# 3.6: The Invertible Matrix Theorem

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##### Objectives
1. 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:

1. $$A$$ is invertible.
2. $$A$$ has $$n$$ pivots.
3. $$\text{Nul}(A) = \{0\}$$.
4. The columns of $$A$$ are linearly independent.
5. The columns of $$A$$ span $$\mathbb{R}^n$$.
6. $$Ax=b$$ has a unique solution for each $$b$$ in $$\mathbb{R}^n$$.
7. $$T$$ is invertible.
8. $$T$$ is one-to-one.
9. $$T$$ 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:

1. invertible matrices, and
2. 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.

##### Note $$\PageIndex{2}$$: Other Conditions for Invertibility

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.

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

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.

##### Example $$\PageIndex{1}$$

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.

##### Example $$\PageIndex{2}$$

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.

##### Example $$\PageIndex{3}$$

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 $$2$$.

##### Example $$\PageIndex{4}$$

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.

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