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13. Characterizations of Invertible Matrices

Dated: 17-04-2025

Solving Linear Equations by Matrix Inversion

Theorem

If \(A^{-1}\) exists and \(b \in \mathbb R^n\) then solution for \(Ax = b\) is

\[x = A^{-1}b\]

Theorem

If \(AB = I\) then \(B = A^{-1}\) and \(A = B^{-1}\).

Example

\[ \begin{align} &x_1 &+ 2x_2 &&+ 3x_3 &= 5\\ &2x_1 &+ 5x_2 &&+ 3x_3 &= 3\\ &x_1 & &&+ 8x_3 &= 17\\ \end{align} \]

We can rewrite it in the \(Ax = b\) form.

\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{bmatrix}, \quad x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, \quad b = \begin{bmatrix} 5 \\ 3 \\ 17 \end{bmatrix} \]
\[ \det(A) = 40 - 2(16 - 3) + 3(0 - 5) = 40 - 2(13) + 3(-5) = 40 - 26 - 15 = -1 \neq 0 \]

Therefore, \(A^{-1}\) exists.
Now our goal is

\[ \begin{bmatrix} A & I \end{bmatrix} \to \begin{bmatrix} I & A^{-1} \end{bmatrix} \]
\[ \left[ \begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 2 & 5 & 3 & 0 & 1 & 0 \\ 1 & 0 & 8 & 0 & 0 & 1 \end{array} \right] \]
\[\Big\downarrow R_2 = R_2 - 2R_1\]
\[\Big\downarrow R_3 = R_3 - R_1\]
\[ \left[ \begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & -3 & -2 & 1 & 0 \\ 0 & -2 & 5 & -1 & 0 & 1 \end{array} \right] \]
\[\Big\downarrow R_3 = R_3 + 2R_2\]
\[ \left[ \begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & -3 & -2 & 1 & 0 \\ 0 & 0 & -1 & -5 & 2 & 1 \end{array} \right] \]
\[\Big\downarrow R_3 = -R_3\]
\[ \left[ \begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & -3 & -2 & 1 & 0 \\ 0 & 0 & 1 & 5 & -2 & -1 \end{array} \right] \]
\[\Big\downarrow R_2 = R_2 + 3R_3\]
\[\Big\downarrow R_1 = R_1 - 3R_3\]
\[ \left[ \begin{array}{ccc|ccc} 1 & 2 & 0 & -14 & 6 & 3 \\ 0 & 1 & 0 & 13 & -5 & -3 \\ 0 & 0 & 1 & 5 & -2 & -1 \end{array} \right] \]
\[\Big\downarrow R_1 = R_1 - 2R_2\]
\[ \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & -40 & 16 & 9 \\ 0 & 1 & 0 & 13 & -5 & -3 \\ 0 & 0 & 1 & 5 & -2 & -1 \end{array} \right] \]
\[ A^{-1} = \begin{bmatrix} -40 & 16 & 9 \\ 13 & -5 & -3 \\ 5 & -2 & -1 \end{bmatrix} \]
\[ x = A^{-1}b = \begin{bmatrix} -40 & 16 & 9 \\ 13 & -5 & -3 \\ 5 & -2 & -1 \end{bmatrix} \begin{bmatrix} 5 \\ 3 \\ 17 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} \]
\[\therefore x_1 = 1, x_2 = -1, x_3 = 2\]
Applicability

It is only applicable when \(A\) is a square matrix1 and fails if \(A^{-1}\) doesn't exist.

Theorem

If \(Ax = 0\) is a homogeneous linear system2 then \(A^{-1}\) exists only iff this system has trivial solution.

Theorem

  • \(A\) is an invertible matrix.3
  • \(A \sim I\).
  • \(A\) has \(n\) pivot positions.
  • The equation \(A\vec x = \vec 0\) has only the trivial solution.
  • The columns of \(A\) form a linearly independent set.4
  • The linear transformation5 \(\vec{x} \to A\vec{x}\) is one-to-one.
  • The equation \(A\vec{x} = \vec{b}\) has at least one solution for each \(\vec{b} \in\mathbb{R}^n\).
  • The columns of \(A\) span6 \(\mathbb{R}^n\).
  • The linear transformation5 \(\vec{x} \to A\vec{x}\) maps \(\mathbb{R}^n\) onto \(\mathbb{R}^n\).
  • There is an \(n \times n\) matrix \(C\) such that \(CA = I\).
  • There is an \(n \times n\) matrix \(D\) such that \(AD = I\).
  • \(A^T\) is an invertible matrix.3

Solving Multiple Linear Systems with a Common Coefficient Matrix

\[ Ax_1 = b_1, \quad Ax_2 = b_2, \quad \dots, \quad Ax_k = b_k \]

If \(A^{-1}\) exists then

\[ x_1 = A^{-1}b_1, \quad x_2 = A^{-1}b_2, \quad \dots, \quad x_k = A^{-1}b_k \]

Find a Matrix1 from Linear Transformation5

Example

Let \(L\) be a linear transformation5 from \(\mathbb R^2 \to \mathbb P_2\) (\(\mathbb P_2\) is set7 of polynomials) defined

\[T(x, y) = xyt + (x + y)t^2\]

Find the matrix1 representing \(T\) with respect to the standard bases.

Solution

Let \(A = \{(1, 0), (0, 1)\}\) be the basis of \(\mathbb R^2\) then

\[T(1, 0) = (1)(0)t + (1 + 0)t^2 = t^2 = (0, 0, 1) \quad \because t^2 = 0 \times 1 + 0 \times t + 1 \times t^2\]
\[T(0, 1) = (0)(1)t + (0 + 1)t^2 = t^2 = (0, 0, 1) \quad \because t^2 = 0 \times 1 + 0 \times t + 1 \times t^2\]
\[A = \begin{bmatrix} 0 & 0\\ 0 & 0\\ 1 & 1 \end{bmatrix} \]

Invertible Linear Transformations

Let \(T\) be a linear transformation5

\[T: \mathbb R^n \to \mathbb R^n\]

And \(A\) be the matrix1 associated with \(T\) then

\[T^{-1} \text{ exists} \iff A^{-1} \text{ exists}\]

Also if \(S(\vec x) = A^{-1} \vec x\) then

\[S(T(x)) = x \forall x \in \mathbb R^n\]
\[T(S(x)) = x \forall x \in \mathbb R^n\]
\(T^{-1}\) is unique.

Proposition

Let \(T: \mathbb R^n \to \mathbb R^m\) be a linear transformation5 \(T(\vec x) = A_{m \times n} \vec x, \forall x \in \mathbb R^n\).
\(T^{-1}\) exists if \(y = A \vec x\) has a unique solution.

Case 1

If \(m < n\) then \(y = A \vec x\) has either no solution or infinitely many solutions for any \(y \in \mathbb R^m\).
Therefore, \(y = A \vec x\) is not invertible.3

Case 2

If \(m = n\), then the system \(A \vec x = y\) has a unique solution if and only if \(\text{Rank} (A) = n\).

Case 3

If \(m > n\), then the transformation5 \(y = A \vec x\) is non-invertible3 because we can find a vector8 \(y \in \mathbb R^m\) such that \(y = A \vec x\) is inconsistent.

References

Read more about notations and symbols.

  1. Read more about matrices. ↩↩↩↩

  2. Read more about homogeneous linear system. ↩

  3. Read more about invertible matrices. ↩↩↩↩

  4. Read more about linear independence. ↩

  5. Read more about linear transformations. ↩↩↩↩↩↩↩

  6. Read more about spanning. ↩

  7. Read more about sets. ↩

  8. Read more about vectors. ↩