# Unexpected results in vector spaces intersection

Basing on the answer to this question I tried to solve an exercise which asks to find the intersections of four vectorial subspaces. I have the subspaces $X_O=\{(1,0,0)\}$, $X_{NO}=\{(0,1,0),(0,0,1)\}$, $X_R=\{(1,1,0),(0,0,1)\}$, $X_{NR}=\{(1,-1,0)\}$ and I have to find:

• $X_1=X_R \cap X_{NO}$
• $X_2=X_R \cap X_O$
• $X_3=X_{NR} \cap X_{NO}$
• $X_4=X_{NR} \cap X_O$

I dont't think (and hope) to have done any algebric error, but the results aren't - probably - correct, because I find out that the three latter intersections are empty and $X_1=\{(0, 0, 1)\}$ while the direct sum of the four intersections has to make up the whole space $X=R^3$.

Any suggestion?

Edit1: as reported in comments I found a possible solution coherent to the hypotesys that the direct sum of the four intersections is the whole space, but which leads to results I can't explain. I'll try to describe it here; if there is any incorrectness feel free to point it out.

I can say that $X_3=X_{NR} \cap X_{NO}=X_{NO} \cap (X_{NR}+X_O)=X_{NO} \cap {X_1}^\perp$

(because, for the second equality, $X_{NO} \cap (X_{NR}+X_O)$ is equal to $X_{NO} \cap X_{NR} + X_{NO} \cap X_O$ which is equal to $X_{NO} \cap X_{NR}$ and for the third because stands $(A \cap B)^{\perp}=A^{\perp}+ B^{\perp}$)

So, with $X_1=\{(0,0,1)\}$, I find $X_3=\{(0,1,0)\}$.

But if, for identical reasons, I substitute $X_{NO}$ instead of $X_{NR}$, I find $X_3=X_{NR} \cap X_{NO}=X_{NR} \cap (X_R + X_{NO})$. The term in parentheses is ${X_4}^{\perp}$. $X_4$ can be found empty with any method, so its orthogonal subspace is the entire space; then the intersection between $X_{NR}$ and the entire space is simply $X_{NR}=\{(1,-1,0)\}$.

To me, it appears that the two one-vector bases found for $X_3$ are not the same: how can it be possible? Moreover, computing also $X_2$ with this method realizes the "condition" that the sum of the four subspaces forms the entire space, condition that is not respected with the "classic" resolution, which finds only $X_1$ to be non empty. I explained the problem as crearly as I could: if you have any problem I'll try to make it clearer.

Edit2: as explained in "Systems and Control Theory An Introduction" by A.Astolfi the following properties are valid:

• $X_1 + X_2 = X_R$

• $X_1 + X_3 = X_{NO}$

• $X_1 + X_2 + X_3 + X_4 = X = {\mathbb{R}}^n$

• $n1+n2+n3+n4=n$, where $n1, n2, n3, n4$ are the dimensions of, respectively, $X_1, X_2, X_3, X_4$

Edit3 (hopefully the last one): I found in other resources online a different definition for the coordinate transformation matrix T needed for the Kalman decomposition (which from the previous definition was formed by the column vectors composing the bases of $X_1, X_2, X_3, X_4$, in this order):

in this document by Perry Li of the University of Minnesota is stated that $T$ = ( $t1$ $t2$ $t3$ $t4$ ) where

• t2 -> ($X_{R} \setminus X_{NO}$): t2 is a basis for $X_R \cap X_{NO}$

• t1 -> ($X_{R} \setminus X_{O}$): t1 $\cup$ t2 is a basis for $X_R$

• t4 -> ($X_{NR} \setminus X_{NO}$): t2 $\cup$ t4 is a basis for $X_{NO}$

• t3 -> ($X_{NR} \setminus X_{O}$): t1 $\cup$ t2 $\cup$ t3 $\cup$ t4 is a basis for $R^3$

while Wikipedia defines it in a similar but slightly different way: (for simplicity I use the same notation with $t_i$) $T$ = ( $t1$ $t2$ $t3$ $t4$ ) where

• t1 is a matrix whose columns span the subspace of states which are both reachable and unobservable.

• t2 is chosen so that the columns of [t1 t2] are a basis for the reachable subspace.

• t3 is chosen so that the columns of [t1 t3] are a basis for the unobservable subspace.

• t4 is chosen so that [t1 t2 t3 t4] is invertible.

Maybe these are the correct definitions: I think that these don't corresponde to those from Astolfi's document. Thanks a lot to @amd who patiently has managed to trace the origin of the errors in my statements.

• Typo (?): You have two $X_1$ in your question. And: are you sure that the direct sum of the four intersections is $\mathbb{R}^3$? – Emilio Novati Jan 19 '17 at 15:49
• My fault, I'll correct. And yes, I'm sure. There was another error on the fourth subspace which I also corrected. – M-elman Jan 19 '17 at 15:51
• It seems to me that $X_2=X_3=X_4=\{(0,0,0)\}$. Maybe that you have some other error ? – Emilio Novati Jan 19 '17 at 16:12
• It seems to me too, but this couldn't be possible. I tried another resolution: $X_2=X_R \cap X_O=X_R \cap (X_{NR}+X_O)=X_R \cap {X_1}^\perp$ and in this way I found a different solution for $X_2$ and $X_3$. Could it be possible? And why? – M-elman Jan 19 '17 at 16:20
• Perhaps you misunderstood a problem statement somewhere along the way. Likely you’re meant to examine the four direct sums of the same pairs of subspaces that are being intersected. – amd Jan 20 '17 at 1:34

You can determine by inspection that most of the intersections are trivial. Both of the vectors in $X_{NO}$ have $0$ as their first component, so there’s no linear combination of them that can be a multiple of $(1,0,0)$. Turning to the next intersection, there has to be a non-zero multiple of $(1,1,0)$ in the intersection to get a non-zero first component, but there’s no way to eliminate the second component of the resulting vector by adding a multiple of $(0,0,1)$. The only one that’s non-trivial is $X_1$, and that’s equal to $\operatorname{span}\{(0,0,1)\}$, as you’ve found.

The sum of these four intersections obviously doesn’t span all of $\mathbb R^3$. If they’re supposed to for some reason, then you’ve likely made some error in computing the original four subspaces in the first place. Since you haven’t shown that work, I can’t help you with that.

As to the derivations in your addition to the question, intersection is not distributive over vector space sum, i.e., it’s not true in general that $V\cap(U+W)=(V\cap U)+(V\cap W)$. This is because $U+W$ can contain vectors that are in neither $U$ nor $W$. For example, let $V=\operatorname{span}\{(1,1)\}$, $U=\operatorname{span}\{(1,0)\}$ and $W=\operatorname{span}\{0,1\}$. We then have $V\cap(U+W)=V$, but $V\cap U=V\cap W=\{0\}$, so $(V\cap U)+(V\cap W)=\{0\}$. This is exactly the situation with $X_O$, $X_{NO}$ and $X_{NR}$: Although $X_{NO} \cap (X_{NR}+X_O)$ is non-trivial, both $X_{NO} \cap X_{NR}$ and $X_{NO} \cap X_O$ are, which makes their sum trivial as well.

Update: It appears that you’re making several fundamental errors with respect to set and vector space operations that are leading to the seemingly contradictory results in your question. I’ve covered one of them above. Another is the confounding of the difference of two sets with the quotient of two vector spaces.

If we have a subspace $W$ of some vector space $V$, then the set difference $V\setminus W$ cannot be a vector space: by definition, $0\in W$, so it cannot be in $V\setminus W$. On the other hand, if we have proper subspaces $U$ and $W$ of $V$ such that $U\oplus W=V$, then there are elements of $V$ that are neither in $U$ nor $W$, so $U\cup W\ne V$. This can be visualized in $\mathbb R^2$: the union of a pair of distinct lines is just those two lines, but their sum is the entire plane.

To put this in concrete terms relative to Astolfi’s book, we have the set of reachable states $\mathcal R=\operatorname{im}R$, which I believe corresponds to $X_R$ in your question. This happens to be a vector space, but the set of unreachable states $\mathbb R^n\setminus\mathcal R$ is not. Additionally, $X_R^\perp\ne\mathbb R^n\setminus\mathcal R$, that is, your $X_{NR}$ is not in fact the set of unreachable states. Similar considerations hold for the sets of observable and non-observable states.

Turning now to the Kalman decomposition, we have $\mathcal X_1=\mathcal R\cap\ker O$, the set of reachable but unobservable states, which, being the intersection of two vectors spaces, is itself a vector space. However, $\mathcal X_2$, the vector space complement of $\mathcal X_1$ within $\mathcal R$, is not the set of states that are both reachable and observable. The Kalman decomposition is not a partition of $\mathbb R^n$ into the four types of state. In fact, if all four subspaces in this decomposition are proper, then there are states that belong to none of them.

• Therefore are you saying that the three empty subspaces are correct? And what about the condition that their direct sum must be the entire space? (Since I don't want to act like it is something murky: for reasons not depending on the algebra, every state - i.e. vector - in the whole space X can be reachable or not and observable or not; follows that every vector in the space must belong to one of the subspaces and so their direct sum must compose the entire X) – M-elman Jan 19 '17 at 21:20
• @M-elman All four intersections consist of the zero vector only. The sum of the four subspaces clearly spans the entire space since all three standard basis vectors are included. In fact, for this collection of subspaces, all four possible sums of one-dimensional and two-dimensional subspaces span $\mathbb R^3$. – amd Jan 19 '17 at 22:24
• Excuse me, but I don't understand: you said that the four intersections (from $X_1$ to $X_4$) consist of the zero vector only; how can their sum span the entire space? – M-elman Jan 19 '17 at 22:36
• @M-elman I was talking about the direct sum of the original four subspaces. The sum of the four intersections is uninteresting. Perhaps you misunderstood a problem statement somewhere along the way. The intersections, being trivial, obviously can’t span anything no matter how many of them you combine. On the other hand, if you take the direct sums of each pair of subspaces that are being intersected, then each of those sums does span the entire space. – amd Jan 20 '17 at 1:33
• Of course the direct sum of the original four subspaces spans the entire X. But from theory, and also from a logical point of view (what else a vector can be if not reachable and observable, unreachable and unobservable, reachable and unobservable or unreachable and observable?) also the direct sum of the requested four subspaces has to span the whole X. – M-elman Jan 20 '17 at 6:43