Fast solutions to the Simultaneous Sylvester and Derivation Equations

These slides are a record keeping tool and will not be used at for any actual presentations, hence the tiny font size and color choices at times

Simultaneous Sylvester Equation

Let \(A,B,C,R,S \) be vector spaces of dimensions \((a,b,c,r,s)\)

Given
  \(\rho : \textcolor{blue}{R} \times B \times C \rightarrowtail K \),
  \( \sigma : A \times \textcolor{green}{S} \times C \rightarrowtail K \),
  \( \tau : A \times B \times C \rightarrowtail K \)

Find
  \( x : A \rightarrow \textcolor{blue}{R},  y : B \rightarrow \textcolor{green}{S} \) such that,

\[ \rho(x(-),-,-) + \sigma(-,y(-),-) = \tau(-,-,-)\]

as trilinear maps \(A \times B \times C \rightarrowtail K \)

Objective: Given data for this problem in coordinates, report the system is inconsistent or find a solution faster than solving a linear system with \(abc\) constraints and \(ar + bs\) variables.

Conventions:

• We will use \((\rho,\sigma,\tau)\) to denote an instance of the Simultaneous Sylvester problem
• When there's no ambiguity what slot each operator belong to, we will write \(\rho(x) + \sigma(y) = \tau\)

Observation: (all finite dimensional vector spaces)

\(R \times B \times C \rightarrowtail K = \text{Mult} (R,B,C;K) \cong \text{Hom}(R, (B \otimes C)^{\ast}) \)

Throughout, assume we are given an isomorphism from a vector space to its dual. So we may take \( (B \otimes C)^{\ast} \cong (B \otimes C) \)

Thus \(\rho: R \times B \times C \rightarrowtail K \) can be thought of as an element \( \rho \in \text{Hom}(R, B \otimes C) \). 

We assume \(\rho\) is injective. The rationale is if not, split \( R = \text{ker}(\rho)\oplus R'\). Any solution \(x: A \rightarrow R\) splits as \(x_{\text{ker}(\rho)} \oplus x_{R'} \), where the first term can be chosen without any restrictions. Thus we may as well start with \(\rho': R' \times B \times C \rightarrowtail K \) and keep track of \(\text{ker}(\rho)\) separately

Terminology: We call the system \((\tilde{\rho}, \tilde{\sigma}, \tilde{\tau})\) restricted, and to differentiate, call the full system \((\rho,\sigma,\tau)\) global.

Suppose we have a subspace \(B' \leq B\) such that \( \tilde{\rho} \coloneqq ((\pi_{B'} \otimes I)\circ \rho) \in \text{Hom}(R, B' \otimes C) \) is still injective.

Depending on context, we also view \(\tilde{\rho}\) as a trilinear map in \(R \times B' \times C \rightarrowtail K\)

Apply the same logic for \(\sigma: A \times S \times C \rightarrowtail K\), and subspace \(A' \leq A\), getting \( \tilde{\sigma}: A' \times S \times C \rightarrowtail K\), and project onto both \(A'\) and \(B'\) for \(\tau\), getting \(\tilde{\tau}: A' \times B' \times C \rightarrowtail K\)

Solve for \(\tilde{x}: A' \rightarrow R, \tilde{y}: B' \rightarrow S\) such that the equation \(\tilde{\rho}(\tilde{x}) + \tilde{\sigma}(\tilde{y}) = \tilde{\tau}\) is satisfied. Each term is a multilinear map \(A' \times B' \times C \rightarrowtail K\)

Steps: \( \text{Mult}(R,B,C;K) \cong \text{Hom}(R \otimes B \otimes C, K) \cong \text{Hom}(R, \text{Hom}(B \otimes C, K)) \cong \text{Hom}(R, (B \otimes C)^{\ast})\) 

Choose complements \(A = A' \oplus U\) and \(B = B' \oplus V\)

Projecting to \(U \leq A\) and \(B' \leq B\) for \(\rho,\sigma,\tau\) results in \( \rho_{R \times B' \times C}(x_U) + \sigma_{U \times S \times C}(\tilde{y}) = \tau_{U \times B' \times C} \)

Notice the first term is the only one with unknowns (being \(x_U\)) and \(\rho_{R \times B' \times C}: R \times B' \times C \rightarrowtail K \cong \text{Hom}(R, B'\otimes C)\) is left invertible

Hence solving for \(x_U\) does not require solving a linear system, but rather, just an application of the left inverse of \(\rho_{R \times B' \times C} \) to all terms, thinking of each as an element in \( \text{Hom}(R, B' \otimes C)\)

Projecting to \(A' \leq A\) and \(V \leq B\) finds \(y_V\) without solving a system of equations

Finally, projecting to \(U \leq A\) and \(V \leq B\) is necessary to verify the restricted solution, \( \tilde{x}, \tilde{y}\), extends to a global solution

We call \(x \coloneqq \tilde{x} \oplus x_U\) and \(y \coloneqq \tilde{y} \oplus y_V\) an extended candidate solution

So the final step is verifying \(\rho_{R \times V \times C}(x_U) + \sigma_{U \times S \times C}(y_V) = \tau_{U \times V \times C}\) holds

Assume now we have some \(\tilde{x}: A' \rightarrow R\) and \(\tilde{y}: B' \rightarrow S\)

What if the restricted system \( (\tilde{\rho},\tilde{\sigma},\tilde{\tau}) \) is already inconsistent?

Then we can immediately report the overall system \( (\rho,\sigma,\tau) \) is inconsistent, as any solution of the overall system must also solve the restricted one.

What if there is not a unique solution to the restricted system? 

That is, the \( (\tilde{x}, \tilde{y}) \) we found satisfying the restricted system is not unique. This is very common. For example, when \(\tau=0\) the Sylvester system is homogeneous meaning all non-zero solutions are subspaces of \( \text{Hom}(A,R)\oplus \text{Hom}(B,S)\).

Even if \( a'r + b's \leq a'b'c \), the restricted system may still be underdetermined - such as when the constraints themselves have relations between them.

Hence we need to search for a solution in the solution space of \( (\tilde{\rho}, \tilde{\sigma}, \tilde{\tau}) \) that extends to a global solution of \((\rho,\sigma,\tau)\)

The correctness of this method is because the global system \((\rho,\sigma,\tau)\) has a superset of the constraints of the restricted system \( (\tilde{\rho}, \tilde{\sigma}, \tilde{\tau}) \)

Furthermore, it is not guaranteed all solutions of the restricted system extend to a solution of the global system.

Another common possibility is if the restricted system is underconstrained. That is, suppose \(\text{dim}(A') = a', \text{dim}(B') = b'\), then the restricted system \( (\tilde{\rho}, \tilde{\sigma}, \tilde{\tau}) \) has \(a'r + b's \) variables, and \(a'b'c \) constraints.

If \( a'r + b's \gt a'b'c \) then the restricted system must have free variables.

The search for a valid solution to the global system

Let \(x \coloneqq \tilde{x} \oplus x_U\) and \(y \coloneqq \tilde{y} \oplus y_V\) be an extended candidate solution

Define the residual data \(\mathcal{E}_{\tilde{x}, \tilde{y}}: U \times V \times C \rightarrowtail K\) as \( \mathcal{E}_{\tilde{x},\tilde{y}}(u,v,c) = \rho(x(u),v,c) + \sigma (u,y(v),c) - \tau(u,v,c) \)

When \(\mathcal{E}_{\tilde{x},\tilde{y}} = 0\), the restricted solution for \((\tilde{\rho}, \tilde{\sigma}, \tilde{\tau})\) correctly extends to a solution of the global system \((\rho,\sigma,\tau)\)

If the restricted system has a unique solution \( (\tilde{x}, \tilde{y}) \), then either the global system has a unique solution, or the global system is inconsistent.

Checking \(\mathcal{E}_{\tilde{x},\tilde{y}}\) solves our problem conclusively.

If the restricted system has multiple solutions, the situation is more complicated. See the next two slides.

Restricted system \((\tilde{\rho}, \tilde{\sigma}, \tilde{\tau})\) has multiple solutions

The solution space of \((\tilde{\rho},\tilde{\sigma},\tilde{\tau})\) is given by a particular solution to \((\tilde{\rho}, \tilde{\sigma},\tilde{\tau}) \), and a vector subspace consisting of solutions to the homogeneous restricted system \((\tilde{\rho}, \tilde{\sigma}, 0)\). 

Suppose the vector subspace is dimension \(d\) and \((\tilde{x}_i, \tilde{y}_i)_{i=1}^{d}\) is a basis of the homogeneous restricted system and \( (\tilde{x}_p, \tilde{y}_p) \) is a particular solution. 

To set what we need to solve, define the notation: \( \mathcal{E}_i \coloneqq \mathcal{E}_{\tilde{x}_i + \tilde{x}_p,\tilde{y}_i+\tilde{y}_p}  - \mathcal{E}_{\tilde{x}_p,\tilde{y}_p} \). 

For \( \lambda = (\lambda_i)_{i=1}^{d} \), define \((\tilde{x}(\lambda), \tilde{y}(\lambda)) = (\tilde{x}_p, \tilde{y}_p) + \sum_i \lambda_i (\tilde{x}_i,\tilde{y}_i) \). For a valid global solution, we need \( \mathcal{E}_{\tilde{x}(\lambda), \tilde{y}(\lambda)} = 0\)

Indeed, suppose we have the above equality. Then, first we observe

\( \mathcal{E}_i = \mathcal{E}_{\tilde{x}_i + \tilde{x}_p,\tilde{y}_i+\tilde{y}_p}  - \mathcal{E}_{\tilde{x}_p,\tilde{y}_p} = \rho(x_i+x_p) + \sigma(y_i + y_p) - \tau - (\rho(x_p) + \sigma(y_p) - \tau) \)

By linearity, \(\rho(x_i + x_p) = \rho(x_i) + \rho(x_p)\)

Hence the above equation simplifes as \(\mathcal{E}_i = \rho(x_i) + \sigma(y_i)\)

If \(\sum_i \lambda_i \mathcal{E}_{i} + \mathcal{E}_{\tilde{x}_p, \tilde{y}_p} = 0\), then \(\sum_i \lambda_i (\rho(x_i) + \sigma(y_i)) + (\rho(x_p) + \sigma(y_p) - \tau) = 0\)

Using linearity, we collect terms as  \(\rho(\sum_i \lambda_i x_i + x_p) + \sigma(\sum_i \lambda_i y_i + y_p) - \tau = 0\), precisely implying \( \mathcal{E}_{\tilde{x}(\lambda),\tilde{y}(\lambda)} = 0\)

Now we claim our problem is to solve for \(\lambda_i\) such that \(\sum_i \lambda_i \mathcal{E}_{i} + \mathcal{E}_{\tilde{x}_p, \tilde{y}_p} = 0\) - we are interested in this problem because there are only \(d\) variables to solve for.

With \(\lambda_i\) as the unknowns, and \(\mathcal{E}_i, \mathcal{E}_{\tilde{x}, \tilde{y}}\) as the knowns, this is a linear algebra problem with \(uvc \) constraints and \(d\) variables. The core idea of this approach to solving the Simultaneous Sylvester System is that d will frequently be small relative to the initial number of variables \(ar + bs\).

We've shown from the last slide if \(\sum_i \lambda_i \mathcal{E}_{i} + \mathcal{E}_{\tilde{x}_p, \tilde{y}_p} = 0\), then \( \mathcal{E}_{\tilde{x}(\lambda), \tilde{y}(\lambda)} = 0\).

In many cases, \(d\) is both small and any extension is valid, so choosing a random element from the restricted system and extending it gives a valid global solution.

Solving the Sylvester System in coordinates

Let \([\rho] \in K^{bc \times r}, [\sigma] \in K^{ac \times s}, [\tau]\in K^{a \times b \times c}\) be the input to the Simultaneous Sylvester system with the appropriate permutations applied to them.

The naive \(abc\) rows by \(as + bt\) columns system is laid out via \( (I_a \otimes [\rho]) \text{Vec}(X) + \Pi_{a,b}(I_b \otimes [\sigma]) \text{Vec}(Y) = \text{Vec}(T)\), where \( \Pi_{a,b}\) permutes an element in \(K^{bac \times bs}\) to \(K^{abc \times bs}\). 

This gives \(abc\) constraints and \(ar + bs\) variables.

Arithmetic complexity to solve: \(O((abc)(ar+bs)^2)\) when \(abc \geq ar + bs\)

TODO: Old graphic, should replace A,B,C with \([\rho], [\sigma], [\tau] \)

We assume for subspaces \(A' \leq A\) and \(B' \leq B\) of dimension \(a'\) and \(b'\), respectively that

• If \( b'c \geq r \), then \( \tilde{\rho}: \text{Hom}(R, B' \otimes C) \) is generically injective, hence has a left inverse
• If \( a'c \geq s \), then \( \tilde{\sigma}: \text{Hom}(S, A' \otimes C) \) is generically injective, hence has a left inverse
• (TODO: Justify that this is a reasonable assumption) If \(a'b'c \geq a'r + b's \), then all restricted solutions extend to global solutions

Costs:

• Restricted system: Solve linear system with \(a'b'c\) constraints, \(a'r + b's\) variables, costing \(O((a'b'c)(a'r+b's)^2)\)
• Left inverses: Compute for \(b'c \times r\) and \(a'c \times s\) matrix corresponding to \(\tilde{\rho}\) and \(\tilde{\sigma}\). This can be done by RREF, costing \(O(b'cr^2 + a'cs^2)\)
• Computing extended solution: TODO: These need more details/pictures 
  • To solve for \(x_U\): [\( \rho_{R \times B' \times C}(x_U) + \sigma_{U \times S \times C}(\tilde{y}) = \tau_{U \times B' \times C} \) in coordinates]
    • Multiple \(uc \times s\) matrix by \(s \times b'\) matrix for \(\sigma_{U \times S \times C}(\tilde{y})\): \(O(ucsb')\), or \(O(b'csu)\) alphabetized
    • Subtract one \(b'c \times u\) matrix from another to move all knowns to one side: \(O(b'cu)\)
    • Multiply \(r \times b'c \) matrix by \(b'c \times u\) matrix to apply left inverse to \(\rho_{R \times B' \times C}\): \(O(rb'cu)\), or \(O(b'cru)\) alphabetized
  • To solve for \(y_V\): [\( \rho_{R \times V \times C}(\tilde{x}) + \sigma_{A' \times S \times C}(\tilde{y}) = \tau_{A' \times V \times C} \) in coordinates]
    • Multiply \(vc \times r\) matrix by \(r \times a'\) matrix: \( O(vcra')\), or \(O(a'crv)\) alphabetized
    • Subtract one \(a'c \times v\) matrix from another: \(O(a'cv)\)
    • Multiply \(s \times a'c \) matrix by \(a'c \times v\) matrix: \(O(sa'cv)\), or \(O(a'csv)\) alphabetized
  • Summary: \(O(b'cu(r+s) + a'cv(r+s)) \)
• Find restricted solution that extends to global solution: No cost by our assumption.

Total: \(O((a'b'c)(a'r+b's)^2 + (b'cr^2 + a'cs^2) + (b'cu(r+s) + a'cv(r+s))) \)

Full verification omitted - randomized check gives expoentially low probability of error, but complexity to be analyzed.

Restricted system

Compute left inverses

Extend solution

Total: \(O((a'b'c)(a'r+b's)^2 + (b'cr^2 + a'cs^2) + (b'cu(r+s) + a'cv(r+s))) \)

An example fast regime: when \( a,b,c,r,s \) are all \(O(n)\), meaning \(a',b'\) are \(O(1)\)

Total: \(O((c)(r+s)^2 + (cr^2 + cs^2) + (cu(r+s) + cv(r+s))) \)

Plug in \(n\) for \(a,b,c,r,s\), we get the total as \(O((n)(n+n)^2 + (nn^2 + nn^2) + (nn(n+n) + nn(n+n))) = O(n^3) \)

Compared to solving a system of \(n^3\) equations with \(n^2\) variables: \(O(n^3(n^2)^2) = O(n^7)\)

If we add code to check solutions deterministically, need to multiply a \(n^2 \times n\) matrix by a \(n \times n\) matrix, which is \(O(n^4)\) naively. But randomly checking a few linear combinations gives expoentially good probability of correct verification

Derivation Equation

Let \(A,B,C,R,S, T \) be vector spaces with dimensions \(a,b,c,r,s,t\)

Given
  \(\rho : \textcolor{blue}{R} \times B \times C \rightarrowtail K \),
  \( \sigma : A \times \textcolor{green}{S} \times C \rightarrowtail K \),
  \( \tau : A \times B \times \textcolor{orange}{T} \rightarrowtail K \)

Find
  \( x : A \rightarrow \textcolor{blue}{R}, y : B \rightarrow \textcolor{green}{S}, z : C \rightarrow \textcolor{orange}{T} \) such that

\[ r(x(-),-,-) + s(-,y(-),-) = t(-,-,z(-)) \]

as trilinear maps \(A \times B \times C \rightarrowtail K\)

Objective: Given data for this problem in coordinates, report the system is inconsistent or find a non-zero solution faster than solving a linear system with \(abc\) constraints and \(ar + bs + ct \) variables.

Conention:

• We will use \((\rho,\sigma,\tau)\) to denote an instance of the derivation system
• When there's no ambiguity what slot the operators belong to, we will write \(\rho(x) + \sigma(y) = \tau(z)\)

Observation: (very similar to Simultaneous Sylvester)

Same as before, \(\rho: R \times B \times C \rightarrowtail K \) can be thought of as an element \( \text{Hom}(R, B \otimes C) \). Similar for \(\sigma \in \text{Hom}(S, A \otimes C)\) and \(\tau \in \text{Hom}(T, A \otimes B)\)

Let us assume \(\rho,\sigma,\tau\) are all injective maps. Same as the Sylvester case, this can be done by preprocessing.

Terminology: We call the system \((\tilde{\rho}, \tilde{\sigma}, \tilde{\tau})\) restricted, and to differentiate, and call the full system \((\rho,\sigma,\tau)\) global.

Suppose we have subspaces \(A' \leq A, B' \leq B, C' \leq C\) such that

• \( \tilde{\rho} \coloneqq ((\pi_{B'} \otimes \pi_{C'})\circ r) \in \text{Hom}(R, B' \otimes C') \) is still injective.
• \( \tilde{\sigma} \coloneqq ((\pi_{A'} \otimes \pi_{C'})\circ s) \in \text{Hom}(S, A' \otimes C') \) is still injective.
• \( \tilde{\tau} \coloneqq ((\pi_{A'} \otimes \pi_{B'})\circ t) \in \text{Hom}(T, A' \otimes B') \) is still injective.

We solve the restricted system for \(\tilde{x}: A' \rightarrow R, \tilde{y}: B' \rightarrow S, \tilde{z}: C' \rightarrow T \). This means we want the equation \(\tilde{\rho}(\tilde{x}) + \tilde{\sigma}(\tilde{y}) = \tilde{\tau}(\tilde{z})\) to be satisfied.

Choose complements \(A = A' \oplus U\), \(B = B' \oplus V\), and \(C = C' \oplus W\)

Projecting to \(U \leq A\), \(B' \leq B\), and \(C' \leq C\) for \(r,s,t\) results in the equation \( \rho_{R \times B'\times C'}(x_U) + \sigma_{A' \times S \times C'}(\tilde{y}) = \tau_{A' \times B' \times T}(\tilde{z}) \)

Notice the first term is the only one with unknowns (being \(x_U\)) and \(\rho_{R \times B \times C'}: R \times B' \times C' \rightarrowtail K \cong \text{Hom}(R, B'\otimes C')\) is left invertible

Hence solving for \(x_U\) does not require solving a linear system, but rather, just an application of the left inverse, \(\rho_{R \times B' \times C'}^{\#}\), to all terms.

Projecting to \(A' \leq A\), \(V \leq B\), \(C' \leq C\) finds us \(y_V\) without solving a system of equations.

And projecting to \(A' \leq A, B' \leq B, W \leq C\) finds us \(z_W\) without solving a system of equations.

We call \(x \coloneqq \tilde{x} \oplus x_U\), \(y \coloneqq \tilde{y} \oplus y_V\), and \(z \coloneqq \tilde{z} \oplus z_W\) an extended candidate solution

So the final step is verifying the unchecked parts of our global system are satisfied by the extended candidate solution. Explicitly, we verify on the spaces

• \(A' \times V \times W\)
• \(U \times B' \times W\)
• \(U \times V \times C'\)
• \(U \times V \times W\)

Notation: Denote the direct sum of these spaces as \(M\)

What if the restricted system \( (\tilde{\rho},\tilde{\sigma},\tilde{\tau}) \) is already inconsistent?

Then we can immediately report the overall system \( (\rho,\sigma,\tau) \) is inconsistent, as any solution of the overall system must also solve the restricted one.

On the other hand, if the restricted system has a solution, it will not be unique. So we handle it similarly to the homogeneous case of the Sylvester system.

We will need to search for a solution in the solution space of \( (\tilde{\rho}, \tilde{\sigma}, \tilde{\tau}) \) that extends to a global solution of \((\rho,\sigma,\tau)\)

The correctness of this method is because the the global system \((\rho,\sigma,\tau)\) has a superset of the constraints of the restricted system \( (\tilde{\rho}, \tilde{\sigma}, \tilde{\tau}) \)

Same as the Sylvester case, let \(x \coloneqq \tilde{x} \oplus x_U\), \(y \coloneqq \tilde{y} \oplus y_V\), and \(z \coloneqq \tilde{z} \oplus z_W\) be the extended candidate solution

Define the residual \(\mathcal{E}_{\tilde{x}, \tilde{y}, \tilde{z}}: M \rightarrowtail K\) as \( \mathcal{E}_{\tilde{x},\tilde{y}, \tilde{z}} = \rho(x) + \sigma(y) - \tau(z)\)

We want an extended candidate solution \((\tilde{x}, \tilde{y},\tilde{z})\) whose residual \(\mathcal{E}_{\tilde{x},\tilde{y},\tilde{z}}\) is zero.

Suppose the solution space of \((\tilde{\rho},\tilde{\sigma},\tilde{\tau})\) is given by a vector subspace of dimension \(d\), and \((\tilde{x}_i, \tilde{y}_i, \tilde{z}_i)_{i=1}^{d}\) is a basis of solutions to the restricted system.

To set this up, we need the notation: \( \mathcal{E}_i \coloneqq \mathcal{E}_{\tilde{x}_i,\tilde{y}_i, \tilde{z}_i}  \). We'll define \(x_i \coloneqq \tilde{x}_i \oplus x_{i,U}\), \(y_i \coloneqq \tilde{y}_i \oplus y_{i,V}\), and \(z_i \coloneqq \tilde{z}_i \oplus z_{i,W} \) as the extended candidate solution corresponding to each basis element.

Now our problem is to solve for \(\lambda_i\) such that \(\sum_i \lambda_i \mathcal{E}_{i} = 0 \). This is a linear algebra problem with \(M\) constraints and \(d\) variables. The core idea of this approach to solving the Derivation System is that d will frequently be small relative to the initial number of variables \(ar + bs + ct\)

A successful solve means the extended candidate solution, \(x \coloneqq \tilde{x} \oplus x_U\), \(y \coloneqq \tilde{y} \oplus y_V\), and \(z \coloneqq \tilde{z} \oplus z_W\) must be global solution to the system \((\rho,\sigma,\tau)\).

For \((\lambda_i)_{i=1}^{d}\), define \((\tilde{x}(\lambda), \tilde{y}(\lambda), \tilde{z}(\lambda))\) as \((\sum_i \lambda_i \tilde{x}_i, \sum_i \lambda_i \tilde{y}_i, \sum_i \lambda_i \tilde{z}_i )\). We solve for \((\lambda_i)\) giving a valid global solution

Solving the Derivation System in coordinates

Let \([\rho] \in K^{bc \times r}, [\sigma] \in K^{ac \times s}, [\tau]\in K^{ab \times t}\) be the input to the Derivaiton system with the appropriate permutations applied to them.

The naive \(abc\) rows by \(ar + bs + ct\) variables system is laid out by \( (I_a \otimes [\rho]) \text{Vec}(X) + \Pi_{a,b}(I_b \otimes [\sigma]) \text{Vec}(Y) = ([\tau] \otimes I_c) \text{Vec}(Z)\), where \( \Pi_{a,b}\) permutes an element in \(K^{bac \times bs}\) to \(K^{abc \times bs}\).

TODO: Old graphic, should replace A,B,C with \([\rho], [\sigma], [\tau] \)

Average case model - we should be able to assume for random subspaces \(A' \leq A\), \(B' \leq B\) and \(C' \leq C\) of dimension \(a', b', c'\) respectively that

• If \( a'c' \geq r \), then \( \tilde{\rho} \in \text{Hom}(R, A' \otimes C') \) is generically injective
• If \( b'c' \geq s \), then \( \tilde{\sigma} \in \text{Hom}(S, B' \otimes C') \) is generically injective
• If \( a'b' \geq t \), then \( \tilde{\tau} \in \text{Hom}(T, A' \otimes B') \) is generically injective
• (TODO: need more justification/thinking) if \(a'b'c' \geq a'r + b's + c't \), then all restricted solutions extend to global solutions

Costs:

• Restricted system: Solve linear system with \(a'b'c'\) constraints, \(a'r + b's + c't\) variables, costing \(O((a'b'c')(a'r+b's + c't)^2)\)
• Left inverses: Compute for \(b'c' \times r\), \(a'c' \times s\), and \(a'b' \times t\) matrix. This can be done by RREF, costing \(O(b'c'r^2 + a'c's^2 + a'b't^2)\)
• Computing extended solution: TODO: These need more details/pictures and split to multiple slides 
  • To solve for \(x_U\):
    • Multiply \(uc' \times s\) matrix by \(s \times b'\) matrix: \(O(uc'sb')\), or \(O(b'c'su)\) alphabetized
    • Multiply \(ub' \times t\) matrix by \(t \times c'\) matrix: \( O(ub'tc') \), or \( O(b'c'tu)\) alphabetized
    • Subtract one \(b'c' \times u\) matrix from another twice: \(O(b'c'u)\)
    • Multiply \(r \times b'c' \) matrix by \(b'c' \times u\) matrix: \(O(rb'c'u)\), or \(O(b'c'ru)\) alphabetized
    • Summary: \(O(b'c'u(r+s+t) )\)
  • To solve for \(y_V\):
    • Multiply \(vc' \times r\) matrix by \(r \times a'\) matrix: \( O(vc'ra')\), or \(O(a'c'rv)\) alphabetized
    • Multiply \(a'v \times t\) matrix by \(t \times c'\) matrix: \( O(a'vtc')\), or \(O(a'c'tv)\) alphabetized 
    • Subtract one \(a'c' \times v\) matrix from another twice: \(O(a'c'v)\)
    • Multiply \(s \times a'c' \) matrix by \(a'c' \times v\) matrix: \(O(sa'c'v)\), or \(O(a'c'sv)\) alphabetized
    • Summary: \(O(a'c'v(r+s+t) )\)
  • To solve for \(z_W\):
    • Multiply \(b'w \times r\) matrix by \(r \times a'\) matrix: \( O(b'wra')\), or \(O(a'b'rw)\) alphabetized
    • Multiply \(a'w \times s\) matrix by \(s \times b'\) matrix: \( O(a'wsb') \), or \( O(a'b'sw) \) alphabetized
    • Subtract one \(a'b' \times w\) matrix from another twice: \(O(a'b'w)\)
    • Multiply \(t \times a'b' \) matrix by \(a'b' \times w\) matrix: \(O(ta'b'w)\), or \(O(a'b'tw)\) alphabetized
    • Summary: \(O(a'b'w(r+s+t)) \)
  • Overall Summary: \(O((a'b'c'(a'r+b's+c't)^2) + (b'c'r^2 + a'c's^2 + a'b't^2) + (b'c'u+a'c'v+a'b'w)(r+s+t)) \)
• Find restricted solution that extends to global solution: No cost by our assumption.

Full verification omitted - randomized check gives expoentially low probability of error, but to be analyzed.

Restricted system

Compute left inverses

Extend solution

Total: \(O((a'b'c'(a'r+b's+c't)^2) + (b'c'r^2 + a'c's^2 + a'b't^2) + (b'c'u+a'c'v+a'b'w)(r+s+t)) \)

An example fast regime: when \( a,b,c,r,s,t \) are all \(O(n)\), meaning \(a',b',c'\) are \(O(\sqrt{n})\)

Plug in \(n\) for \(a,b,c,r,s,t\), and \(\sqrt{n}\) for \(a',b',c'\), and we get the total as \(O((\sqrt{n}\sqrt{n}\sqrt{n}(\sqrt{n}n+\sqrt{n}n+\sqrt{n}n)^2) + (\sqrt{n}\sqrt{n}n^2 + \sqrt{n}\sqrt{n}n^2 + \sqrt{n}\sqrt{n}n^2) + (\sqrt{n}\sqrt{n}n+\sqrt{n}\sqrt{n}n+\sqrt{n}\sqrt{n}n)(n+n+n)) \)

Compared to solving a system of \(n^3\) equations with \(n^2\) variables: \(O(n^3(n^2)^2) = O(n^7)\)

Total: \(O((a'b'c'(a'r+b's+c't)^2) + (b'c'r^2 + a'c's^2 + a'b't^2) + (b'c'u+a'c'v+a'b'w)(r+s+t)) \)

Simplifies to \(O(n^{1.5}(n^{1.5})^2 + (n^3 + n^3 + n^3) + (n^2 + n^2 + n^2)(n+n+n)\) = \(O(n^{4.5}) \)