MATH 240

Mon. September 30th, 2019


Section 2.3 – Invertible Matrix Theorem

Let AA be a n×nn\times n (square) matrix. The following are equivalent (TFAE):

  1. AA is invertible.
  2. AA is row equivalent to the identity matrix.
  3. AA has nn pivot positions.
  4. The equation Ax=0Ax=0 has only the trivial solution.
  5. The columns of AA are linearly independent.
  6. The linear transformation xAxx \rightarrow Ax is one-to-one.
  7. The equation Ax=bAx=b has at least one solution for each bb in Rn\R^n.
  8. The columns of AA span Rn\R^n.
  9. The linear transformation xAxx \rightarrow Ax maps Rn\R^n onto Rn\R^n.
  10. There is an n×nn\times n matrix CC such that CA=ICA=I.
  11. There is an n×nn\times n matrix DD such that AD=IAD=I.
  12. ATA^T is an invertible matrix.

Remember – the fact that these statements are equivalent means that if one of the is true, all of them are true, and if one of them is false, all of them are false.


Application


  1. Suppose that AA and BB are square matrices, and AB=IAB=I.
    Now we know that AA is invertible and A1=BA^{-1}=B, and that BB is invertible and B1=AB^{-1}=A.

A=(100011111).A=\begin{pmatrix} 1&0&0\\0&1&1\\1&1&1 \end{pmatrix}. Is AA invertible?
(100011111)(100011000)\begin{pmatrix} 1&0&0\\0&1&1\\1&1&1 \end{pmatrix}\rightarrow \begin{pmatrix} 1&0&0\\0&1&1\\0&0&0 \end{pmatrix}
No, because it only has two pivot positions.


A=(102312519).A=\begin{pmatrix} 1&0&-2\\3&1&-2\\-5&-1&9 \end{pmatrix}. Is AA invertible?
(102312519)(102014011)(102014003)\begin{pmatrix} 1&0&-2\\3&1&-2\\-5&-1&9 \end{pmatrix} \rightarrow \begin{pmatrix} 1&0&-2\\0&1&4\\0&-1&-1 \end{pmatrix} \rightarrow \begin{pmatrix} 1&0&-2\\0&1&4\\0&0&3 \end{pmatrix}
Yes, because it has three pivot positions.


Invertible Functions

Definition.
Let T:RnRnT: \R^n \rightarrow \R^n be a linear transformation. TT is invertible if there is some linear transformation S:RnRnS: \R^n \rightarrow \R^n such that T(S(x))=xT(S(x))=x and S(T(x))=xS(T(x))=x. We usually denote SS by T1T^{-1}.

Theorem.
Let T:RnRnT: \R^n \rightarrow \R^n be a linear transformation, and let AA be the standard matrix for TT. TT is invertible if and only if AA is invertible.


True or False:

Let TT be a linear transformation from Rn\R^n to Rn\R^n.

  1. TT can be one-to-one but not onto.
    • False. Let AA be the standard matrix for TT. If TT is one-to-one, then TT is also onto by the 6th and 9th entry in the Invertible Matrix Theorem.
  2. TT can be onto but not one-to-one.
    • False. Similarly, if TT onto, then TT is also one-to-one by the same theorem.

Note that these are only false because we know TT is from Rn\R^n to Rn\R^n, i.e. its standard matrix is square.