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Understanding Linear Algebra: Data Science Edition

Randall Pruim

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Section 4.1 Invertibility

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Up to this point, we have used the Gaussian elimination algorithm to find solutions to linear systems, which is equivalent to solving matrix equations of the form Ax=b, where b is a vector of constants, A a matrix of coefficients, and x a vector of variables, or unknowns. We now investigate another way to find solutions to the equation Ax=b when the matrix A is square. To get started, let’s look at some familiar examples.
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Preview Activity 4.1.1.

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  1. 🔗
    Explain how you would solve the equation 3x=5 using multiplication rather than division.
  2. 🔗
    Find the 2×2 matrix A that rotates vectors counterclockwise by 90∘.
  3. 🔗
    Find the 2×2 matrix B that rotates vectors clockwise by 90∘.
  4. 🔗
    What do you expect the product AB to be? Explain the reasoning behind your expectation and then compute AB to verify it.
  5. 🔗
    Solve the equation Ax=[3−2] using Gaussian elimination.
  6. 🔗
    Explain why your solution may also be found by computing x=B[3−2].
Solution.
  1. Dividing by 3 is the same as multiplying by 13, the multiplicative inverse 13 of 3. We have
    13(3x)=135(133)x=531x=53x=53.
  2. As we have seen a few times, the matrix is A=[0−110].
  3. Here, the matrix is B=[01−10].
  4. We should expect that AB=I since the effect of rotating by 90∘ clockwise followed by rotating 90∘ counterclockwise is to leave a vector unchanged. We can verify this by performing the matrix multiplication.
  5. We have
    [0−1310−2]∼[10−201−3]
    so the solution is x=[−2−3].
  6. The equation Ax=[3−2] is asking us to find the vector that becomes [3−2] after being rotated by 90∘. If we rotate [3−2] by 90∘ in the opposite direction, it will have this property. That is, if x=B[−2−3], then
    Ax=A(Bx)=(AB)[3−2]=I[3−2]=[3−2].
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Subsection 4.1.1 Invertible matrices

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The preview activity began with a familiar type of equation, 3x=5, and asked for a strategy to solve it. One possible response is to divide both sides by 3. Instead, let’s rephrase this as multiplying by 3−1=13, the multiplicative inverse of 3.
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Now that we are interested in solving equations of the form Ax=b, we might try to find a similar approach. Is there a matrix A−1 that plays the role of the multiplicative inverse of A? Of course, the real number 0 does not have a multiplicative inverse so we probably shouldn’t expect every matrix to have a multiplicative inverse. We will see, however, that many do.
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Definition 4.1.1.

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Let A be an m×n matrix. If LA=In, then we call L a left inverse of A. If AR=Im, then we call R a right inverse of A.
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Activity 4.1.2.

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  1. 🔗
    If A is m×n and has a left inverse L, what shape must L be?
  2. 🔗
    If A is m×n and has a right inverse R, what shape must R be?
Answer.
In both cases, n×m, the reverse of the shape of A.
Solution.
In both cases, n×m, the reverse of the shape of A. For example, a left inverse L must have m columns because A has m rows. And the product LA will have n columns because A has n coloumns. Since an idenity matrix is square, the product also has n rows. This implies the L has n rows.
A similar argument shows that R must be n×m.
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Activity 4.1.3.

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  1. 🔗
    Show that if a matrix A has left inverse L and right inverse R, then L=R. Hint: Consider the product LAR.
  2. 🔗
    Show that if an m×n matrix A has right inverse R, then n≥m
  3. 🔗
    Show that if an m×n matrix A has left inverse L, then m≥n. Hint: What does LAx=Lb tell you about the equation Ax=b?
  4. 🔗
    Show that if an m×n matrix A has both a left and a right inverse, then m=n, so A is a square matrix.
Solution.
  1. If LA=I and AR=I, then
    LAR=(LA)R=IR=RLAR=L(AR)=LI=L
    so L=R
  2. If AR=I, then for any b∈Rm, ARb=Ib=b. This means that Ax=b is consistent for every b∈Rm. That means every row of the RREF of A has a pivot position. So there are m pivots. There can be at most one pivot in each column, so there must be at least as many columns as pivots: n≥m.
  3. If A has a left inverse L, then L has a right inverse (namely A). So L must have at least as many columns as rows, which means A must have at least as many rows as columns: m≥n
  4. Combine the two previous parts.
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Activity 4.1.3 tells us two important pieces of information about matrices that have both a left and a right inverse.
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This leads to the following definition
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Definition 4.1.3.

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An n×n square matrix A is called invertible if there is a matrix B that is both a left and a right inverse for A. That is BA=In and AB=In. The matrix B is called the inverse of A and denoted A−1.
If B is the inverse of A, then AB=I and BA=I. But this means that A is the inverse of B as well.
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We have seen that if a matrix has a left and a right inverse, then these must be the same matrix. In fact, more is true. For square matrices, any right inverse is also a left inverse, and any left inverse is also a right inverse.
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It is important to remember that the product of two matrices depends on the order in which they are multiplied. That is, if C and D are matrices, then it sometimes happens that CD≠DC, even if the matrices are square. So it is not immediate that RA=I just becuase AR=I. For matrices that are not square, AR and RA won’t even be the same shape, and at most one of the two products will be an identity matrix. So Proposition 4.1.5 establishes something special about one-sided inverses of square matrices.
Suppose A is n×n and has a right inverse R. Then for any b∈Rn, ARb=b. So Ax=b is consistent for every b. This implies that the RREF of A has a pivot in each row, and therefore also in each column. In other words, A∼I.
Notice that if Rx=Ry, then x=ARx=ARy=y. This means the matrix transformation associated with R is one-to-one, so the RREF for R has a pivot in every column, and therefore in every row. In other words, R∼I.
Now consider RAx for any x∈Rn. Because R∼I, there must be a y with Ry=x. So
RAx=RARy=Ry=x
But this means that RA is the identity matrix since the transformation associated with it is the identity transformation. In other words, RA=I and R is also a left inverse of A
Now suppose that L is a left inverse for A. That means A is a right inverse for L, and we just show that this means A is also a left inverse for L. So AL=I and L is a right inverse for A.
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Proposition 4.1.5 means that to show that matrix is invertible, it suffices to find a one-sided inverse -- it will always also be a full inverse.
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Example 4.1.6.

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Consider the following two matrices
A=[102221]B=[10−10.500].
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Then
AB=[1001]=I2BA=[10201−1.5000]≠I3
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So B is a right inverse of A, but B is not a left inverse of A.
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Now consider
B2=[−3−423.522].
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Multiplication shows that
AB2=I2,
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So A has more than one right inverse. In fact, there are infinitely many and none of them can be a left inverse, since if L had both a left and a right inverse, they would have to be the same by Proposition 4.1.2. But that means a left inverse would have to be the same as both B and B2, which is not posssible.
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Example 4.1.7.

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Suppose that A is the matrix that rotates two-dimensional vectors counterclockwise by 90∘ and that B rotates vectors by −90∘. We have
A=[0−110],   B=[01−10].
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We can check that
AB=[0−110][01−10]=[1001]=I
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which shows that B is a right inverse of A.
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If we multiply the matrices in the opposite order, we find that
BA=[01−10][0−110]=[1001]=I,
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which says that (a) B is also the left inverse of A, and (b) A is the inverse of B as well. Inverses always come in pairs like this. If A is invertible with inverse B, then B is also invertible and its inverse is A. In other words, A and B are inverses of each other.
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If we think about the matrix transformations associated with A and B, this makes geometric sense. Rotating clockwise "undoes" a counterclockwise rotation. Rotating counterclockwise "undoes" a clockwise rotation.
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Subsection 4.1.2 Solving equations with an inverse

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If A is an invertible matrix, then for any b∈Rn,
A(A−1b)=(AA−1)b=b,
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so Ax=b is consistent for every b. Furthermore,
Ab=\xvcec⇒A−1Ab=A−1x⇒b=Ax,
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so solutions are unique.
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This result is import enough to capture in a proposition.
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Notice that this is similar to saying that the solution to 3x=5 is x=13⋅5, as we saw in the preview activity.
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Proposition 4.1.8 also implies that A∼I for any invertible matrix A since A must have a pivot position in each row and in each column.
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You may have noticed that Proposition 4.1.8 says that the solution to the equation Ax=b is x=A−1b. Indeed, we know that this equation has a unique solution because A has a pivot position in every column.
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Proposition 4.1.8 shows us how to use A−1 to solve equations of the form Ax=b. In Subsection 4.1.3 we will learn how to find an inverse for any square matrix that has an inverse. But before we get to that, let’s see some examples of how to use an inverse once we have it.
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Activity 4.1.4.

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We’ll begin by considering the square matrix
A=[102221111].
  1. 🔗
    Describe the solution space to the equation Ax=[343] by augmenting A and finding the reduced row echelon form.
  2. 🔗
    Show that
    A−1=[12−4−1−130−12]
  3. 🔗
    Now use A−1 to solve the equation Ax=[343] and verify that your result agrees with what you found in part a.
  4. 🔗
    We’ll learn a way to compute the inverse of a matrix shortly. NumPy knows how to do this, of course. If you have defined a matrix B in Python, you can find it’s inverse as np.linalg.inv(B). Use Python to find the inverse of the matrix
    B=[1−2−1−1565−46]
    and use the inverse to solve the equation Bx=[8336].
  5. 🔗
    If A and B are the two matrices defined in this activity, compute their product AB.
  6. 🔗
    Compute the products A−1B−1 and B−1A−1. One of these is (AB)−1. Which one?
  7. 🔗
    Explain your finding by considering the product
    (AB)(B−1A−1)
    and using associativity to regroup the products so that the middle two terms are multiplied first.
Solution.
  1. We can demonstrate that the matrix is an inverse by multiplication.
    [12−4−1−130−12][102221111]=I3.
    It suffices to show we have a left inverse, but you can multiply the other way around to see that it also a right inverse.
  2. We see that A−1[343]=[−122].
  3. Python tells us that B−1[8336]=[4−12].
  4. Python helps us see that (AB)∼I, which tells us that AB is invertible.
  5. We find that (AB)−1=B−1A−1.
  6. We see that
    (AB)(B−1A−1)=A(BB−1)A−1=AIA−1=AA−1=I.
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Subsection 4.1.3 Finding inverses

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Now that we have seen one use of inverses -- solving equations of the form Ax=b -- we turn our attention to the task of computing the inverse of a matrix, or demonstrating that no inverse exists.
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Activity 4.1.5.

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This activity demonstrates a procedure for finding the (right) inverse of a matrix A.
  1. 🔗
    Suppose that A=[3−21−1]. To find a right inverse B, we write its columns as B=[b1b2] and require that
    AB=I[Ab1Ab2]=[1001].
    In other words, we can find the columns of B by solving the equations
    Ab1=[10],Ab2=[01].
    Solve these equations to find b1 and b2. Then write the matrix B and verify that AB=I.
  2. 🔗
    By Proposition 4.1.5 this is enough for us to conclude that B is the inverse of A. But le’ts compute the product BA just to confirm.
  3. 🔗
    What happens when you try to find the inverse of C=[−214−2]?
  4. 🔗
    We now develop a condition that must be satisfied by an invertible matrix. Suppose that A is an invertible n×n matrix with inverse B and suppose that b is any n-dimensional vector. Since AB=I, we have
    A(Bb)=(AB)b=Ib=b.
    This says that the equation Ax=b is consistent and that x=Bb is a solution.
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    Since we know that Ax=b is consistent for any vector b, what does this say about the span of the columns of A?
  5. 🔗
    Since A is a square matrix, what does this say about the pivot positions of A? What is the reduced row echelon form of A?
  6. 🔗
    In this activity, we have studied the matrices
    A=[3−21−1],C=[−214−2].
    Find the reduced row echelon form of each and explain how those forms enable us to conclude that one matrix is invertible and the other is not.
Solution.
  1. Solving the two equations for b1 and b2 gives B=[1−21−3]. We can verify that, as we expect, AB=I.
  2. We find that BA=I, which is the condition that tells us that B is invertible.
  3. Seeking the first column of C−1, we see that the equation Cx=[10] is not consistent. This means that C is not invertible.
  4. Since the equation Ax=b is consistent for every b, we know that the span of the columns of A is Rn.
  5. Because the span of the columns of A is Rn, there is a pivot position in every row. Since A is square, there is also a pivot position in every column. This means that the reduced row echelon form of A must be the identity matrix In.
  6. We see that
    A∼[1001],C∼[1−1200],
    which shows that A is invertible and C is not.
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Example 4.1.9.

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We can reformulate this procedure for finding the inverse of a matrix. For the sake of convenience, suppose that A is a 2×2 invertible matrix with inverse B=[b1b2]. Rather than solving the equations
Ab1=[10],Ab2=[01]
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separately, we can solve them at the same time by augmenting A by both vectors [10] and [01] and finding the reduced row echelon form.
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For example, if A=[1211], we form
[12101101]∼[10−12011−1].
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This shows that the matrix B=[−1211] is the inverse of A.
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In other words, beginning with A, we augment by the identify and find the reduced row echelon form to determine A−1:
[AI]∼[IA−1].
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This reformulation will always work.
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The next proposition summarizes much of what we have found about invertible matrices.
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Remark 4.1.12. Formulas for inverse matrices.

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There is a simple formula for finding the inverse of a 2×2 matrix:
[abcd]−1=1ad−bc[d−b−ca],
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which can be easily checked. The condition that A be invertible is, in this case, reduced to the condition that ad−bc≠0. We will understand this condition better once we have explored determinants in Section 4.5. There is a similar formula for the inverse of a 3×3 matrix, but there is not a good reason to write it here.
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Subsection 4.1.4 Summary

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In this section, we found conditions guaranteeing that a matrix has an inverse. When these conditions hold, we also found an algorithm for finding the inverse.
  • 🔗
    A square matrix is invertible if there is a matrix B, known as the inverse of A, such that AB=I. We usually write A−1=B.
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    The n×n matrix A is invertible if and only if it is row equivalent to In, the n×n identity matrix.
  • 🔗
    If a matrix A is invertible, we can use Gaussian elimination to find its inverse:
    [AI]∼[IA−1].
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    If a matrix A is invertible, then the solution to the equation Ax=b is x=A−1b.
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Exercises 4.1.5 Exercises

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1.

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Consider the matrix
A=[3−1140231−210−23012].
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  1. 🔗
    Explain why A has an inverse.
  2. 🔗
    Find the inverse of A by augmenting by the identity I to form [AI].
  3. 🔗
    Use your inverse to solve the equation Ax=[32−3−1].
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2.

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In this exercise, we will consider 2×2 matrices as defining matrix transformations.
  1. 🔗
    Write the matrix A that performs a 45∘ rotation. What geometric operation undoes this rotation? Find the matrix that perform this operation and verify that it is A−1.
  2. 🔗
    Write the matrix A that performs a 180∘ rotation. Verify that A2=I so that A−1=A, and explain geometrically why this is the case.
  3. 🔗
    Find three more matrices A that satisfy A2=I.
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3.

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Inverses for certain types of matrices can be found in a relatively straightforward fashion.
  1. 🔗
    The matrix D=[2000−1000−4] is called diagonal since the only nonzero entries are on the diagonal of the matrix.
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    1. 🔗
      Find D−1 by augmenting D by the identity and finding its reduced row echelon form.
    2. 🔗
      Under what conditions is a diagonal matrix invertible?
    3. 🔗
      Explain why the inverse of an invertible diagonal matrix is also diagonal and explain the relationship between the diagonal entries in D and D−1.
  2. 🔗
    Consider the lower triangular matrix L=[100−2103−41].
    1. 🔗
      Find L−1 by augmenting L by the identity and finding its reduced row echelon form.
    2. 🔗
      Explain why the inverse of an invertible lower triangular matrix is also lower triangular.
    3. 🔗
      How can we tell whether a triangular matrix is invertible?
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4.

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Our definition of an invertible matrix requires that A be a square n×n matrix. Let’s examine what happens when A is not square. For instance, suppose that
A=[−1−1−2−130],B=[−2211−2−1].
  1. 🔗
    Verify that BA=I2. In this case, we say that B is a left inverse of A.
  2. 🔗
    If A has a left inverse B, we can still use it to find solutions to linear equations. If we know there is a solution to the equation Ax=b, we can multiply both sides of the equation by B to find x=Bb.
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    Suppose you know there is a solution to the equation Ax=[−1−36]. Use the left inverse B to find x and verify that it is a solution.
  3. 🔗
    Now consider the matrix
    C=[1−10−210]
    and verify that C is also a left inverse of A. This shows that the matrix A may have more than one left inverse.
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5.

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Suppose that A is an n×n matrix.
  1. 🔗
    Suppose that A2=AA is invertible with inverse B. This means that A2B=AAB=I. Explain why A must be invertible with inverse AB.
  2. 🔗
    Suppose that A100 is invertible with inverse B. Explain why A is invertible. What is A−1 in terms of A and B?
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6.

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Determine whether the following statements are true or false and explain your reasoning.
  1. 🔗
    If A is invertible, then the columns of A are linearly independent.
  2. 🔗
    If A is a square matrix whose diagonal entries are all nonzero, then A is invertible.
  3. 🔗
    If A is an invertible n×n matrix, then span of the columns of A is Rn.
  4. 🔗
    If A is invertible, then there is a nonzero solution to the homogeneous equation Ax=0.
  5. 🔗
    If A is an n×n matrix and the equation Ax=b has a solution for every vector b, then A is invertible.
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7.

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Provide a justification for your response to the following questions.
  1. 🔗
    Suppose that A is a square matrix with two identical columns. Can A be invertible?
  2. 🔗
    Suppose that A is a square matrix with two identical rows. Can A be invertible?
  3. 🔗
    Suppose that A is an invertible matrix and that AB=AC. Can you conclude that B=C?
  4. 🔗
    Suppose that A is an invertible n×n matrix. What can you say about the span of the columns of A−1?
  5. 🔗
    Suppose that A is an invertible matrix and that B is row equivalent to A. Can you guarantee that B is invertible?
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8.

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We say that two square matrices A and B are similar if there is an invertible matrix P such that B=PAP−1.
  1. 🔗
    If A and B are similar, explain why A2 and B2 are similar as well. In particular, if B=PAP−1, explain why B2=PA2P−1.
  2. 🔗
    If A and B are similar and A is invertible, explain why B is also invertible.
  3. 🔗
    If A and B are similar and both are invertible, explain why A−1 and B−1 are similar.
  4. 🔗
    If A is similar to B and B is similar to C, explain why A is similar to C. To begin, you may wish to assume that B=PAP−1 and C=QBQ−1.
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9.

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Suppose that A and B are two n×n matrices and that AB is invertible. We would like to explain why both A and B are invertible.
  1. 🔗
    We first explain why B is invertible.
    1. 🔗
      Since AB is invertible, explain why any solution to the homogeneous equation ABx=0 is x=0.
    2. 🔗
      Use this fact to explain why any solution to Bx=0 must be x=0.
    3. 🔗
      Explain why B must be invertible.
  2. 🔗
    Now we explain why A is invertible.
    1. 🔗
      Since AB is invertible, explain why the equation ABx=b is consistent for every vector b.
    2. 🔗
      Using the fact that ABx=A(Bx)=b is consistent for every b, explain why every equation Ax=b is consistent.
    3. 🔗
      Explain why A must be invertible.
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