# 3.6: The Invertible Matrix Theorem

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*Theorem:*the invertible matrix theorem.

This section consists of a single important theorem containing many equivalent conditions for a matrix to be invertible. This is one of the most important theorems in this textbook. We will append two more criteria in Section 5.1.

Theorem \(\PageIndex{1}\): Invertible Matrix Theorem

Let \(A\) be an \(n\times n\) matrix, and let \(T\colon\mathbb{R}^n \to\mathbb{R}^n \) be the matrix transformation \(T(x) = Ax\). The following statements are equivalent:

- \(A\) is invertible.
- \(A\) has \(n\) pivots.
- \(\text{Nul}(A) = \{0\}\).
- The columns of \(A\) are linearly independent.
- The columns of \(A\) span \(\mathbb{R}^n \).
- \(Ax=b\) has a unique solution for each \(b\) in \(\mathbb{R}^n \).
- \(T\) is invertible.
- \(T\) is one-to-one.
- \(T\) is onto.

**Proof**-
\((1\iff 2)\text{:}\) The matrix \(A\) has \(n\) pivots if and only if its reduced row echelon form is the identity matrix \(I_n\). This happens exactly when the procedure in Section 3.5, Theorem 3.5.1, to compute the inverse succeeds.

\((2\iff 3)\text{:}\) The null space of a matrix is \(\{0\}\) if and only if the matrix has no free variables, which means that every column is a pivot column, which means \(A\) has \(n\) pivots. See Recipe: Compute a Spanning Set for a Null Space in Section 2.6.

\((2\iff 4,\,2\iff 5)\text{:}\) These follow from the Recipe: Checking Linear Independence in Section 2.5 and Theorem 2.3.1, in Section 2.3, respectively, since \(A\) has \(n\) pivots if and only if has a pivot in every row/column.

\((4+5\iff 6)\text{:}\) We know \(Ax=b\) has at least one solution for every \(b\) if and only if the columns of \(A\) span \(\mathbb{R}^n \) by Theorem 3.2.2 in Section 3.2, and \(Ax=b\) has at most one solution for every \(b\) if and only if the columns of \(A\) are linearly independent by Theorem 3.2.1 in Section 3.2. Hence \(Ax=b\) has exactly one solution for every \(b\) if and only if its columns are linearly independent and span \(\mathbb{R}^n \).

\((1\iff 7)\text{:}\) This is the content of Theorem 3.5.3 in Section 3.5.

\((7\implies 8+9)\text{:}\) See Proposition 3.5.2 in Section 3.5.

\((8\iff 4,\,9\iff 5)\text{:}\) See Theorem 3.2.2 in Section 3.2 and Theorem 3.2.1 in Section 3.2.

To reiterate, the invertible matrix theorem means:

Note \(\PageIndex{1}\)

There are two kinds of *square* matrices:

- invertible matrices, and
- non-invertible matrices.

For invertible matrices, all of the statements of the invertible matrix theorem are true.

For non-invertible matrices, all of the statements of the invertible matrix theorem are false.

The reader should be comfortable translating any of the statements in the invertible matrix theorem into a statement about the pivots of a matrix.

The following conditions are also equivalent to the invertibility of a square matrix \(A\). They are all simple restatements of conditions in the invertible matrix theorem.

- The reduced row echelon form of \(A\) is the identity matrix \(I_n\).
- \(Ax=0\) has no solutions other than the trivial one.
- \(\text{nullity}(A) = 0\).
- The columns of \(A\) form a basis for \(\mathbb{R}^n \).
- \(Ax=b\) is consistent for all \(b\) in \(\mathbb{R}^n \).
- \(\text{Col}(A) = \mathbb{R}^n .\)
- \(\dim\text{Col}(A) = n.\)
- \(\text{rank}(A) = n.\)

Now we can show that to check \(B = A^{-1}\text{,}\) it's enough to show \(AB=I_n\) *or* \(BA=I_n\).

Corollary \(\PageIndex{1}\): A Left or Right Inverse Suffices

Let \(A\) be an \(n\times n\) matrix, and suppose that there exists an \(n\times n\) matrix \(B\) such that \(AB=I_n\) *or* \(BA=I_n\). Then \(A\) is invertible and \(B = A^{-1}\).

**Proof**-
Suppose that \(AB = I_n\). We claim that \(T(x)=Ax\) is onto. Indeed, for any \(b\) in \(\mathbb{R}^n \text{,}\) we have

\[ b = I_nb = (AB)b = A(Bb), \nonumber \]

so \(T(Bb) = b\text{,}\) and hence \(b\) is in the range of \(T\). Therefore, \(A\) is invertible by the Theorem \(\PageIndex{1}\). Since \(A\) is invertible, we have

\[ A^{-1} = A^{-1} I_n = A^{-1} (AB) = (A^{-1} A)B = I_n B = B, \nonumber \]

so \(B = A^{-1}.\)

Now suppose that \(BA = I_n\). We claim that \(T(x) = Ax\) is one-to-one. Indeed, suppose that \(T(x) = T(y)\). Then \(Ax = Ay\text{,}\) so \(BAx = BAy\). But \(BA = I_n\text{,}\) so \(I_nx = I_ny\text{,}\) and hence \(x=y\). Therefore, \(A\) is invertible by the Theorem \(\PageIndex{1}\). One shows that \(B = A^{-1}\) as above.

We conclude with some common situations in which the invertible matrix theorem is useful.

Is this matrix invertible?

\[ A = \left(\begin{array}{ccc}1&2&-1\\2&4&7\\-2&-4&1\end{array}\right) \nonumber \]

**Solution**

The second column is a multiple of the first. The columns are linearly dependent, so \(A\) does not satisfy condition 4 of the Theorem \(\PageIndex{1}\). Therefore, \(A\) is not invertible.

Let \(A\) be an \(n\times n\) matrix and let \(T(x) = Ax\). Suppose that the range of \(T\) is \(\mathbb{R}^n \). Show that the columns of \(A\) are linearly independent.

**Solution**

The range of \(T\) is the column space of \(A\text{,}\) so \(A\) satisfies condition 5 of the Theorem \(\PageIndex{1}\). Therefore, \(A\) also satisfies condition 4, which says that the columns of \(A\) are linearly independent.

Let \(A\) be a \(3\times 3\) matrix such that

\[ A\left(\begin{array}{c}1\\7\\0\end{array}\right) = A\left(\begin{array}{c}2\\0\\-1\end{array}\right). \nonumber \]

Show that the rank of \(A\) is at most \(2\).

**Solution**

If we set

\[ b = A\left(\begin{array}{c}1\\7\\0\end{array}\right) = A\left(\begin{array}{c}2\\0\\-1\end{array}\right), \nonumber \]

then \(Ax=b\) has multiple solutions, so it does not satisfy condition 6 of the Theorem \(\PageIndex{1}\). Therefore, it does not satisfy condition 5, so the columns of \(A\) do not span \(\mathbb{R}^3 \). Therefore, the column space has dimension strictly less than 3, the rank is at most \(2\).

Suppose that \(A\) is an \(n\times n\) matrix such that \(Ax=b\) is inconsistent some vector \(b\). Show that \(Ax=b\) has infinitely many solutions for some (other) vector \(b\).

**Solution**

By hypothesis, \(A\) does not satisfy condition 6 of the Theorem \(\PageIndex{1}\). Therefore, it does not satisfy condition \(3\text{,}\) so \(\text{Nul}(A)\) is an infinite set. If we take \(b=0\text{,}\) then the equation \(Ax=b\) has infinitely many solutions.