Intersection of Orthogonal Subspaces

[CormJack]

Hello. I am a British economics student, advised to come here by a friend. She said this is a very helpful and engaged forum, but she recommended i emphasise that i am an economics student, so that people know to be patient with my maths hahah! Thanks in advance!

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

- I believe this is because $N(A^T) = RS(A^T)^\bot = R(A)^\bot$ 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 both the geometric intuition that these subspaces can only share the zero vector? I'm also struggling with the Linear Algebra Intuition, what is it about the Null Space and the Range that implies this property.

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?
- My 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.

3) Visual Confusion: What if i had two perpendicular planes in $\mathbb{R^4}$, i.e. Like two pieces of paper, but living in $\mathbb{R^4}$ (or even in $\mathbb{R^3}$). 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? And yet Row Space and Null space are orthogonal compliments?

5) Question: Building on the previous questions, the fact that $S \cap S^\bot = \{0\}$ 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?

Thanks!

[Martino]

Hello! Welcome.

Of course by $A^T$ you mean the transpose of $A$. About your first question, I am assuming that you are talking about real spaces (over the complex numbers things are different). An easy way to think about it is the following: if $u in R(A) nn N(A^T)$ then $u=Av$ for some vector $v$ (by definition of range) and $A^Tu=0$ (by definition of kernel), so $A^TAv=0$ and we want to prove that this implies $u=Av=0$. We can do this by multiplying on the left by $v^T$ obtaining $v^TA^TAv=0$ which can be written as $(Av)^TAv=0$. This implies that $Av=0$ because if $w$ is any vector then $w^Tw$ is the scalar product of $w$ with itself, so it is the norm of $w$ squared (here I am using that the coefficients are real numbers). If a vector has zero norm, then it is the zero vector.

About your other questions, it seems to me that you are always imagining things in a $3$-dimensional space, where of course two 2-dimensional things usually intersect in a line. But in fact in your situation the space is 4-dimensional and two planes can intersect trivially (i.e. the intersection can be a point). Think of points as 4-tuples $(x,y,z,w)$, then take the two planes defined by $x=y=0$ (first plane) and $z=w=0$ (second plane). They intersect trivially because the intersection is given by $x=y=z=w=0$. Probably trying to imagine this happening is not helpful because we human beings are not able to imagine a $4$-dimensional world. In our 3-dimensional world two planes cannot possibly intersect in a point.

I don't know what you mean by $RS$.

PS. I have removed two identical copies of your message. I have also moved your message to the correct section.

[gabriella127]

@CormJack Welcome to Matematicamente!

[CormJack]

Hi thank you so much for your quick response! Apologies, i somehow didn't spot the Linear Algebra and Geometry section when i was looking through the options...

By RS i meant Row Space, so i was referring to the fact that the Row Space and Null Space of $A$ are orthogonal compliments.

For your second answer that's so interesting, and that makes sense. Initially i was struggling to see why $x = y = 0$ would define a plane. Because i was imagining $\mathbb{R^3}$ where of course this only leaves $z$ to vary and so is a line. But in $\mathbb{R^4}$ we still have two variables varying $w$ and $z$ so it's a plane. And then as you point out those two planes intersect only at $0$. Amazing!

Regarding the first answer, thank you for the algebraic explanation that is very helpful. I realise now i had seen that proof but hadn't connected the dots for it's use here. Going further, are you able to elaborate with a kind of qualitative or geometric intuition. I suppose if someone asked me to explain in words an answer to the following question:

"You have the Null Space, in which lives all the vectors that this transformation $A^T$ converts to the Zero vector, and you have the range of $A$ which contains all the vectors that the transformation A spits out. The only vector they share in common is $0$. Why must this be true, based on the geometric meaning of the range and null space, and transpose."

How would i make sense of that? As if i was trying to explain to a friend who didn't know linear algebra. But can understand the idea of lines (vectors) being squished and extended a moved around etc. I.e. they a have a model of 3D space. Are there any visual intuitions or otherwise you can point me to?

Anyway thanks a lot, and i look forward to your response if you find ant extra time, and thanks for the warm welcome @gabriella127

Grazie molto!

[Martino]

A vector in the kernel of $A$ is a vector orthogonal to all the rows of $A$ (by definition of matrix product). So a vector in the kernel of $A^T$ is a vector orthogonal to all the rows of $A^T$, which are precisely the columns of $A$. The range of $A$ is the space generated by the columns of $A$. So being in the range of $A$ and also in the kernel of $A^T$ means being in the range of $A$ and, at the same time, being orthogonal to all the vectors in the range of $A$. In particular, every such vector is orthogonal to itself. Since only the zero vector is orthogonal to itself we obtain $R(A) nn N(A^T)=0$.

If you think about it, this statement (after a suitable reformulation) does not have anything to do with matrices. In other words, if you call $W$ the range of $A$, the equality $R(A) nn N(A^T) =0$ is $W nn W^\bot =0$. This means that you can forget about matrices and only think about subspaces and orthogonal complements.

We also have $R(A^T) nn N(A)=0$ yes, to see this you can apply the equality $R(A) nn N(A^T)=0$ replacing $A$ with $A^T$ and recalling that $(A^T)^T=A$.

[CormJack]

Thank you very much again Martino! Apologies for my delay in replying. You're post is very helpful, i'm feeling much closer to satisfaction now. I've attached a below a passage from Gilbert Strang, 1993, The Fundamental Theorem of Linear Algebra, that was recommended to me! Once i have read through it, if i have more questions i will post again. But hopefully that will be it for this one! https://home.engineering.iastate.edu/~j ... gpaper.pdf