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3.6: The Invertible Matrix Theorem

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    70200
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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.


    3.6: The Invertible Matrix Theorem is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Dan Margalit & Joseph Rabinoff via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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