## SectionMINMMatrix Inverses and Nonsingular Matrices

We saw in TheoremÂ CINM that if a square matrix $$A$$ is nonsingular, then there is a matrix $$B$$ so that $$AB=I_n\text{.}$$ In other words, $$B$$ is halfway to being an inverse of $$A\text{.}$$ We will see in this section that $$B$$ automatically fulfills the second condition ($$BA=I_n$$). ExampleÂ MWIAA showed us that the coefficient matrix from ArchetypeÂ A had no inverse. Not coincidentally, this coefficient matrix is singular. We will make all these connections precise now. Not many examples or definitions in this section, just theorems.

### SubsectionNMINonsingular Matrices are Invertible

We need a couple of technical results for starters. Some books would call these minor, but essential, results lemmas. We'll just call 'em theorems. See Proof TechniqueÂ LC for more on the distinction.

The first of these technical results is interesting in that the hypothesis says something about a product of two square matrices and the conclusion then says the same thing about each individual matrix in the product. This result has an analogy in the algebra of complex numbers: suppose $$\alpha,\,\beta\in\complexes\text{,}$$ then $$\alpha\beta\neq 0$$ if and only if $$\alpha\neq 0$$ and $$\beta\neq 0\text{.}$$ We can view this result as suggesting that the term ânonsingularâ for matrices is like the term ânonzeroâ for scalars. Consider too that we know singular matrices, as coefficient matrices for systems of equations, will sometimes lead to systems with no solutions, or systems with infinitely many solutions (TheoremÂ NMUS). What do linear equations with zero look like? Consider $$0x=5\text{,}$$ which has no solution, and $$0x=0\text{,}$$ which has infinitely many solutions. In the algebra of scalars, zero is exceptional (meaning different, not better), and in the algebra of matrices, singular matrices are also the exception. While there is only one zero scalar, and there are infinitely many singular matrices, we will see that singular matrices are a distinct minority.

##### (â)Â

For this portion of the proof we will form the logically-equivalent contrapositive and prove that statement using two cases. â$$AB$$ is nonsingular implies $$A$$ and $$B$$ are both nonsingularâ becomes â$$A$$ or $$B$$ is singular implies $$AB$$ is singular.â (Be sure to undertstand why the âandâ became an âorâ, see Proof TechniqueÂ CP.)

Case 1. Suppose $$B$$ is singular. Then there is a nonzero vector $$\vect{z}$$ that is a solution to $$\homosystem{B}\text{.}$$ So

\begin{align*} (AB)\vect{z} &=A(B\vect{z})&& \knowl{./knowl/theorem-MMA.html}{\text{Theorem MMA}}\\ &=A\zerovector&& \knowl{./knowl/theorem-SLEMM.html}{\text{Theorem SLEMM}}\\ &=\zerovector&& \knowl{./knowl/theorem-MMZM.html}{\text{Theorem MMZM}}\text{.} \end{align*}

With TheoremÂ SLEMM we can translate this vector equality to the statement that $$\vect{z}$$ is a nonzero solution to $$\homosystem{AB}\text{.}$$ Thus $$AB$$ is singular (DefinitionÂ NM), as desired.

Case 2. Suppose $$A$$ is singular, and $$B$$ is not singular. In other words, with Case 1 complete, we can be more precise about this remaining case and assume that $$B$$ is nonsingular. Because $$A$$ is singular, there is a nonzero vector $$\vect{y}$$ that is a solution to $$\homosystem{A}\text{.}$$ Now consider the linear system $$\linearsystem{B}{\vect{y}}\text{.}$$ Since $$B$$ is nonsingular, the system has a unique solution (TheoremÂ NMUS), which we will denote as $$\vect{w}\text{.}$$ We first claim $$\vect{w}$$ is not the zero vector either. Assuming the opposite, suppose that $$\vect{w}=\zerovector$$ (Proof TechniqueÂ CD). Then

\begin{align*} \vect{y} &=B\vect{w}&& \knowl{./knowl/theorem-SLEMM.html}{\text{Theorem SLEMM}}\\ &=B\zerovector&&\text{Hypothesis}\\ &=\zerovector&& \knowl{./knowl/theorem-MMZM.html}{\text{Theorem MMZM}}\\ \end{align*}

contrary to $$\vect{y}$$ being nonzero. So $$\vect{w}\neq\zerovector\text{.}$$ The pieces are in place, so here we go,

\begin{align*} (AB)\vect{w} &=A(B\vect{w})&& \knowl{./knowl/theorem-MMA.html}{\text{Theorem MMA}}\\ &=A\vect{y}&& \knowl{./knowl/theorem-SLEMM.html}{\text{Theorem SLEMM}}\\ &=\zerovector&& \knowl{./knowl/theorem-SLEMM.html}{\text{Theorem SLEMM}}\text{.} \end{align*}

With TheoremÂ SLEMM we can translate this vector equality to the statement that $$\vect{w}$$ is a nonzero solution to $$\homosystem{AB}\text{.}$$ Thus $$AB$$ is singular (DefinitionÂ NM), as desired. And this conclusion holds for both cases.

##### (â)Â

Now assume that both $$A$$ and $$B$$ are nonsingular. Suppose that $$\vect{x}\in\complex{n}$$ is a solution to $$\homosystem{AB}\text{.}$$ Then

\begin{align*} \zerovector &=\left(AB\right)\vect{x}&& \knowl{./knowl/theorem-SLEMM.html}{\text{Theorem SLEMM}}\\ &=A\left(B\vect{x}\right)&& \knowl{./knowl/theorem-MMA.html}{\text{Theorem MMA}}\text{.} \end{align*}

By TheoremÂ SLEMM, $$B\vect{x}$$ is a solution to $$\homosystem{A}\text{,}$$ and by the definition of a nonsingular matrix (DefinitionÂ NM), we conclude that $$B\vect{x}=\zerovector\text{.}$$ Now, by an entirely similar argument, the nonsingularity of $$B$$ forces us to conclude that $$\vect{x}=\zerovector\text{.}$$ So the only solution to $$\homosystem{AB}$$ is the zero vector and we conclude that $$AB$$ is nonsingular by DefinitionÂ NM.

This is a powerful result in the âforwardâ direction, because it allows us to begin with a hypothesis that something complicated (the matrix product $$AB$$) has the property of being nonsingular, and we can then conclude that the simpler constituents ($$A$$ and $$B$$ individually) then also have the property of being nonsingular. If we had thought that the matrix product was an artificial construction, results like this would make us begin to think twice.

The contrapositive of this entire result is equally interesting. It says that $$A$$ or $$B$$ (or both) is a singular matrix if and only if the product $$AB$$ is singular. (See Proof TechniqueÂ CP.)

The matrix $$I_n$$ is nonsingular (since it row-reduces easily to $$I_n\text{,}$$ TheoremÂ NMRRI). So $$A$$ and $$B$$ are nonsingular by TheoremÂ NPNT, so in particular $$B$$ is nonsingular. We can therefore apply TheoremÂ CINM to assert the existence of a matrix $$C$$ so that $$BC=I_n\text{.}$$ This application of TheoremÂ CINM could be a bit confusing, mostly because of the names of the matrices involved. $$B$$ is nonsingular, so there must be a âright-inverseâ for $$B\text{,}$$ and we are calling it $$C\text{.}$$

Now

\begin{align*} BA &=(BA)I_n&& \knowl{./knowl/theorem-MMIM.html}{\text{Theorem MMIM}}\\ &=(BA)(BC)&& \knowl{./knowl/theorem-CINM.html}{\text{Theorem CINM}}\\ &=B(AB)C&& \knowl{./knowl/theorem-MMA.html}{\text{Theorem MMA}}\\ &=BI_nC&&\text{Hypothesis}\\ &=BC&& \knowl{./knowl/theorem-MMIM.html}{\text{Theorem MMIM}}\\ &=I_n&& \knowl{./knowl/theorem-CINM.html}{\text{Theorem CINM}} \end{align*}

which is the desired conclusion.

So TheoremÂ OSIS tells us that if $$A$$ is nonsingular, then the matrix $$B$$ guaranteed by TheoremÂ CINM will be both a âright-inverseâ and a âleft-inverseâ for $$A\text{,}$$ so $$A$$ is invertible and $$\inverse{A}=B\text{.}$$

So if you have a nonsingular matrix, $$A\text{,}$$ you can use the procedure described in TheoremÂ CINM to find an inverse for $$A\text{.}$$ If $$A$$ is singular, then the procedure in TheoremÂ CINM will fail as the first $$n$$ columns of $$M$$ will not row-reduce to the identity matrix. However, we can say a bit more. When $$A$$ is singular, then $$A$$ does not have an inverse (which is very different from saying that the procedure in TheoremÂ CINM fails to find an inverse). This may feel like we are splitting hairs, but it is important that we do not make unfounded assumptions. These observations motivate the next theorem.

##### (â)Â

Since $$A$$ is invertible, we can write $$I_n=A\inverse{A}$$ (DefinitionÂ MI). Notice that $$I_n$$ is nonsingular (TheoremÂ NMRRI) so TheoremÂ NPNT implies that $$A$$ (and $$\inverse{A}$$) is nonsingular.

##### (â)Â

Suppose now that $$A$$ is nonsingular. By TheoremÂ CINM we find $$B$$ so that $$AB=I_n\text{.}$$ Then TheoremÂ OSIS tells us that $$BA=I_n\text{.}$$ So $$B$$ is $$A$$'s inverse, and by construction, $$A$$ is invertible.

So for a square matrix, the properties of having an inverse and of having a trivial null space are one and the same. Cannot have one without the other.

In the case that $$A$$ is a nonsingular coefficient matrix of a system of equations, the inverse allows us to very quickly compute the unique solution, for any vector of constants.

By TheoremÂ NMUS we know already that $$\linearsystem{A}{\vect{b}}$$ has a unique solution for every choice of $$\vect{b}\text{.}$$ We need to show that the expression stated is indeed a solution (the solution). That is easy, just âplug it inâ to the vector equation representation of the system (TheoremÂ SLEMM)

\begin{align*} A\left(\inverse{A}\vect{b}\right) &=\left(A\inverse{A}\right)\vect{b}&& \knowl{./knowl/theorem-MMA.html}{\text{Theorem MMA}}\\ &=I_n\vect{b}&& \knowl{./knowl/definition-MI.html}{\text{Definition MI}}\\ &=\vect{b}&& \knowl{./knowl/theorem-MMIM.html}{\text{Theorem MMIM}}\text{.} \end{align*}

Since $$A\vect{x}=\vect{b}$$ is true when we substitute $$\inverse{A}\vect{b}$$ for $$\vect{x}\text{,}$$ $$\inverse{A}\vect{b}$$ is a (the!) solution to $$\linearsystem{A}{\vect{b}}\text{.}$$

#### SageMI.Matrix Inverse.

Now we know that invertibility is equivalent to nonsingularity, and that the procedure outlined in TheoremÂ CINM will always yield an inverse for a nonsingular matrix. But rather than using that procedure, Sage implements a .inverse()method. In the following, we compute the inverse of a $$3\times 3$$ matrix, and then purposely convert it to a singular matrix by replacing the last column by a linear combination of the first two.

A = matrix(QQ,[[ 3,  7, -6],
[ 2,  5, -5],
[-3, -8,  8]])
A.is_singular()

Ainv = A.inverse(); Ainv

A*Ainv

col_0 = A.column(0)
col_1 = A.column(1)
C = column_matrix([col_0, col_1, 2*col_0 - 4*col_1])
C.is_singular()

C.inverse()


Notice how the failure to invert Cis explained by the matrix being singular.

Systems with nonsingular coefficient matrices can be solved easily with the matrix inverse. We will recycle Aas a coefficient matrix, so be sure to execute the code above.

const = vector(QQ, [2, -1, 4])
A.solve_right(const)

A.inverse()*const

A.solve_right(const) == A.inverse()*const


If you find it more convenient, you can use the same notation as the text for a matrix inverse.

A^-1


#### SageNME3.Nonsingular Matrix Equivalences, Round 3.

For square matrices, Sage has the methods .is_singular()and .is_invertible(). By TheoremÂ NI we know these two functions to be logical opposites. One way to express this is that these two methods will always return different values. Here we demonstrate with a nonsingular matrix and a singular matrix. The comparison !=is ânot equal.â

nonsing = matrix(QQ, [[ 0, -1,  1, -3],
[ 1,  1,  0, -3],
[ 0,  4, -3,  8],
[-2, -4,  1,  5]])
nonsing.is_singular() != nonsing.is_invertible()

sing = matrix(QQ, [[ 1, -2, -6,  8],
[-1,  3,  7, -8],
[ 0, -4, -3, -2],
[ 0,  3,  1,  4]])
sing.is_singular() != sing.is_invertible()


We could test other properties of the matrix inverse, such as TheoremÂ SS.

A = matrix(QQ, [[ 3, -5, -2,  8],
[-1,  2,  0, -1],
[-2,  4,  1, -4],
[ 4, -5,  0,  8]])
B = matrix(QQ, [[ 1,  2,  4, -1],
[-2, -3, -8, -2],
[-2, -4, -7,  5],
[ 2,  5,  7, -8]])
A.is_invertible() and B.is_invertible()

(A*B).inverse() == B.inverse()*A.inverse()


### SubsectionUMUnitary Matrices

Recall that the adjoint of a matrix is $$\adjoint{A}=\transpose{\left(\conjugate{A}\right)}$$ (DefinitionÂ A).

#### DefinitionUM.Unitary Matrices.

Suppose that $$U$$ is a square matrix of size $$n$$ such that $$\adjoint{U}U=I_n\text{.}$$ Then we say $$U$$ is unitary.

This condition may seem rather far-fetched at first glance. Would there be any matrix that behaved this way? Well, yes, here is one.

Let

\begin{equation*} U= \begin{bmatrix} \frac{1 + i }{{\sqrt{5}}} & \frac{3 + 2\,i }{{\sqrt{55}}} & \frac{2+2i}{\sqrt{22}} \\ \frac{1 - i }{{\sqrt{5}}} & \frac{2 + 2\,i }{{\sqrt{55}}} & \frac{-3 + i }{{\sqrt{22}}} \\ \frac{i }{{\sqrt{5}}} & \frac{3 - 5\,i }{{\sqrt{55}}} & -\frac{2}{\sqrt{22}} \end{bmatrix}\text{.} \end{equation*}

The computations get a bit tiresome, but if you work your way through the computation of $$\adjoint{U}U\text{,}$$ you will arrive at the $$3\times 3$$ identity matrix $$I_3\text{.}$$

Unitary matrices do not have to look quite so gruesome. Here is a larger one that is a bit more pleasing.

The matrix

\begin{equation*} P= \begin{bmatrix} 0&1&0&0&0\\ 0&0&0&1&0\\ 1&0&0&0&0\\ 0&0&0&0&1\\ 0&0&1&0&0 \end{bmatrix} \end{equation*}

is unitary as can be easily checked. Notice that it is just a rearrangement of the columns of the $$5\times 5$$ identity matrix, $$I_5$$ (DefinitionÂ IM).

An interesting exercise is to build another $$5\times 5$$ unitary matrix, $$R\text{,}$$ using a different rearrangement of the columns of $$I_5\text{.}$$ Then form the product $$PR\text{.}$$ This will be another unitary matrix (ExerciseÂ MINM.T10). If you were to build all $$5!=5\times 4\times 3\times 2\times 1=120$$ matrices of this type you would have a set that remains closed under matrix multiplication. It is an example of another algebraic structure known as a group since together the set and the one operation (matrix multiplication here) is closed, associative, has an identity ($$I_5$$), and inverses (TheoremÂ UMI). Notice though that the operation in this group is not commutative!

If a matrix $$A$$ has only real number entries (we say it is a real matrix) then the defining property of being unitary simplifies to $$\transpose{A}A=I_n\text{.}$$ In this case we, and everybody else, call the matrix orthogonal, so you may often encounter this term in your other reading when the complex numbers are not under consideration.

Unitary matrices have easily computed inverses. They also have columns that form orthonormal sets. Here are the theorems that show us that unitary matrices are not as strange as they might initially appear.

By DefinitionÂ UM, we know that $$\adjoint{U}U=I_n\text{.}$$ The matrix $$I_n$$ is nonsingular (since it row-reduces easily to $$I_n\text{,}$$ TheoremÂ NMRRI). So by TheoremÂ NPNT, $$U$$ and $$\adjoint{U}$$ are both nonsingular matrices.

The equation $$\adjoint{U}U=I_n$$ gets us halfway to an inverse of $$U\text{,}$$ and TheoremÂ OSIS tells us that then $$U\adjoint{U}=I_n$$ also. So $$U$$ and $$\adjoint{U}$$ are inverses of each other (DefinitionÂ MI).

The proof revolves around recognizing that a typical entry of the product $$\adjoint{A}A$$ is an inner product of columns of $$A\text{.}$$ Here are the details to support this claim.

\begin{align*} \matrixentry{\adjoint{A}A}{ij} &=\sum_{k=1}^{n}\matrixentry{\adjoint{A}}{ik}\matrixentry{A}{kj}&& \knowl{./knowl/theorem-EMP.html}{\text{Theorem EMP}}\\ &=\sum_{k=1}^{n}\matrixentry{\transpose{\conjugate{A}}}{ik}\matrixentry{A}{kj}&& \knowl{./knowl/definition-A.html}{\text{Definition A}}\\ &=\sum_{k=1}^{n}\matrixentry{\,\conjugate{A}\,}{ki}\matrixentry{A}{kj}&& \knowl{./knowl/definition-TM.html}{\text{Definition TM}}\\ &=\sum_{k=1}^{n}\conjugate{\matrixentry{A}{ki}}\matrixentry{A}{kj}&& \knowl{./knowl/definition-CCM.html}{\text{Definition CCM}}\\ &=\sum_{k=1}^{n}\conjugate{\vectorentry{\vect{A}_i}{k}}\vectorentry{\vect{A}_j}{k}\\ &=\innerproduct{\vect{A}_i}{\vect{A}_j}&& \knowl{./knowl/definition-IP.html}{\text{Definition IP}} \end{align*}

We now employ this equality in a chain of equivalences,

\begin{align*} &S=\set{\vectorlist{A}{n}}\text{ is an orthonormal set}\\ &\iff \innerproduct{\vect{A}_i}{\vect{A}_j}= \begin{cases} 0 &\text{if }i\neq j\\ 1 &\text{if }i=j \end{cases}&& \knowl{./knowl/definition-ONS.html}{\text{Definition ONS}}\\ &\iff \matrixentry{\adjoint{A}A}{ij}= \begin{cases} 0 &\text{if }i\neq j\\ 1 &\text{if }i=j \end{cases}\\ &\iff \matrixentry{\adjoint{A}A}{ij}=\matrixentry{I_n}{ij},\ 1\leq i\leq n,\ 1\leq j\leq n&& \knowl{./knowl/definition-IM.html}{\text{Definition IM}}\\ &\iff \adjoint{A}A=I_n&& \knowl{./knowl/definition-ME.html}{\text{Definition ME}}\\ &\iff A\text{ is a unitary matrix}&& \knowl{./knowl/definition-UM.html}{\text{Definition UM}} \end{align*}

The matrix

\begin{equation*} U= \begin{bmatrix} \frac{1 + i }{{\sqrt{5}}} & \frac{3 + 2\,i }{{\sqrt{55}}} & \frac{2+2i}{\sqrt{22}} \\ \frac{1 - i }{{\sqrt{5}}} & \frac{2 + 2\,i }{{\sqrt{55}}} & \frac{-3 + i }{{\sqrt{22}}} \\ \frac{i }{{\sqrt{5}}} & \frac{3 - 5\,i }{{\sqrt{55}}} & -\frac{2}{\sqrt{22}} \end{bmatrix} \end{equation*}

from ExampleÂ UM3 is a unitary matrix. By TheoremÂ CUMOS, its columns

\begin{equation*} \set{ \colvector{ \frac{1 + i }{{\sqrt{5}}}\\ \frac{1 - i }{{\sqrt{5}}}\\ \frac{i }{{\sqrt{5}}} },\, \colvector{ \frac{3 + 2\,i }{{\sqrt{55}}}\\ \frac{2 + 2\,i }{{\sqrt{55}}}\\ \frac{3 - 5\,i }{{\sqrt{55}}} },\, \colvector{ \frac{2+2i}{\sqrt{22}}\\ \frac{-3 + i }{{\sqrt{22}}}\\ -\frac{2}{\sqrt{22}} } } \end{equation*}

form an orthonormal set. You might find checking the six inner products of pairs of these vectors easier than doing the matrix product $$\adjoint{U}U\text{.}$$ Or, because the inner product is anti-commutative (TheoremÂ IPAC) you only need check three inner products (see ExerciseÂ MINM.T12).

When using vectors and matrices that only have real number entries, orthogonal matrices are those matrices with inverses that equal their transpose. Similarly, the inner product is the familiar dot product. Keep this special case in mind as you read the next theorem.

\begin{align*} \innerproduct{U\vect{u}}{U\vect{v}} &=\transpose{\left(\conjugate{U\vect{u}}\right)}U\vect{v}&& \knowl{./knowl/theorem-MMIP.html}{\text{Theorem MMIP}}\\ &=\transpose{\left(\conjugate{U}\conjugate{\vect{u}}\right)}U\vect{v}&& \knowl{./knowl/theorem-MMCC.html}{\text{Theorem MMCC}}\\ &=\transpose{\conjugate{\vect{u}}}\transpose{\conjugate{U}}U\vect{v}&& \knowl{./knowl/theorem-MMT.html}{\text{Theorem MMT}}\\ &=\transpose{\conjugate{\vect{u}}}\adjoint{U}U\vect{v}&& \knowl{./knowl/definition-A.html}{\text{Definition A}}\\ &=\transpose{\conjugate{\vect{u}}}I_n\vect{v}&& \knowl{./knowl/definition-UM.html}{\text{Definition UM}}\\ &=\transpose{\conjugate{\vect{u}}}\vect{v}&& \knowl{./knowl/theorem-MMIM.html}{\text{Theorem MMIM}}\\ &=\innerproduct{\vect{u}}{\vect{v}}&& \knowl{./knowl/theorem-MMIP.html}{\text{Theorem MMIP}} \end{align*}

The second conclusion is just a specialization of the first conclusion.

\begin{align*} \norm{U\vect{v}} &=\sqrt{\norm{U\vect{v}}^2}\\ &=\sqrt{\innerproduct{U\vect{v}}{U\vect{v}}}&& \knowl{./knowl/theorem-IPN.html}{\text{Theorem IPN}}\\ &=\sqrt{\innerproduct{\vect{v}}{\vect{v}}}\\ &=\sqrt{\norm{\vect{v}}^2}&& \knowl{./knowl/theorem-IPN.html}{\text{Theorem IPN}}\\ &=\norm{\vect{v}}&& \end{align*}

Aside from the inherent interest in this theorem, it makes a bigger statement about unitary matrices. When we view vectors geometrically as directions or forces, then the norm equates to a notion of length. If we transform a vector by multiplication with a unitary matrix, then the length (norm) of that vector stays the same. If we consider column vectors with two or three slots containing only real numbers, then the inner product of two such vectors is just the dot product, and this quantity can be used to compute the angle between two vectors. When two vectors are multiplied (transformed) by the same unitary matrix, their dot product is unchanged and their individual lengths are unchanged. This results in the angle between the two vectors remaining unchanged.

A unitary transformation (matrix-vector products with unitary matrices) thus preserve geometrical relationships among vectors representing directions, forces, or other physical quantities. In the case of a two-slot vector with real entries, this is simply a rotation. These sorts of computations are exceedingly important in computer graphics such as games and real-time simulations, especially when increased realism is achieved by performing many such computations quickly. We will see unitary matrices again in subsequent sections (especially TheoremÂ OD) and in each instance, consider the interpretation of the unitary matrix as a sort of geometry-preserving transformation. Some authors use the term isometry to highlight this behavior. We will speak loosely of a unitary matrix as being a sort of generalized rotation.

#### SageUM.Unitary Matrices.

No surprise about how we check if a matrix is unitary. Here is ExampleÂ UM3,

A = matrix(QQbar, [
[(1+I)/sqrt(5), (3+2*I)/sqrt(55), (2+2*I)/sqrt(22)],
[(1-I)/sqrt(5), (2+2*I)/sqrt(55),  (-3+I)/sqrt(22)],
[    I/sqrt(5), (3-5*I)/sqrt(55),    (-2)/sqrt(22)]
])
A.is_unitary()

A.conjugate_transpose() == A.inverse()


We can verify TheoremÂ UMPIP, where the vectors uand vare created randomly. Try evaluating this compute cell with your own choices.

u = random_vector(QQ, 3) + QQbar(I)*random_vector(QQ, 3)
v = random_vector(QQ, 3) + QQbar(I)*random_vector(QQ, 3)
(A*u).hermitian_inner_product(A*v) == u.hermitian_inner_product(v)


If you want to experiment with permutation matrices, Sage has these too. We can create a permutation matrix from a list that indicates for each column the row with a one in it. Notice that the product here of two permutation matrices is again a permutation matrix.

sigma = Permutation([2,3,1])
S = sigma.to_matrix(); S

tau = Permutation([1,3,2])
T = tau.to_matrix(); T

S*T

rho = Permutation([2, 1, 3])
S*T == rho.to_matrix()


A final reminder: the terms âdot product,â âsymmetric matrixâ and âorthogonal matrixâ used in reference to vectors or matrices with real number entries are special cases of the terms âinner product,â âHermitian matrixâ and âunitary matrixâ that we use for vectors or matrices with complex number entries, so keep that in mind as you read elsewhere.

#### 1.

Compute the inverse of the coefficient matrix of the system of equations below and use the inverse to solve the system.

\begin{align*} 4x_1 + 10x_2 &= 12\\ 2x_1 + 6x_2 &= 4 \end{align*}

#### 2.

In the reading questions for SectionÂ MISLE you were asked to find the inverse of the $$3\times 3$$ matrix below.

\begin{equation*} \begin{bmatrix} 2 & 3 & 1\\ 1 & -2 & -3\\ -2 & 4 & 6 \end{bmatrix} \end{equation*}

Because the matrix was not nonsingular, you had no theorems at that point that would allow you to compute the inverse. Explain why you now know that the inverse does not exist (which is different than not being able to compute it) by quoting the relevant theorem's acronym.

#### 3.

Is the matrix $$A$$ unitary? Why?

\begin{equation*} A=\begin{bmatrix} \frac{1}{\sqrt{22}}\left(4+2i\right) & \frac{1}{\sqrt{374}}\left(5+3i\right) \\ \frac{1}{\sqrt{22}}\left(-1-i\right) & \frac{1}{\sqrt{374}}\left(12+14i\right) \\ \end{bmatrix} \end{equation*}

### ExercisesMINMExercises

#### C20.

Verify that $$AB$$ is nonsingular.

\begin{align*} A&= \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 1\\ 1 & 0 & 2 \end{bmatrix} & B &= \begin{bmatrix} -1 & 1 & 0\\ 1 & 2 & 1\\ 0 & 1 & 1 \end{bmatrix} \end{align*}

#### C40.

Solve the system of equations below using the inverse of a matrix.

\begin{align*} x_1+x_2+3x_3+x_4&=5\\ -2x_1-x_2-4x_3-x_4&=-7\\ x_1+4x_2+10x_3+2x_4&=9\\ -2x_1-4x_3+5x_4&=9 \end{align*}
Solution.

The coefficient matrix and vector of constants for the system are

\begin{align*} \begin{bmatrix} 1 & 1 & 3 & 1\\ -2 & -1 & -4 & -1\\ 1 & 4 & 10 & 2\\ -2 & 0 & -4 & 5 \end{bmatrix}&& \vect{b}=\colvector{5\\-7\\9\\9}\text{.} \end{align*}

$$\inverse{A}$$ can be computed by using a calculator, or by the method of TheoremÂ CINM. Then TheoremÂ SNCM says the unique solution is

\begin{equation*} \inverse{A}\vect{b}= \begin{bmatrix} 38 & 18 & -5 & -2\\ 96 & 47 & -12 & -5\\ -39 & -19 & 5 & 2\\ -16 & -8 & 2 & 1 \end{bmatrix} \colvector{5\\-7\\9\\9} = \colvector{1\\-2\\1\\3}\text{.} \end{equation*}

#### M10.

Find values of $$x\text{,}$$ $$y\text{,}$$ $$z$$ so that matrix $$A$$ is invertible.

\begin{equation*} A = \begin{bmatrix} 1 & 2 & x\\ 3 & 0 & y \\ 1 & 1 & z \end{bmatrix} \end{equation*}
Solution.

There are an infinite number of possible answers. We want to find a vector $$\colvector{x \\ y \\ z}$$ so that the set

\begin{gather*} S = \set{ \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} x \\ y \\ z \end{bmatrix} } \end{gather*}

is a linearly independent set. We need a vector not in the span of the first two columns, which geometrically means that we need it to not be in the same plane as the first two columns of $$A\text{.}$$ We can choose any values we want for $$x$$ and $$y\text{,}$$ and then choose a value of $$z$$ that makes the three vectors independent.

I will (arbitrarily) choose $$x = 1\text{,}$$ $$y = 1\text{.}$$ Then, we have

\begin{align*} A = \begin{bmatrix} 1 & 2 & 1\\ 3 & 0 & 1 \\ 1 & 1 & z \end{bmatrix} &\rref \begin{bmatrix} \leading{1} & 0 & 2z-1 \\ 0 & \leading{1} & 1-z \\ 0 & 0 & 4 - 6z \end{bmatrix} \end{align*}

which is invertible if and only if $$4-6z \ne 0\text{.}$$ Thus, we can choose any value as long as $$z \ne \frac{2}{3}\text{,}$$ so we choose $$z = 0\text{,}$$ and we have found a matrix $$A = \begin{bmatrix} 1 & 2 & 1\\ 3 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}$$ that is invertible.

#### M11.

Find values of $$x\text{,}$$ $$y$$ $$z$$ so that matrix $$A$$ is singular.

\begin{equation*} A = \begin{bmatrix} 1 & x & 1\\ 1 & y & 4 \\ 0 & z & 5 \end{bmatrix} \end{equation*}
Solution.

There are an infinite number of possible answers. We need the set of vectors

\begin{align*} S &= \set{ \begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix}x \\ y \\ z \end{bmatrix}, \begin{bmatrix} 1 \\ 4 \\ 5 \end{bmatrix} } \end{align*}

to be linearly dependent. One way to do this by inspection is to have$$\colvector{x \\ y \\ z} =\colvector{1 \\ 4 \\ 5}\text{.}$$ Thus, if we let $$x = 1\text{,}$$ $$y = 4\text{,}$$ $$z = 5\text{,}$$ then the matrix $$A = \begin{bmatrix} 1 & 1 & 1\\ 1 & 4 & 4 \\ 0 & 5 & 5 \end{bmatrix}$$ is singular.

#### M15.

If $$A$$ and $$B$$ are $$n \times n$$ matrices, $$A$$ is nonsingular, and $$B$$ is singular, show directly that $$AB$$ is singular, without using TheoremÂ NPNT.

Solution.

If $$B$$ is singular, then there exists a vector $$\vect{x}\ne\zerovector$$ so that $$\vect{x}\in \nsp{B}\text{.}$$ Thus, $$B\vect{x} = \vect{0}\text{,}$$ so $$A(B\vect{x}) = (AB)\vect{x} = \zerovector\text{,}$$ so $$\vect{x}\in\nsp{AB}\text{.}$$ Since the null space of $$AB$$ is not trivial, $$AB$$ is a singular matrix.

#### M20.

Construct an example of a $$4\times 4$$ unitary matrix.

Solution.

The $$4\times 4$$ identity matrix, $$I_4\text{,}$$ would be one example (DefinitionÂ IM). Any of the 23 other rearrangements of the columns of $$I_4$$ would be a simple, but less trivial, example. See ExampleÂ UPM.

#### M80.

Matrix multiplication interacts nicely with many operations. But not always with transforming a matrix to reduced row-echelon form. Suppose that $$A$$ is an $$m\times n$$ matrix and $$B$$ is an $$n\times p$$ matrix. Let $$P$$ be a matrix that is row-equivalent to $$A$$ and in reduced row-echelon form, $$Q$$ be a matrix that is row-equivalent to $$B$$ and in reduced row-echelon form, and let $$R$$ be a matrix that is row-equivalent to $$AB$$ and in reduced row-echelon form. Is $$PQ=R\text{?}$$ (In other words, with nonstandard notation, is $$\text{rref}(A)\text{rref}(B)=\text{rref}(AB)\text{?}$$)

Construct a counterexample to show that, in general, this statement is false. Then find a large class of matrices where if $$A$$ and $$B$$ are in the class, then the statement is true.

Solution.

Take

\begin{align*} A&= \begin{bmatrix} 1&0\\0&0 \end{bmatrix} & B&= \begin{bmatrix} 0&0\\1&0 \end{bmatrix}\text{.} \end{align*}

Then $$A$$ is already in reduced row-echelon form, and by swapping rows, $$B$$ row-reduces to $$A\text{.}$$ So the product of the row-echelon forms of $$A$$ is $$AA=A\neq\zeromatrix\text{.}$$ However, the product $$AB$$ is the $$2\times 2$$ zero matrix, which is in reduced-echelon form, and not equal to $$AA\text{.}$$ When you get there, TheoremÂ PEEF or TheoremÂ EMDRO might shed some light on why we would not expect this statement to be true in general.

If $$A$$ and $$B$$ are nonsingular, then $$AB$$ is nonsingular (TheoremÂ NPNT), and all three matrices $$A\text{,}$$ $$B$$ and $$AB$$ row-reduce to the identity matrix (TheoremÂ NMRRI). By TheoremÂ MMIM, the desired relationship is true.

#### T10.

Suppose that $$Q$$ and $$P$$ are unitary matrices of size $$n\text{.}$$ Prove that $$QP$$ is a unitary matrix.

#### T11.

Prove that Hermitian matrices (DefinitionÂ HM) have real entries on the diagonal. More precisely, suppose that $$A$$ is a Hermitian matrix of size $$n\text{.}$$ Then $$\matrixentry{A}{ii}\in\reals\text{,}$$ $$1\leq i\leq n\text{.}$$

#### T12.

Suppose that we are checking if a square matrix of size $$n$$ is unitary. Show that a straightforward application of TheoremÂ CUMOS requires the computation of $$n^2$$ inner products when the matrix is unitary, and fewer when the matrix is not orthogonal. Then show that this maximum number of inner products can be reduced to $$\frac{1}{2}n(n+1)$$ in light of TheoremÂ IPAC.

#### T25.

The notation $$A^k$$ means a repeated matrix product between $$k$$ copies of the square matrix $$A\text{.}$$

1. Assume $$A$$ is an $$n\times n$$ matrix where $$A^2=\zeromatrix$$ (which does not imply that $$A=\zeromatrix\text{.}$$) Prove that $$I_n-A$$ is invertible by showing that $$I_n+A$$ is an inverse of $$I_n-A\text{.}$$

2. Assume that $$A$$ is an $$n\times n$$ matrix where $$A^3=\zeromatrix\text{.}$$ Prove that $$I_n-A$$ is invertible.

3. Form a general theorem based on your observations from parts (1) and (2) and provide a proof.