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3.6 Invertible Matrix Theorem

Invertible Matrix Theorem (IMT). Let \(A\) be an \(n\times n\) matrix, and let \(T:\mathbb R^n\rightarrow \mathbb R^n\) be the matrix transformation \(T(\vec x)=A\vec x\). The following statements are equivalent:

  1. \(A\) is invertible.

  2. \(A\) has \(n\) pivots.

  3. \(\text{Null}(A)={\vec 0}\).

  4. The columns of \(A\) are linearly independent.

  5. The columns of \(A\) span \(\mathbb R^n\).

  6. \(A\vec x=\vec b\) has a unique solution for each \(b\) in \(\mathbb R^n\).

  7. The linear transformation \(T\) is invertible, specifically:

    • \(T\) is one-to-one.

    • \(T\) is onto.

Using the IMT

We use the IMT all the time. Is a set of vectors a basis for \(\mathbb R^n\)? Combine them into a matrix and check to see if the matrix is invertible. Is the linear transformation one-to-one? Verify the null space contains only the \(\vec 0\). Does \(A\) have an inverse? Row reduce and check how many pivots the matrix has.

Examples

Example 1

Determine if the linear transformation \(T\) is one-to-one. Is it onto? The standard matrix associated with \(T\) is

\[\begin{split}A = \left[\begin{array}{rrrr}0&-2&-3&9\\0&0&0&1\\-2&2&5&-15\\1&-1&-2&7\\\end{array}\right]\end{split}\]
A = [0 -2 -3 9 ; 0 0 0 1 ; -2 2 5 -15 ; 1 -1 -2 7 ];
rref(A)

ans =

     1     0     0     0
     0     1     0     0
     0     0     1     0
     0     0     0     1

With 4 pivots, we know the dimension of the null is zero and thus that the transformation is both one-to-one and onto (invertible).

Example 2

Is the set of vectors a basis for \(\mathbb R^5\)?

\[\begin{split}\left\{\vec v_1 =\left[\begin{array}{r}1\\-2\\2\\-1\\3\\\end{array}\right],\vec v_2=\left[\begin{array}{r}2\\0\\5\\4\\5\\\end{array}\right],\vec v_3=\left[\begin{array}{r}-2\\-1\\3\\3\\4\\\end{array}\right],\vec v_4=\left[\begin{array}{r}2\\1\\2\\4\\3\\\end{array}\right],\vec v_5=\left[\begin{array}{r}2\\0\\3\\1\\2\\\end{array}\right]\right\}\end{split}\]

We create the matrix \(B = [\vec b_1,\vec b_2, \vec b_3 , \vec b_4 , \vec b_5 ]\) and row reduce.

B = [1 2 -2 2 2 ; -2 0 -1 1 0 ; 2 5 3 2 3 ; -1 4 3 4 1 ; 3 5 4 3 2 ];
inv(B)

ans =

    1.0000   -4.0000    0.0000    2.0000   -2.0000
  -25.0000  110.0000   -7.0000  -61.0000   66.0000
   10.0000  -45.0000    3.0000   25.0000  -27.0000
   12.0000  -52.0000    3.0000   29.0000  -31.0000
   23.0000 -101.0000    7.0000   56.0000  -61.0000

Since \(B\) is invertible, we know the columns are linearly independent and thus form a basis for \(\mathbb R^5\).

3.5.1 Lab 6 Chapter 4