证明实矩阵可逆并推导其列向量张成ℝ³的技术问询

Alright, let's break this down step by step to tackle both parts of the problem: first proving the given matrix is invertible, then showing the three vectors span $\mathbb{R}^3$.

1. Prove Matrix $A$ is Invertible #

First, here's the matrix we're working with:

A = \begin{pmatrix} 2 & -2 & 1 \\ 1 & -2 & 1 \\ -2 & 3 & -1 \\ \end{pmatrix}

Option 1: Using the Determinant #

A square matrix is invertible if and only if its determinant is non-zero. Let's calculate $\det(A)$ by expanding along the first row:

$$
\begin{align*}
\det(A) &= 2 \cdot \det\begin{pmatrix}-2 & 1 \ 3 & -1\end{pmatrix} - (-2) \cdot \det\begin{pmatrix}1 & 1 \ -2 & -1\end{pmatrix} + 1 \cdot \det\begin{pmatrix}1 & -2 \ -2 & 3\end{pmatrix} \
&= 2 \cdot [(-2)(-1) - (1)(3)] + 2 \cdot [(1)(-1) - (1)(-2)] + 1 \cdot [(1)(3) - (-2)(-2)] \
&= 2 \cdot (2 - 3) + 2 \cdot (-1 + 2) + 1 \cdot (3 - 4) \
&= 2(-1) + 2(1) + 1(-1) \
&= -2 + 2 - 1 = -1
\end{align*}
$$

Since $\det(A) = -1 \neq 0$, matrix $A$ is invertible.

Option 2: Explicitly Finding Inverse Matrix $B$ (as requested) #

To find $B$ such that $AB = I_3$, we'll use row operations on the augmented matrix $[A | I_3]$:

Start with the augmented matrix:

\left(\begin{array}{ccc|ccc}
2 & -2 & 1 & 1 & 0 & 0 \\
1 & -2 & 1 & 0 & 1 & 0 \\
-2 & 3 & -1 & 0 & 0 & 1
\end{array}\right)

1. Swap Row 1 and Row 2 to get a leading 1 in the top-left corner:

   \left(\begin{array}{ccc|ccc}
   1 & -2 & 1 & 0 & 1 & 0 \\
   2 & -2 & 1 & 1 & 0 & 0 \\
   -2 & 3 & -1 & 0 & 0 & 1
   \end{array}\right)
   

2. Row 2 = Row 2 - 2*Row 1 and Row 3 = Row 3 + 2*Row 1 to eliminate the first column below the leading 1:

   \left(\begin{array}{ccc|ccc}
   1 & -2 & 1 & 0 & 1 & 0 \\
   0 & 2 & -1 & 1 & -2 & 0 \\
   0 & -1 & 1 & 0 & 2 & 1
   \end{array}\right)
   

3. Swap Row 2 and Row 3, then multiply Row 2 by -1 to get a leading 1:

   \left(\begin{array}{ccc|ccc}
   1 & -2 & 1 & 0 & 1 & 0 \\
   0 & 1 & -1 & 0 & -2 & -1 \\
   0 & 2 & -1 & 1 & -2 & 0
   \end{array}\right)
   

4. Row 3 = Row 3 - 2*Row 2 to eliminate the second column below the leading 1:

   \left(\begin{array}{ccc|ccc}
   1 & -2 & 1 & 0 & 1 & 0 \\
   0 & 1 & -1 & 0 & -2 & -1 \\
   0 & 0 & 1 & 1 & 2 & 2
   \end{array}\right)
   

5. Back-substitute to get zeros above the leading 1s:

   • Row 2 = Row 2 + Row 3
   • Row 1 = Row 1 - Row 3
   • Row 1 = Row 1 + 2*Row 2

   This gives us the reduced row-echelon form:

   \left(\begin{array}{ccc|ccc}
   1 & 0 & 0 & 1 & -1 & 0 \\
   0 & 1 & 0 & 1 & 0 & 1 \\
   0 & 0 & 1 & 1 & 2 & 2
   \end{array}\right)
   

The right half of the augmented matrix is our inverse $B$:

B = \begin{pmatrix} 1 & -1 & 0 \\ 1 & 0 & 1 \\ 1 & 2 & 2 \\ \end{pmatrix}

Let's verify $AB = I_3$:
$$
AB = \begin{pmatrix} 2 & -2 & 1 \ 1 & -2 & 1 \ -2 & 3 & -1 \ \end{pmatrix} \begin{pmatrix} 1 & -1 & 0 \ 1 & 0 & 1 \ 1 & 2 & 2 \ \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \ \end{pmatrix} = I_3
$$

Perfect, this confirms $A$ is invertible.

2. Show the Given Vectors Span $\mathbb{R}^3$ #

The three vectors in question are exactly the column vectors of matrix $A$:
$$
\mathbf{v}_1 = \begin{pmatrix} 2 \ 1 \ -2 \ \end{pmatrix}, \quad \mathbf{v}_2 = \begin{pmatrix} -2 \ -2 \ 3 \ \end{pmatrix}, \quad \mathbf{v}_3 = \begin{pmatrix} 1 \ 1 \ -1 \ \end{pmatrix}
$$

To prove they span $\mathbb{R}^3$, we need to show that any vector $\mathbf{b} \in \mathbb{R}^3$ can be written as a linear combination of $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$. In matrix terms, this means the equation $A\mathbf{x} = \mathbf{b}$ has a solution for every $\mathbf{b} \in \mathbb{R}^3$.

Since we already proved $A$ is invertible, we can explicitly write the solution as:
$$
\mathbf{x} = A^{-1}\mathbf{b}
$$

This solution exists (and is unique) for every $\mathbf{b} \in \mathbb{R}^3$. That means every vector in $\mathbb{R}^3$ can be expressed as a linear combination of $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$, so they span $\mathbb{R}^3$.

As an extra check: since $A$ is invertible, its rank is 3 (equal to the dimension of $\mathbb{R}^3$). A set of 3 linearly independent vectors in $\mathbb{R}^3$ must span the entire space—this is another way to see the result.

内容的提问来源于stack exchange，提问作者Simbörg