Note: This is a Linear Algebra Question. I'm posting here because I find this community more helpful than maths stack! And ofc linear algebra is fundamental to econometrics, as well!</em></p>
<ol>
<li>Martin Anthony Linear Algebra p.g. 371, notes that  <span class="math-container">$R(A) \cap N(A^T) = \{0\}$</span>  where R = Range and N = Null</li>
</ol>
<ul>
<li>I believe this is because <span class="math-container">$N(A^T) = RS(A^T)^\perp = R(A)^\perp$</span> and the intersection of two orthogonal compliments is the zero vector.</li>
<li><strong>Question:</strong> Would we also have <span class="math-container">$R(A^T) \cap N(A) = \{0\}$</span>?</li>
<li><strong>Question:</strong> I'm struggling with the kind of geometric or Linear Algebra intuition that these subspaces can only share the zero vector?</li>
</ul>
<ol start="2">
<li><strong>Question:</strong> can we have two subspaces be orthogonal and their intersection be vectors other than the zero vector?</li>
</ol>
<ul>
<li>I.e. is there is a difference between subspaces which are orthogonal compliments and orthogonal subspaces?</li>
<li><strong>Answer:</strong> I assume that there is no difference and that for two subspaces <span class="math-container">$A$</span> &amp; <span class="math-container">$B$</span> to be orthogonal, every element in <span class="math-container">$A$</span> must be orthogonal to every element in B. In which case if they contained the same vectors, not equal to zero, the vector is not orthogonal to itself. Hence the subspaces cannot be orthogonal?</li>
</ul>
<ol start="3">
<li><p><strong>Visual Confusion:</strong> What  if i had two perpendicular planes in R4, i.e. Like two pieces of paper, but living in R4 (or even in R3). If i had them 90 degrees to one another, so colloquially they look orthogonal, forming a cross. These would intersect in a line. And so their intersection is a line i.e. not the zero vector? So they are not orthogonal compliments? So our visual analogy fails us here?</p>
</li>
<li><p><strong>Question:</strong>  What if we had a 4x4 Matrix A, And what if the <span class="math-container">$Dim(R(A)) = 2$</span>, therefore <span class="math-container">$Dim(RS(A)) = 2$</span></p>
<ol>
<li>Then using the Rank Nulity Theorem we must have <span class="math-container">$Dim(N(A)) = 2$</span></li>
<li>But if Both Row Space and Null Space have dimension two, how can their intersection be the Zero vector? Surely they must intersect along a line?</li>
</ol>
</li>
<li><p><strong>Question:</strong> Building on the previous questions, the fact that <span class="math-container">$S \cap S^\perp = \{0\}$</span> implies to me that one of these subspaces must always be one dimensional, otherwise how else can they intersect at a single point - where this single point is in fact the origin? I.e. we always have a line and then something else?</p>
</li>
</ol>

CC BY-SA 4.0
2023-06-11T13:02:43.437Z
Most of your questions have been answered by Gilbert Strang, 1993, The Fundamental Theorem of Linear Algebra, The American Mathematical Monthly, Vol. 100, No. 9, pp. 848-855
Bertrand
CC BY-SA 4.0
2023-06-11T13:14:50.367Z
Okay great thanks! I will see if I can access a copy of that anywhere
CormJack
CC BY-SA 4.0
2023-06-11T15:42:15.097Z
@Bertrand can i confirm this is what you were referring to, thanks: chrome-extension://efaidnbmnnnibpcajpcglclefindmkaj/https://home.engineering.iastate.edu/~julied/classes/CE570/Notes/strangpaper.pdf
CormJack
CC BY-SA 4.0
2023-06-11T18:52:27.663Z
yes this is the right article, I found the figures and explanations useful. Sorry for not being able to help more.
Bertrand
CC BY-SA 4.0
2023-06-11T20:50:17.237Z
No problem! Some good pointers is a lot better than nothing! If i'm able to point my owner answer together, perhaps you would have a moment to check it!
CormJack
CC BY-SA 4.0
2023-06-12T08:25:27.643Z
Your "visual analogy" fails because nobody can visualize a 4- or higher-dimensional space. E.g. two 2-dim planes cannot intersect only in the origin in 3-dim space, but it's easy in 4-dim space. Just think of the plane consisting of all vectors of the form $(a,b,0,0)$ and the plane consisting of all vectors of the form $(0,0,c,d)$.
VARulle
CC BY-SA 4.0
2023-06-12T08:31:03.853Z
As to the difference between orthogonal subspaces and orthogonal complements: The latter are just orthogonal subspaces that, together, span the whole vector space they are embedded in. E.g. the x-axis and the y-axis of the 3-dim x-y-z-space are orthogonal subspaces, but not orthogonal complements. But the x-axis and the y-z-plane are both.
VARulle
<econometrics><mathematical-economics><linear-algebra>
Orthogonality of two Subspaces
CormJack
# Orthogonality of two Subspaces

(Asked by: username_0 on 2023-06-10)

*Note: This is a Linear Algebra Question. I'm posting here because I find this community more helpful than maths stack! And ofc linear algebra is fundamental to econometrics, as well!*

1. Martin Anthony Linear Algebra p.g. 371, notes that $R(A) \cap N(A^T) = \{0\}$ where R = Range and N = Null

* I believe this is because $N(A^T) = RS(A^T)^\perp = R(A)^\perp$ and the intersection of two orthogonal compliments is the zero vector.
* **Question:** Would we also have $R(A^T) \cap N(A) = \{0\}$?
* **Question:** I'm struggling with the kind of geometric or Linear Algebra intuition that these subspaces can only share the zero vector?

2. **Question:** can we have two subspaces be orthogonal and their intersection be vectors other than the zero vector?

* I.e. is there is a difference between subspaces which are orthogonal compliments and orthogonal subspaces?
* **Answer:** I assume that there is no difference and that for two subspaces $A$ & $B$ to be orthogonal, every element in $A$ must be orthogonal to every element in B. In which case if they contained the same vectors, not equal to zero, the vector is not orthogonal to itself. Hence the subspaces cannot be orthogonal?

3. **Visual Confusion:** What if i had two perpendicular planes in R4, i.e. Like two pieces of paper, but living in R4 (or even in R3). If i had them 90 degrees to one another, so colloquially they look orthogonal, forming a cross. These would intersect in a line. And so their intersection is a line i.e. not the zero vector? So they are not orthogonal compliments? So our visual analogy fails us here?
4. **Question:** What if we had a 4x4 Matrix A, And what if the $Dim(R(A)) = 2$, therefore $Dim(RS(A)) = 2$

   1. Then using the Rank Nulity Theorem we must have $Dim(N(A)) = 2$
   2. But if Both Row Space and Null Space have dimension two, how can their intersection be the Zero vector? Surely they must intersect along a line?
5. **Question:** Building on the previous questions, the fact that $S \cap S^\perp = \{0\}$ implies to me that one of these subspaces must always be one dimensional, otherwise how else can they intersect at a single point - where this single point is in fact the origin? I.e. we always have a line and then something else?

**Tags:** econometrics, mathematical-economics, linear-algebra
> **Comments:**
> **username_1**: Most of your questions have been answered by Gilbert Strang, 1993, The Fundamental Theorem of Linear Algebra, The American Mathematical Monthly, Vol. 100, No. 9, pp. 848-855  
> **username_0**: Okay great thanks! I will see if I can access a copy of that anywhere  
> **username_0**: @Bertrand can i confirm this is what you were referring to, thanks: chrome-extension://efaidnbmnnnibpcajpcglclefindmkaj/https://home.engineering.iastate.edu/~julied/classes/CE570/Notes/strangpaper.pdf  
> **username_1**: yes this is the right article, I found the figures and explanations useful. Sorry for not being able to help more.  
> **username_0**: No problem! Some good pointers is a lot better than nothing! If i'm able to point my owner answer together, perhaps you would have a moment to check it!  
> **username_2**: Your "visual analogy" fails because nobody can visualize a 4- or higher-dimensional space. E.g. two 2-dim planes cannot intersect only in the origin in 3-dim space, but it's easy in 4-dim space. Just think of the plane consisting of all vectors of the form $(a,b,0,0)$ and the plane consisting of all vectors of the form $(0,0,c,d)$.  
> **username_2**: As to the difference between orthogonal subspaces and orthogonal complements: The latter are just orthogonal subspaces that, together, span the whole vector space they are embedded in. E.g. the x-axis and the y-axis of the 3-dim x-y-z-space are orthogonal subspaces, but not orthogonal complements. But the x-axis and the y-z-plane are both.  

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