Efficient Computation for Invertibility Sequence of Banded Toeplitz Matrices

Let matrices $P$ and $Q$ be defined as follows:

We have the following equation, where $I_{k}$ is the $n$-order identity matrix.

First, we prove the necessity: if $M_{n}$ is an invertible matrix, then when it is multiplied by a column full-rank matrix, the result is a column full-rank matrix. The rank of $P$ is $k$, so $P$ is invertible.

Next, we prove the sufficiency: if $W_{n}$ is invertible, then $P$ is also invertible. Multiplying both sides of the Eq. 5 by $P^{-1}$ on the right, there exist $k$ vectors $U_{n-k+1},U_{n-k+2},\cdots U_{n}$ that make:

At this point, we can construct a unique matrix $U=[U_{1},U_{2},\cdots,U_{n}$] such that $M_{n}U=I_{n}$, where for $i>n$, $U_{i}=0$ and $E_{i}$ is the vector of order $i$ of the canonical basis of $\mathbb{K}^{n}$.

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In this case, $k=1$, which means the invertibility determining matrix $W$ consists of a single element. We only need to determine whether it is zero. The cost of calculating $W_{1}$ to $W_{n}$ using Eq. 2 is $3n$. ∎

In this case, $k=2$, which means the invertibility determining matrix $W$ is a $2*2$ matrix. We only need to determine if the ratio of the two elements in each row of $W$ is equal to that in the next row, which costs $n$. The cost of calculating $W_{1}$ to $W_{n}$ using Eq. 2 is $10n$. Therefore, the total cost is $11n$. ∎

Calculating the invertibility sequence for banded Toeplitz matrices involves determining the invertibility of $W_{1},W_{2},\cdots W_{n}$, where each $W$ is a $k*k$ matrix. If Gaussian elimination is used for each matrix to row-echelon form, the time complexity could reach $O(k^{3}n)$. Since $k-1$ rows are the same between $W_{i}$ and $W_{i+1}$, this means that the computations performed on $W_{i}$ can be utilized for $W_{i+1}$. This reuse of calculations can optimize the overall process and reduce the computational cost.

Consider the initial situation, set $W_{0}=Y_{0}=I$, where $I$ is the identity matrix, a row-echelon form of a full-rank matrix. By removing the first row of $W_{0}$ and adding the bottom row of $W_{1}$, a new matrix $Y_{1}$ is formed. It can be seen that the invertibility of $W_{1}$ and $Y_{1}$ is the same. At this point, the first $k-1$ rows of $Y_{1}$ are quasi-row-echelon which can be transformed into row-echelon form through row swapping. Assuming that $index[a]$ is the index of the first non-zero element from the left in the $a$-th row of each $Y$, there are two scenarios:

• 1.

  If $\forall i\in\mathbb{Z},1\leq i\leq k$, $index[k]\neq index[i]$, this indicates that $Y_{i}$ is quasi-row-echelon. In this scenario, we can quickly determine the invertibility of $Y_{1}$, whether there exists a row that is entirely zero.

• 2.

  If $\exists i\in\mathbb{Z},1\leq i\leq k$, $index[k]=index[i]$, then use the $k$-th row to perform Gaussian elimination on the $i$-th row ($k>i$). After Gaussian elimination, $index[i]$ will grow and be updated. If at this point, $\exists j\in\mathbb{Z},1\leq j\leq k$, $index[j]=index[i]$, then use the lower row to perform Gaussian elimination on the upper one. Continue this process until $Y_{1}$ achieves a quasi-row-echelon form. Then, we can quickly determine the invertibility of $Y_{1}$.

We continuously remove the first row of $Y_{i}$ and add the last row of $W_{i+1}$ to construct $Y_{i+1}$. Since the first $k-1$ rows of $Y_{i+1}$ are already in quasi-row-echelon form, we need at most $k$ Gaussian eliminations to transform $Y_{i+1}$ into quasi-row-echelon form, rather than $k^{2}$ eliminations. Since we always use the larger-index rows to perform Gaussian elimination on the small-index ones during the Gaussian elimination, the $j$-th row of $Y_{i+1}$ is only linearly expressed by the $j$-th to $k$-th rows of $W_{i+1}$. Therefore, when we use $W_{i+1}$ and $Y_{i}$ to construct $Y_{i+1}$, we can remove the first row of $Y_{i}$ and ensure that $Y_{i+1}$ and $W_{i+1}$ have the same invertibility, as shown in Eq. 8, where $V_{l}$ is the $l$-th row of $W_{i}$ and function $L$ represents linear combination.

Algorithm 1 shows the calculations in detail.

The time cost analysis of this algorithm is as follows:

• 1.

  Step 1 (lines 1 to 17) is for initialization and involves no computation.

• 2.

  Step 2 (lines 19 to 24) calculates each element with a time cost of $2k+1$. Since there are $kn$ elements, the total time cost for this step is $2k^{2}n+kn$.

• 3.

  Step 3 (lines 25 to 36) requires at most $k$ Gaussian eliminations to transform each $Y$ into quasi-row-echelon form, with a time cost of $k$ for each cycle. As this while-loop is repeated $n$ times, the total cost for this step is $k^{2}n/2$.

In summary, the total time cost of the program is $5k^{2}n/2+kn$. This algorithm is more space-efficient: the size of $W$ is $2k*k$, and the size of $Y$ is $k*k$, therefore, the total space consumption of the algorithm is only $3k^{2}$.

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