Section NM  Nonsingular Matrices

From A First Course in Linear Algebra
Version 2.10
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/

In this section we specialize and consider matrices with equal numbers of rows and columns, which when considered as coefficient matrices lead to systems with equal numbers of equations and variables. We will see in the second half of the course (Chapter D, Chapter E Chapter LT, Chapter R) that these matrices are especially important.

Subsection NM: Nonsingular Matrices

Our theorems will now establish connections between systems of equations (homogeneous or otherwise), augmented matrices representing those systems, coefficient matrices, constant vectors, the reduced row-echelon form of matrices (augmented and coefficient) and solution sets. Be very careful in your reading, writing and speaking about systems of equations, matrices and sets of vectors. A system of equations is not a matrix, a matrix is not a solution set, and a solution set is not a system of equations. Now would be a great time to review the discussion about speaking and writing mathematics in Technique L.

Definition SQM
Square Matrix
A matrix with m rows and n columns is square if m = n. In this case, we say the matrix has size n. To emphasize the situation when a matrix is not square, we will call it rectangular.

We can now present one of the central definitions of linear algebra.

Definition NM
Nonsingular Matrix
Suppose A is a square matrix. Suppose further that the solution set to the homogeneous linear system of equations ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ) is \left \{0\right \}, i.e. the system has only the trivial solution. Then we say that A is a nonsingular matrix. Otherwise we say A is a singular matrix.

We can investigate whether any square matrix is nonsingular or not, no matter if the matrix is derived somehow from a system of equations or if it is simply a matrix. The definition says that to perform this investigation we must construct a very specific system of equations (homogeneous, with the matrix as the coefficient matrix) and look at its solution set. We will have theorems in this section that connect nonsingular matrices with systems of equations, creating more opportunities for confusion. Convince yourself now of two observations, (1) we can decide nonsingularity for any square matrix, and (2) the determination of nonsingularity involves the solution set for a certain homogeneous system of equations.

Notice that it makes no sense to call a system of equations nonsingular (the term does not apply to a system of equations), nor does it make any sense to call a 5 × 7 matrix singular (the matrix is not square).

Example S
A singular matrix, Archetype A
Example HISAA shows that the coefficient matrix derived from Archetype A, specifically the 3 × 3 matrix,

A = \left [\array{ 1&−1&2 \cr 2& 1 &1 \cr 1& 1 &0 } \right ]

is a singular matrix since there are nontrivial solutions to the homogeneous system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ).

Example NM
A nonsingular matrix, Archetype B
Example HUSAB shows that the coefficient matrix derived from Archetype B, specifically the 3 × 3 matrix,

B = \left [\array{ −7&−6&−12 \cr 5 & 5 & 7 \cr 1 & 0 & 4 } \right ]

is a nonsingular matrix since the homogeneous system, ℒS\kern -1.95872pt \left (B,\kern 1.95872pt 0\right ), has only the trivial solution.

Notice that we will not discuss Example HISAD as being a singular or nonsingular coefficient matrix since the matrix is not square.

The next theorem combines with our main computational technique (row-reducing a matrix) to make it easy to recognize a nonsingular matrix. But first a definition.

Definition IM
Identity Matrix
The m × m identity matrix, {I}_{m}, is defined by

\eqalignno{ {\left [{I}_{m}\right ]}_{ij} & = \left \{\array{ 1\quad &i = j \cr 0\quad &i\mathrel{≠}j } \right .\quad \quad 1 ≤ i,\kern 1.95872pt j ≤ m & & }

(This definition contains Notation IM.)

Example IM
An identity matrix
The 4 × 4 identity matrix is

{ I}_{4} = \left [\array{ 1&0&0&0 \cr 0&1&0&0 \cr 0&0&1&0 \cr 0&0&0&1 } \right ].

Notice that an identity matrix is square, and in reduced row-echelon form. So in particular, if we were to arrive at the identity matrix while bringing a matrix to reduced row-echelon form, then it would have all of the diagonal entries circled as leading 1’s.

Theorem NMRRI
Nonsingular Matrices Row Reduce to the Identity matrix
Suppose that A is a square matrix and B is a row-equivalent matrix in reduced row-echelon form. Then A is nonsingular if and only if B is the identity matrix.

Proof   () Suppose B is the identity matrix. When the augmented matrix \left [\left .A\kern 1.95872pt \right \vert \kern 1.95872pt 0\right ] is row-reduced, the result is \left [\left .B\kern 1.95872pt \right \vert \kern 1.95872pt 0\right ] = \left [\left .{I}_{n}\kern 1.95872pt \right \vert \kern 1.95872pt 0\right ]. The number of nonzero rows is equal to the number of variables in the linear system of equations ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ), so n = r and Theorem FVCS gives n − r = 0 free variables. Thus, the homogeneous system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ) has just one solution, which must be the trivial solution. This is exactly the definition of a nonsingular matrix.

() If A is nonsingular, then the homogeneous system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ) has a unique solution, and has no free variables in the description of the solution set. The homogeneous system is consistent (Theorem HSC) so Theorem FVCS applies and tells us there are n − r free variables. Thus, n − r = 0, and so n = r. So B has n pivot columns among its total of n columns. This is enough to force B to be the n × n identity matrix {I}_{n}.

Notice that since this theorem is an equivalence it will always allow us to determine if a matrix is either nonsingular or singular. Here are two examples of this, continuing our study of Archetype A and Archetype B.

Example SRR
Singular matrix, row-reduced
The coefficient matrix for Archetype A is

A = \left [\array{ 1&−1&2 \cr 2& 1 &1 \cr 1& 1 &0 } \right ]

which when row-reduced becomes the row-equivalent matrix

B = \left [\array{ \text{1}&0& 1 \cr 0&\text{1}&−1 \cr 0&0& 0 } \right ].

Since this matrix is not the 3 × 3 identity matrix, Theorem NMRRI tells us that A is a singular matrix.

Example NSR
Nonsingular matrix, row-reduced
The coefficient matrix for Archetype B is

A = \left [\array{ −7&−6&−12 \cr 5 & 5 & 7 \cr 1 & 0 & 4 } \right ]

which when row-reduced becomes the row-equivalent matrix

B = \left [\array{ \text{1}&0&0 \cr 0&\text{1}&0 \cr 0&0&\text{1}} \right ].

Since this matrix is the 3 × 3 identity matrix, Theorem NMRRI tells us that A is a nonsingular matrix.

Subsection NSNM: Null Space of a Nonsingular Matrix

Nonsingular matrices and their null spaces are intimately related, as the next two examples illustrate.

Example NSS
Null space of a singular matrix
Given the coefficient matrix from Archetype A,

A = \left [\array{ 1&−1&2 \cr 2& 1 &1 \cr 1& 1 &0 } \right ]

the null space is the set of solutions to the homogeneous system of equations ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ) has a solution set and null space constructed in Example HISAA as

N\kern -1.95872pt \left (A\right ) = \left \{\left [\array{ −{x}_{3} \cr {x}_{3} \cr {x}_{3} } \right ]\mathrel{∣}{x}_{3} ∈ {ℂ}^{}\right \}

Example NSNM
Null space of a nonsingular matrix
Given the coefficient matrix from Archetype B,

A = \left [\array{ −7&−6&−12 \cr 5 & 5 & 7 \cr 1 & 0 & 4 } \right ]

the homogeneous system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ) has a solution set constructed in Example HUSAB that contains only the trivial solution, so the null space has only a single element,

N\kern -1.95872pt \left (A\right ) = \left \{\left [\array{ 0 \cr 0 \cr 0 } \right ]\right \}

These two examples illustrate the next theorem, which is another equivalence.

Theorem NMTNS
Nonsingular Matrices have Trivial Null Spaces
Suppose that A is a square matrix. Then A is nonsingular if and only if the null space of A, N\kern -1.95872pt \left (A\right ), contains only the zero vector, i.e. N\kern -1.95872pt \left (A\right ) = \left \{0\right \}.

Proof   The null space of a square matrix, A, is equal to the set of solutions to the homogeneous system, ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ). A matrix is nonsingular if and only if the set of solutions to the homogeneous system, ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ), has only a trivial solution. These two observations may be chained together to construct the two proofs necessary for each half of this theorem.

The next theorem pulls a lot of big ideas together. Theorem NMUS tells us that we can learn much about solutions to a system of linear equations with a square coefficient matrix by just examining a similar homogeneous system.

Theorem NMUS
Nonsingular Matrices and Unique Solutions
Suppose that A is a square matrix. A is a nonsingular matrix if and only if the system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) has a unique solution for every choice of the constant vector b.

Proof   () The hypothesis for this half of the proof is that the system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) has a unique solution for every choice of the constant vector b. We will make a very specific choice for b: b = 0. Then we know that the system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ) has a unique solution. But this is precisely the definition of what it means for A to be nonsingular (Definition NM). That almost seems too easy! Notice that we have not used the full power of our hypothesis, but there is nothing that says we must use a hypothesis to its fullest.

() We assume that A is nonsingular of size n × n, so we know there is a sequence of row operations that will convert A into the identity matrix {I}_{n} (Theorem NMRRI). Form the augmented matrix {A}^{′} = \left [\left .A\kern 1.95872pt \right \vert \kern 1.95872pt b\right ] and apply this same sequence of row operations to {A}^{′}. The result will be the matrix {B}^{′} = \left [\left .{I}_{ n}\kern 1.95872pt \right \vert \kern 1.95872pt c\right ], which is in reduced row-echelon form with r = n. Then the augmented matrix {B}^{′} represents the (extremely simple) system of equations {x}_{i} ={ \left [c\right ]}_{i}, 1 ≤ i ≤ n. The vector c is clearly a solution, so the system is consistent (Definition CS). With a consistent system, we use Theorem FVCS to count free variables. We find that there are n − r = n − n = 0 free variables, and so we therefore know that the solution is unique. (This half of the proof was suggested by Asa Scherer.)

This theorem helps to explain part of our interest in nonsingular matrices. If a matrix is nonsingular, then no matter what vector of constants we pair it with, using the matrix as the coefficient matrix will always yield a linear system of equations with a solution, and the solution is unique. To determine if a matrix has this property (non-singularity) it is enough to just solve one linear system, the homogeneous system with the matrix as coefficient matrix and the zero vector as the vector of constants (or any other vector of constants, see Exercise MM.T10).

Formulating the negation of the second part of this theorem is a good exercise. A singular matrix has the property that for some value of the vector b, the system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) does not have a unique solution (which means that it has no solution or infinitely many solutions). We will be able to say more about this case later (see the discussion following Theorem PSPHS). Square matrices that are nonsingular have a long list of interesting properties, which we will start to catalog in the following, recurring, theorem. Of course, singular matrices will then have all of the opposite properties. The following theorem is a list of equivalences. We want to understand just what is involved with understanding and proving a theorem that says several conditions are equivalent. So have a look at Technique ME before studying the first in this series of theorems. 

Theorem NME1
Nonsingular Matrix Equivalences, Round 1
Suppose that A is a square matrix. The following are equivalent.

  1. A is nonsingular.
  2. A row-reduces to the identity matrix.
  3. The null space of A contains only the zero vector, N\kern -1.95872pt \left (A\right ) = \left \{0\right \}.
  4. The linear system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) has a unique solution for every possible choice of b.

Proof   That A is nonsingular is equivalent to each of the subsequent statements by, in turn, Theorem NMRRI, Theorem NMTNS and Theorem NMUS. So the statement of this theorem is just a convenient way to organize all these results.

Finally, you may have wondered why we refer to a matrix as nonsingular when it creates systems of equations with single solutions (Theorem NMUS)! I’ve wondered the same thing. We’ll have an opportunity to address this when we get to Theorem SMZD. Can you wait that long?

Subsection READ: Reading Questions

  1. What is the definition of a nonsingular matrix?
  2. What is the easiest way to recognize a nonsingular matrix?
  3. Suppose we have a system of equations and its coefficient matrix is nonsingular. What can you say about the solution set for this system?

Subsection EXC: Exercises

In Exercises C30–C33 determine if the matrix is nonsingular or singular. Give reasons for your answer.
C30

\left [\array{ −3& 1 &2& 8 \cr 2 & 0 &3& 4 \cr 1 & 2 &7&−4 \cr 5 &−1&2& 0 } \right ]

 
Contributed by Robert Beezer Solution [245]

C31

\left [\array{ 2 &3&1&4 \cr 1 &1&1&0 \cr −1&2&3&5 \cr 1 &2&1&3 } \right ]

 
Contributed by Robert Beezer Solution [245]

C32

\left [\array{ 9& 3 &2& 4 \cr 5&−6&1& 3 \cr 4& 1 &3&−5 } \right ]

 
Contributed by Robert Beezer Solution [246]

C33

\left [\array{ −1& 2 & 0 &3 \cr 1 &−3&−2&4 \cr −2& 0 & 4 &3 \cr −3& 1 &−2&3 } \right ]

 
Contributed by Robert Beezer Solution [246]

C40 Each of the archetypes below is a system of equations with a square coefficient matrix, or is itself a square matrix. Determine if these matrices are nonsingular, or singular. Comment on the null space of each matrix.
Archetype A
Archetype B
Archetype F
Archetype K
Archetype L  
Contributed by Robert Beezer

C50 Find the null space of the matrix E below.

\eqalignno{ E & = \left [\array{ 2 &1& −1 &−9 \cr 2 &2& −6 &−6 \cr 1 &2& −8 & 0 \cr −1&2&−12&12 } \right ] & & }

 
Contributed by Robert Beezer Solution [246]

M30 Let A be the coefficient matrix of the system of equations below. Is A nonsingular or singular? Explain what you could infer about the solution set for the system based only on what you have learned about A being singular or nonsingular.

\eqalignno{ − {x}_{1} + 5{x}_{2} & = −8 & & \cr − 2{x}_{1} + 5{x}_{2} + 5{x}_{3} + 2{x}_{4} & = 9 & & \cr − 3{x}_{1} − {x}_{2} + 3{x}_{3} + {x}_{4} & = 3 & & \cr 7{x}_{1} + 6{x}_{2} + 5{x}_{3} + {x}_{4} & = 30 & & }

 
Contributed by Robert Beezer Solution [247]

For Exercises M51–M52 say as much as possible about each system’s solution set. Be sure to make it clear which theorems you are using to reach your conclusions.
M51 6 equations in 6 variables, singular coefficient matrix.  
Contributed by Robert Beezer Solution [248]

M52 A system with a nonsingular coefficient matrix, not homogeneous.  
Contributed by Robert Beezer Solution [248]

T10 Suppose that A is a singular matrix, and B is a matrix in reduced row-echelon form that is row-equivalent to A. Prove that the last row of B is a zero row.  
Contributed by Robert Beezer Solution [248]

T30 Suppose that A is a nonsingular matrix and A is row-equivalent to the matrix B. Prove that B is nonsingular.  
Contributed by Robert Beezer Solution [249]

T90 Provide an alternative for the second half of the proof of Theorem NMUS, without appealing to properties of the reduced row-echelon form of the coefficient matrix. In other words, prove that if A is nonsingular, then ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) has a unique solution for every choice of the constant vector b. Construct this proof without using Theorem REMEF or Theorem RREFU.  
Contributed by Robert Beezer Solution [250]

Subsection SOL: Solutions

C30 Contributed by Robert Beezer Statement [240]
The matrix row-reduces to

\left [\array{ \text{1}&0&0&0 \cr 0&\text{1}&0&0 \cr 0&0&\text{1}&0 \cr 0&0&0&\text{1} } \right ]

which is the 4 × 4 identity matrix. By Theorem NMRRI the original matrix must be nonsingular.

C31 Contributed by Robert Beezer Statement [240]
Row-reducing the matrix yields,

\left [\array{ \text{1}&0&0&−2 \cr 0&\text{1}&0& 3 \cr 0&0&\text{1}&−1 \cr 0&0&0& 0 } \right ]

Since this is not the 4 × 4 identity matrix, Theorem NMRRI tells us the matrix is singular.

C32 Contributed by Robert Beezer Statement [241]
The matrix is not square, so neither term is applicable. See Definition NM, which is stated for just square matrices.

C33 Contributed by Robert Beezer Statement [241]
Theorem NMRRI tells us we can answer this question by simply row-reducing the matrix. Doing this we obtain,

\left [\array{ \text{1}&0&0&0 \cr 0&\text{1}&0&0 \cr 0&0&\text{1}&0 \cr 0&0&0&\text{1} } \right ]

Since the reduced row-echelon form of the matrix is the 4 × 4 identity matrix {I}_{4}, we know that B is nonsingular.

C50 Contributed by Robert Beezer Statement [242]
We form the augmented matrix of the homogeneous system ℒS\kern -1.95872pt \left (E,\kern 1.95872pt 0\right ) and row-reduce the matrix,

\eqalignno{ \left [\array{ 2 &1& −1 &−9&0 \cr 2 &2& −6 &−6&0 \cr 1 &2& −8 & 0 &0 \cr −1&2&−12&12&0 } \right ] &\mathop{\longrightarrow}\limits_{}^{\text{RREF}}\left [\array{ \text{1}&0& 2 &−6&0 \cr 0&\text{1}&−5& 3 &0 \cr 0&0& 0 & 0 &0 \cr 0&0& 0 & 0 &0 } \right ] & & }

We knew ahead of time that this system would be consistent (Theorem HSC), but we can now see there are n − r = 4 − 2 = 2 free variables, namely {x}_{3} and {x}_{4} since F = \left \{3, 4, 5\right \} (Theorem FVCS). Based on this analysis, we can rearrange the equations associated with each nonzero row of the reduced row-echelon form into an expression for the lone dependent variable as a function of the free variables. We arrive at the solution set to this homogeneous system, which is the null space of the matrix by Definition NSM,

\eqalignno{ N\kern -1.95872pt \left (E\right ) = \left \{\left [\array{ −2{x}_{3} + 6{x}_{4} \cr 5{x}_{3} − 3{x}_{4} \cr {x}_{3} \cr {x}_{4} } \right ]\mathrel{∣}{x}_{3},\kern 1.95872pt {x}_{4} ∈ ℂ\right \} & & }

M30 Contributed by Robert Beezer Statement [242]
We row-reduce the coefficient matrix of the system of equations,

\eqalignno{ \left [\array{ −1& 5 &0&0 \cr −2& 5 &5&2 \cr −3&−1&3&1 \cr 7 & 6 &5&1 } \right ] &\mathop{\longrightarrow}\limits_{}^{\text{RREF}}\left [\array{ \text{1}&0&0&0 \cr 0&\text{1}&0&0 \cr 0&0&\text{1}&0 \cr 0&0&0&\text{1} } \right ] & & }

Since the row-reduced version of the coefficient matrix is the 4 × 4 identity matrix, {I}_{4} (Definition IM byTheorem NMRRI, we know the coefficient matrix is nonsingular. According to Theorem NMUS we know that the system is guaranteed to have a unique solution, based only on the extra information that the coefficient matrix is nonsingular.

M51 Contributed by Robert Beezer Statement [243]
Theorem NMRRI tells us that the coefficient matrix will not row-reduce to the identity matrix. So if we were to row-reduce the augmented matrix of this system of equations, we would not get a unique solution. So by Theorem PSSLS the remaining possibilities are no solutions, or infinitely many.

M52 Contributed by Robert Beezer Statement [243]
Any system with a nonsingular coefficient matrix will have a unique solution by Theorem NMUS. If the system is not homogeneous, the solution cannot be the zero vector (Exercise HSE.T10).

T10 Contributed by Robert Beezer Statement [243]
Let n denote the size of the square matrix A. By Theorem NMRRI the hypothesis that A is singular implies that B is not the identity matrix {I}_{n}. If B has n pivot columns, then it would have to be {I}_{n}, so B must have fewer than n pivot columns. But the number of nonzero rows in B (r) is equal to the number of pivot columns as well. So the n rows of B have fewer than n nonzero rows, and B must contain at least one zero row. By Definition RREF, this row must be at the bottom of B.

T30 Contributed by Robert Beezer Statement [243]
Since A and B are row-equivalent matrices, consideration of the three row operations (Definition RO) will show that the augmented matrices, \left [\left .A\kern 1.95872pt \right \vert \kern 1.95872pt 0\right ] and \left [\left .B\kern 1.95872pt \right \vert \kern 1.95872pt 0\right ], are also row-equivalent matrices. This says that the two homogeneous systems, ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ) and ℒS\kern -1.95872pt \left (B,\kern 1.95872pt 0\right ) are equivalent systems. ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ) has only the zero vector as a solution (Definition NM), thus ℒS\kern -1.95872pt \left (B,\kern 1.95872pt 0\right ) has only the zero vector as a solution. Finally, by Definition NM, we see that B is nonsingular.

Form a similar theorem replacing “nonsingular” by “singular” in both the hypothesis and the conclusion. Prove this new theorem with an approach just like the one above, and/or employ the result about nonsingular matrices in a proof by contradiction.

T90 Contributed by Robert Beezer Statement [244]
We assume A is nonsingular, and try to solve the system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt b\right ) without making any assumptions about b. To do this we will begin by constructing a new homogeneous linear system of equations that looks very much like the original. Suppose A has size n (why must it be square?) and write the original system as,

\eqalignno{ {a}_{11}{x}_{1} + {a}_{12}{x}_{2} + {a}_{13}{x}_{3} + \mathrel{⋯} + {a}_{1n}{x}_{n} & = {b}_{1} & & \cr {a}_{21}{x}_{1} + {a}_{22}{x}_{2} + {a}_{23}{x}_{3} + \mathrel{⋯} + {a}_{2n}{x}_{n} & = {b}_{2} & & \cr {a}_{31}{x}_{1} + {a}_{32}{x}_{2} + {a}_{33}{x}_{3} + \mathrel{⋯} + {a}_{3n}{x}_{n} & = {b}_{3} & & \cr \mathop{\mathop{⋮}} & &\text{($∗$)} \cr {a}_{n1}{x}_{1} + {a}_{n2}{x}_{2} + {a}_{n3}{x}_{3} + \mathrel{⋯} + {a}_{nn}{x}_{n} & = {b}_{n} & & }

form the new, homogeneous system in n equations with n + 1 variables, by adding a new variable y, whose coefficients are the negatives of the constant terms,

\eqalignno{ {a}_{11}{x}_{1} + {a}_{12}{x}_{2} + {a}_{13}{x}_{3} + \mathrel{⋯} + {a}_{1n}{x}_{n} − {b}_{1}y & = 0 & & \cr {a}_{21}{x}_{1} + {a}_{22}{x}_{2} + {a}_{23}{x}_{3} + \mathrel{⋯} + {a}_{2n}{x}_{n} − {b}_{2}y & = 0 & & \cr {a}_{31}{x}_{1} + {a}_{32}{x}_{2} + {a}_{33}{x}_{3} + \mathrel{⋯} + {a}_{3n}{x}_{n} − {b}_{3}y & = 0 & & \cr \mathop{\mathop{⋮}} & &\text{($ ∗ ∗$)} \cr {a}_{n1}{x}_{1} + {a}_{n2}{x}_{2} + {a}_{n3}{x}_{3} + \mathrel{⋯} + {a}_{nn}{x}_{n} − {b}_{n}y & = 0 & & }

Since this is a homogeneous system with more variables than equations (m = n + 1 > n), Theorem HMVEI says that the system has infinitely many solutions. We will choose one of these solutions, any one of these solutions, so long as it is not the trivial solution. Write this solution as

\eqalignno{ {x}_{1} = {c}_{1} & &{x}_{2} = {c}_{2} & &{x}_{3} = {c}_{3} & &\mathop{\mathop{…}} & &{x}_{n} = {c}_{n} & &y = {c}_{n+1} & & & & & & & & & & & & }

We know that at least one value of the {c}_{i} is nonzero, but we will now show that in particular {c}_{n+1}\mathrel{≠}0. We do this using a proof by contradiction (Technique CD). So suppose the {c}_{i} form a solution as described, and in addition that {c}_{n+1} = 0. Then we can write the i-th equation of system (∗∗) as,

\eqalignno{ {a}_{i1}{c}_{1} + {a}_{i2}{c}_{2} + {a}_{i3}{c}_{3} + \mathrel{⋯} + {a}_{in}{c}_{n} − {b}_{i}(0) & = 0 & & \text{which becomes} \cr {a}_{i1}{c}_{1} + {a}_{i2}{c}_{2} + {a}_{i3}{c}_{3} + \mathrel{⋯} + {a}_{in}{c}_{n} & = 0 & & \cr & & }

Since this is true for each i, we have that {x}_{1} = {c}_{1},\kern 1.95872pt {x}_{2} = {c}_{2},\kern 1.95872pt {x}_{3} = {c}_{3},\mathop{\mathop{…}},\kern 1.95872pt {x}_{n} = {c}_{n} is a solution to the homogeneous system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ) formed with a nonsingular coefficient matrix. This means that the only possible solution is the trivial solution, so {c}_{1} = 0,\kern 1.95872pt {c}_{2} = 0,\kern 1.95872pt {c}_{3} = 0,\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {c}_{n} = 0. So, assuming simply that {c}_{n+1} = 0, we conclude that all of the {c}_{i} are zero. But this contradicts our choice of the {c}_{i} as not being the trivial solution to the system (∗∗). So {c}_{n+1}\mathrel{≠}0.

We now propose and verify a solution to the original system (∗). Set

\eqalignno{ {x}_{1} = {{c}_{1}\over { c}_{n+1}} & &{x}_{2} = {{c}_{2}\over { c}_{n+1}} & &{x}_{3} = {{c}_{3}\over { c}_{n+1}} & &\mathop{\mathop{…}} & &{x}_{n} = {{c}_{n}\over { c}_{n+1}} & & & & & & & & & & }

Notice how it was necessary that we know that {c}_{n+1}\mathrel{≠}0 for this step to succeed. Now, evaluate the i-th equation of system (∗) with this proposed solution, and recognize in the third line that {c}_{1} through {c}_{n+1} appear as if they were substituted into the left-hand side of the i-th equation of system (∗∗),

\eqalignno{ &{a}_{i1} {{c}_{1}\over { c}_{n+1}} + {a}_{i2} {{c}_{2}\over { c}_{n+1}} + {a}_{i3} {{c}_{3}\over { c}_{n+1}} + \mathrel{⋯} + {a}_{in} {{c}_{n}\over { c}_{n+1}} & & \cr & = {1\over { c}_{n+1}}\left ({a}_{i1}{c}_{1} + {a}_{i2}{c}_{2} + {a}_{i3}{c}_{3} + \mathrel{⋯} + {a}_{in}{c}_{n}\right ) & & \cr & = {1\over { c}_{n+1}}\left ({a}_{i1}{c}_{1} + {a}_{i2}{c}_{2} + {a}_{i3}{c}_{3} + \mathrel{⋯} + {a}_{in}{c}_{n} − {b}_{i}{c}_{n+1}\right ) + {b}_{i} & & \cr & = {1\over { c}_{n+1}}\left (0\right ) + {b}_{i} & & \cr & = {b}_{i} & & }

Since this equation is true for every i, we have found a solution to system (∗). To finish, we still need to establish that this solution is unique.

With one solution in hand, we will entertain the possibility of a second solution. So assume system (∗) has two solutions,

\eqalignno{ {x}_{1} = {d}_{1} & &{x}_{2} = {d}_{2} & &{x}_{3} = {d}_{3} & &\mathop{\mathop{…}} & &{x}_{n} = {d}_{n} & & & & & & & & & & \cr {x}_{1} = {e}_{1} & &{x}_{2} = {e}_{2} & &{x}_{3} = {e}_{3} & &\mathop{\mathop{…}} & &{x}_{n} = {e}_{n} & & & & & & & & & & }

Then,

\eqalignno{ &\left ({a}_{i1}({d}_{1} − {e}_{1}) + {a}_{i2}({d}_{2} − {e}_{2}) + {a}_{i3}({d}_{3} − {e}_{3}) + \mathrel{⋯} + {a}_{in}({d}_{n} − {e}_{n})\right ) & & \cr & = \left ({a}_{i1}{d}_{1} + {a}_{i2}{d}_{2} + {a}_{i3}{d}_{3} + \mathrel{⋯} + {a}_{in}{d}_{n}\right ) −\left ({a}_{i1}{e}_{1} + {a}_{i2}{e}_{2} + {a}_{i3}{e}_{3} + \mathrel{⋯} + {a}_{in}{e}_{n}\right ) & & \cr & = {b}_{i} − {b}_{i} & & \cr & = 0 & & }

This is the i-th equation of the homogeneous system ℒS\kern -1.95872pt \left (A,\kern 1.95872pt 0\right ) evaluated with {x}_{j} = {d}_{j} − {e}_{j}, 1 ≤ j ≤ n. Since A is nonsingular, we must conclude that this solution is the trivial solution, and so 0 = {d}_{j} − {e}_{j}, 1 ≤ j ≤ n. That is, {d}_{j} = {e}_{j} for all j and the two solutions are identical, meaning any solution to (∗) is unique.

Notice that the proposed solution ({x}_{i} = {{c}_{i}\over { c}_{n+1}}) appeared in this proof with no motivation whatsoever. This is just fine in a proof. A proof should convince you that a theorem is true. It is your job to read the proof and be convinced of every assertion. Questions like “Where did that come from?” or “How would I think of that?” have no bearing on the validity of the proof.

Annotated Acronyms SLE: Systems of Linear Equations

At the conclusion of each chapter you will find a section like this, reviewing selected definitions and theorems. There are many reasons for why a definition or theorem might be placed here. It might represent a key concept, it might be used frequently for computations, provide the critical step in many proofs, or it may deserve special comment.

These lists are not meant to be exhaustive, but should still be useful as part of reviewing each chapter. We will mention a few of these that you might eventually recognize on sight as being worth memorization. By that we mean that you can associate the acronym with a rough statement of the theorem — not that the exact details of the theorem need to be memorized. And it is certainly not our intent that everything on these lists is important enough to memorize.

Theorem RCLS
We will repeatedly appeal to this theorem to determine if a system of linear equations, does, or doesn’t, have a solution. This one we will see often enough that it is worth memorizing.

Theorem HMVEI
This theorem is the theoretical basis of several of our most important theorems. So keep an eye out for it, and its descendants, as you study other proofs. For example, Theorem HMVEI is critical to the proof of Theorem SSLD, Theorem SSLD is critical to the proof of Theorem G, Theorem G is critical to the proofs of the pair of similar theorems, Theorem ILTD and Theorem SLTD, while finally Theorem ILTD and Theorem SLTD are critical to the proof of an important result, Theorem IVSED. This chain of implications might not make much sense on a first reading, but come back later to see how some very important theorems build on the seemingly simple result that is Theorem HMVEI. Using the “find” feature in whatever software you use to read the electronic version of the text can be a fun way to explore these relationships.

Theorem NMRRI
This theorem gives us one of simplest ways, computationally, to recognize if a matrix is nonsingular, or singular. We will see this one often, in computational exercises especially.

Theorem NMUS
Nonsingular matrices will be an important topic going forward (witness the NMEx series of theorems). This is our first result along these lines, a useful theorem for other proofs, and also illustrates a more general concept from Chapter LT.