对称可逆正元素n阶矩阵逆矩阵零元素个数证明及实例求解

Hey there, let's work through these two linear algebra problems step by step. I'll start with the proof about the inverse matrix's zero entries, then move on to computing the inverse for that specific patterned matrix.

First, let's recap the given info: $A$ is an $n \times n$ (where $n≥2$) symmetric invertible real matrix with all positive entries, and $z_n$ is the number of zero entries in $A^{-1}$. We need to show $z_n$ can't exceed $n^2 - 2n$.

Let's use a proof by contradiction. Suppose $z_n > n^2 - 2n$. That means the number of non-zero entries in $A^{-1}$ is less than $n^2 - (n^2 - 2n) = 2n$.

Now, remember that $AA^{-1} = I$, the identity matrix. For any row $i$ of $A$ and column $j$ of $A^{-1}$, their dot product equals $\delta_{ij}$ (1 if $i=j$, 0 otherwise).

Here's the key point: if any column of $A^{-1}$ had $n-1$ zero entries, that column would have exactly one non-zero entry (since $A^{-1}$ is invertible, no column can be all zeros). Let's say column $k$ of $A^{-1}$ has only $(A^{-1})_{m,k} \neq 0$.

Now, take the dot product of row $i$ (where $i≠k$) of $A$ with column $k$ of $A^{-1}$: this must equal 0. But that dot product simplifies to $A_{i,m} \cdot (A^{-1}){m,k} = 0$. Wait, $A{i,m}$ is positive (all entries of $A$ are positive) and $(A^{-1})_{m,k} ≠0$—their product can't be zero! That's a contradiction.

This means every column of $A^{-1}$ can have at most $n-2$ zero entries. Since there are $n$ columns, the total number of zero entries is at most $n(n-2) = n^2 - 2n$. Exactly what we needed to prove.

Let's first make sure we understand the structure of the given matrix $A$:

• Row 1 is all 1s: $A_{1,j} = 1$ for every $j$
• Row 2 has $A_{2,1}=1$, and $A_{2,j}=2$ for all $j≥2$
• For rows $k≥3$: we copy the previous row up to column $k-1$, then switch to 1 if $k$ is odd, 2 if even, and keep that value for all columns beyond $k$. So row 3 is $[1,2,1,1,...,1]$, row4 is $[1,2,1,2,...,2]$, row5 is $[1,2,1,2,1,...,1]$, and so on.

To find $A^{-1}$, we use the matrix multiplication rule $AB = I$ (where $B = A^{-1}$) and solve for each column of $B$. Let's walk through the pattern we find:

Step 1: Compute Column 1 of $B$ #

We need $\sum_{j=1}^n A_{i,j}b_{j1} = \delta_{i1}$.

• For $i=1$: Sum of all entries in column1 is 1.
• For $i=2$: $b_{11} + 2*(sum of entries from row2 to n) =0$.
  Solving these gives $b_{11}=2$, $b_{21}=-1$, and $b_{j1}=0$ for all $j≥3$.

Step 2: Compute Column 2 of $B$ #

Similarly, $\sum_{j=1}^n A_{i,j}b_{j2} = \delta_{i2}$.

• For $i=1$: Sum of all entries in column2 is 0.
• For $i=2$: $b_{12} +2*(sum of entries from row2 to n)=1$.
  Solving gives $b_{12}=-1$, $b_{32}=1$, and $b_{j2}=0$ for all $j≠1,3$.

Step3: Generalize for Columns $l≥3$ #

For any column $l≥3$, we find that the only non-zero entries are:

• $b_{l-1,l}=1$
• $b_{l+1,l}=-1$ (with adjustments for the last column: $b_{n,n}=1$, $b_{n-1,n}=-1$)

Counting Zero Entries #

From this pattern, we can count the non-zero entries in $B$:

• Row1 has 2 non-zeros: $b_{11}, b_{12}$
• Row2 has 2 non-zeros: $b_{21}, b_{23}$
• Each row $k$ from 3 to $n-1$ has 2 non-zeros: $b_{k,k-1}, b_{k,k+1}$
• Rown has 2 non-zeros: $b_{n,n-1}, b_{n,n}$

Total non-zero entries: $2 + 2 + 2*(n-2) = 2n$.

Therefore, the number of zero entries in $A^{-1}$ is $n^2 - 2n$ (since total entries are $n^2$).

Fun fact: This matrix's inverse achieves the upper bound we proved in the first problem!

内容的提问来源于stack exchange，提问作者Guy Fsone