Section PEE  Properties of Eigenvalues and Eigenvectors

From A First Course in Linear Algebra
Version 2.99
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/

The previous section introduced eigenvalues and eigenvectors, and concentrated on their existence and determination. This section will be more about theorems, and the various properties eigenvalues and eigenvectors enjoy. Like a good 4 × 100  meter relay, we will lead-off with one of our better theorems and save the very best for the anchor leg.

Theorem EDELI
Eigenvectors with Distinct Eigenvalues are Linearly Independent
Suppose that A is an n × n square matrix and S = x1,x2,x3,,xp is a set of eigenvectors with eigenvalues λ1,λ2,λ3,,λp such that λiλj whenever ij. Then S is a linearly independent set.

Proof   If p = 1, then the set S = x1 is linearly independent since eigenvectors are nonzero (Definition EEM), so assume for the remainder that p 2.

We will prove this result by contradiction (Technique CD). Suppose to the contrary that S is a linearly dependent set. Define Si = x1,x2,x3,,xi and let k be an integer such that Sk1 = x1,x2,x3,,xk1 is linearly independent and Sk = x1,x2,x3,,xk is linearly dependent. We have to ask if there is even such an integer k? First, since eigenvectors are nonzero, the set x1 is linearly independent. Since we are assuming that S = Sp is linearly dependent, there must be an integer k, 2 k p, where the sets Si transition from linear independence to linear dependence (and stay that way). In other words, xk is the vector with the smallest index that is a linear combination of just vectors with smaller indices.

Since x1,x2,x3,,xk is linearly dependent there are scalars, a1,a2,a3,,ak, some non-zero (Definition LI), so that

0 = a1x1 + a2x2 + a3x3 + + akxk

Then,

0 = A λkIn 0  Theorem ZVSM = A λkIn a1x1 + a2x2 + a3x3 + + akxk  Definition RLD = A λkIn a1x1 + A λkIn a2x2 + + A λkIn akxk  Theorem MMDAA = a1 A λkIn x1 + a2 A λkIn x2 + + ak A λkIn xk  Theorem MMSMM = a1 Ax1 λkInx1 + a2 Ax2 λkInx2 + + ak Axk λkInxk Theorem MMDAA = a1 Ax1 λkx1 + a2 Ax2 λkx2 + + ak Axk λkxk  Theorem MMIM = a1 λ1x1 λkx1 + a2 λ2x2 λkx2 + + ak λkxk λkxk  Definition EEM = a1 λ1 λk x1 + a2 λ2 λk x2 + + ak λk λk xk  Theorem MMDAA = a1 λ1 λk x1 + a2 λ2 λk x2 + + ak 0 xk  Property AICN = a1 λ1 λk x1 + a2 λ2 λk x2 + + ak1 λk1 λk xk1 + 0  Theorem ZSSM = a1 λ1 λk x1 + a2 λ2 λk x2 + + ak1 λk1 λk xk1  Property Z

This is a relation of linear dependence on the linearly independent set x1,x2,x3,,xk1, so the scalars must all be zero. That is, ai λi λk = 0 for 1 i k 1. However, we have the hypothesis that the eigenvalues are distinct, so λiλk for 1 i k 1. Thus ai = 0 for 1 i k 1.

This reduces the original relation of linear dependence on x1,x2,x3,,xk to the simpler equation akxk = 0. By Theorem SMEZV we conclude that ak = 0 or xk = 0. Eigenvectors are never the zero vector (Definition EEM), so ak = 0. So all of the scalars ai, 1 i k are zero, contradicting their introduction as the scalars creating a nontrivial relation of linear dependence on the set x1,x2,x3,,xk. With a contradiction in hand, we conclude that S must be linearly independent.

There is a simple connection between the eigenvalues of a matrix and whether or not the matrix is nonsingular.

Theorem SMZE
Singular Matrices have Zero Eigenvalues
Suppose A is a square matrix. Then A is singular if and only if λ = 0 is an eigenvalue of A.

Proof   We have the following equivalences:

 A is singular  there exists x0Ax = 0  Definition NSM  there exists x0Ax = 0x  Theorem ZSSM  λ = 0 is an eigenvalue of A  Definition EEM

With an equivalence about singular matrices we can update our list of equivalences about nonsingular matrices.

Theorem NME8
Nonsingular Matrix Equivalences, Round 8
Suppose that A is a square matrix of size n. 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, NA = 0.
  4. The linear system SA,b has a unique solution for every possible choice of b.
  5. The columns of A are a linearly independent set.
  6. A is invertible.
  7. The column space of A is n, CA = n.
  8. The columns of A are a basis for n.
  9. The rank of A is n, r A = n.
  10. The nullity of A is zero, n A = 0.
  11. The determinant of A is nonzero, det A0.
  12. λ = 0 is not an eigenvalue of A.

Proof   The equivalence of the first and last statements is the contrapositive of Theorem SMZE, so we are able to improve on Theorem NME7.

Certain changes to a matrix change its eigenvalues in a predictable way.

Theorem ESMM
Eigenvalues of a Scalar Multiple of a Matrix
Suppose A is a square matrix and λ is an eigenvalue of A. Then αλ is an eigenvalue of αA.

Proof   Let x0 be one eigenvector of A for λ. Then

αAx = α Ax  Theorem MMSMM = α λx  x eigenvector of A = αλx  Property SMAC

So x0 is an eigenvector of αA for the eigenvalue αλ.

Unfortunately, there are not parallel theorems about the sum or product of arbitrary matrices. But we can prove a similar result for powers of a matrix.

Theorem EOMP
Eigenvalues Of Matrix Powers
Suppose A is a square matrix, λ is an eigenvalue of A, and s 0 is an integer. Then λs is an eigenvalue of As.

Proof   Let x0 be one eigenvector of A for λ. Suppose A has size n. Then we proceed by induction on s (Technique I). First, for s = 0,

Asx = A0x = Inx = x  Theorem MMIM = 1x  Property OC = λ0x = λsx

so λs is an eigenvalue of As in this special case. If we assume the theorem is true for s, then we find

As+1x = AsAx = As λx  x eigenvector of A for λ = λ Asx  Theorem MMSMM = λ λsx  Induction hypothesis = λλs x  Property SMAC = λs+1x

So x0 is an eigenvector of As+1 for λs+1, and induction tells us the theorem is true for all s 0.

While we cannot prove that the sum of two arbitrary matrices behaves in any reasonable way with regard to eigenvalues, we can work with the sum of dissimilar powers of the same matrix. We have already seen two connections between eigenvalues and polynomials, in the proof of Theorem EMHE and the characteristic polynomial (Definition CP). Our next theorem strengthens this connection.

Theorem EPM
Eigenvalues of the Polynomial of a Matrix
Suppose A is a square matrix and λ is an eigenvalue of A. Let q(x) be a polynomial in the variable x. Then q(λ) is an eigenvalue of the matrix q(A).

Proof   Let x0 be one eigenvector of A for λ, and write q(x) = a0 + a1x + a2x2 + + a mxm. Then

q(A)x = a0A0 + a 1A1 + a 2A2 + + a mAm x = (a0A0)x + (a 1A1)x + (a 2A2)x + + (a mAm)x Theorem MMDAA = a0(A0x) + a 1(A1x) + a 2(A2x) + + a m(Amx) Theorem MMSMM = a0(λ0x) + a 1(λ1x) + a 2(λ2x) + + a m(λmx)  Theorem EOMP = (a0λ0)x + (a 1λ1)x + (a 2λ2)x + + (a mλm)x  Property SMAC = a0λ0 + a 1λ1 + a 2λ2 + + a mλm x  Property DSAC = q(λ)x

So x0 is an eigenvector of q(A) for the eigenvalue q(λ).

Example BDE
Building desired eigenvalues
In Example ESMS4 the 4 × 4 symmetric matrix

C = 1011 0 1 1 1 1 110 1 1 0 1

is shown to have the three eigenvalues λ = 3,1, 1. Suppose we wanted a 4 × 4 matrix that has the three eigenvalues λ = 4,0, 2. We can employ Theorem EPM by finding a polynomial that converts 3 to 4, 1 to 0, and 1 to 2. Such a polynomial is called an interpolating polynomial, and in this example we can use

r(x) = 1 4x2 + x 5 4

We will not discuss how to concoct this polynomial, but a text on numerical analysis should provide the details or see Section CF. For now, simply verify that r(3) = 4, r(1) = 0 and r(1) = 2.

Now compute

r(C) = 1 4C2 + C 5 4I4 = 1 4 3222 2 3 2 2 2 232 2 2 2 3 + 1011 0 1 1 1 1 110 1 1 0 1 5 4 1000 0 1 0 0 0 010 0 0 0 1 = 1 2 1133 1 1 3 3 3 311 3 3 1 1

Theorem EPM tells us that if r(x) transforms the eigenvalues in the desired manner, then r(C) will have the desired eigenvalues. You can check this by computing the eigenvalues of r(C) directly. Furthermore, notice that the multiplicities are the same, and the eigenspaces of C and r(C) are identical.

Inverses and transposes also behave predictably with regard to their eigenvalues.

Theorem EIM
Eigenvalues of the Inverse of a Matrix
Suppose A is a square nonsingular matrix and λ is an eigenvalue of A. Then 1 λ is an eigenvalue of the matrix A1.

Proof   Notice that since A is assumed nonsingular, A1 exists by Theorem NI, but more importantly, 1 λ does not involve division by zero since Theorem SMZE prohibits this possibility.

Let x0 be one eigenvector of A for λ. Suppose A has size n. Then

A1x = A1(1x)  Property OC = A1(1 λλx)  Property MICN = 1 λA1(λx)  Theorem MMSMM = 1 λA1(Ax)  Definition EEM = 1 λ(A1A)x  Theorem MMA = 1 λInx  Definition MI = 1 λx  Theorem MMIM

So x0 is an eigenvector of A1 for the eigenvalue 1 λ.

The theorems above have a similar style to them, a style you should consider using when confronted with a need to prove a theorem about eigenvalues and eigenvectors. So far we have been able to reserve the characteristic polynomial for strictly computational purposes. However, the next theorem, whose statement resembles the preceding theorems, has an easier proof if we employ the characteristic polynomial and results about determinants.

Theorem ETM
Eigenvalues of the Transpose of a Matrix
Suppose A is a square matrix and λ is an eigenvalue of A. Then λ is an eigenvalue of the matrix At.

Proof   Suppose A has size n. Then

pA x = det A xIn  Definition CP = det A xIn t  Theorem DT = det At xI n t  Theorem TMA = det At xI nt  Theorem TMSM = det At xI n  Definition IM = pAt x  Definition CP

So A and At have the same characteristic polynomial, and by Theorem EMRCP, their eigenvalues are identical and have equal algebraic multiplicities. Notice that what we have proved here is a bit stronger than the stated conclusion in the theorem.

If a matrix has only real entries, then the computation of the characteristic polynomial (Definition CP) will result in a polynomial with coefficients that are real numbers. Complex numbers could result as roots of this polynomial, but they are roots of quadratic factors with real coefficients, and as such, come in conjugate pairs. The next theorem proves this, and a bit more, without mentioning the characteristic polynomial.

Theorem ERMCP
Eigenvalues of Real Matrices come in Conjugate Pairs
Suppose A is a square matrix with real entries and x is an eigenvector of A for the eigenvalue λ. Then x¯ is an eigenvector of A for the eigenvalue λ¯.

Proof  

Ax¯ = A¯x¯  A has real entries = Ax¯  Theorem MMCC = λx¯  x eigenvector of A = λ¯x¯  Theorem CRSM

So x¯ is an eigenvector of A for the eigenvalue λ¯.

This phenomenon is amply illustrated in Example CEMS6, where the four complex eigenvalues come in two pairs, and the two basis vectors of the eigenspaces are complex conjugates of each other. Theorem ERMCP can be a time-saver for computing eigenvalues and eigenvectors of real matrices with complex eigenvalues, since the conjugate eigenvalue and eigenspace can be inferred from the theorem rather than computed.

Subsection ME: Multiplicities of Eigenvalues

A polynomial of degree n will have exactly n roots. From this fact about polynomial equations we can say more about the algebraic multiplicities of eigenvalues.

Theorem DCP
Degree of the Characteristic Polynomial
Suppose that A is a square matrix of size n. Then the characteristic polynomial of A, pA x, has degree n.

Proof   We will prove a more general result by induction (Technique I). Then the theorem will be true as a special case. We will carefully state this result as a proposition indexed by m, m 1.

P(m): Suppose that A is an m × m matrix whose entries are complex numbers or linear polynomials in the variable x of the form c x, where c is a complex number. Suppose further that there are exactly k entries that contain x and that no row or column contains more than one such entry. Then, when k = m, det A is a polynomial in x of degree m, with leading coefficient ± 1, and when k < m, det A is a polynomial in x of degree k or less.

Base Case: Suppose A is a 1 × 1 matrix. Then its determinant is equal to the lone entry (Definition DM). When k = m = 1, the entry is of the form c x, a polynomial in x of degree m = 1 with leading coefficient 1. When k < m, then k = 0 and the entry is simply a complex number, a polynomial of degree 0 k. So P(1) is true.

Induction Step: Assume P(m) is true, and that A is an (m + 1) × (m + 1) matrix with k entries of the form c x. There are two cases to consider.

Suppose k = m + 1. Then every row and every column will contain an entry of the form c x. Suppose that for the first row, this entry is in column t. Compute the determinant of A by an expansion about this first row (Definition DM). The term associated with entry t of this row will be of the form

(c x)(1)1+t det A 1|t

The submatrix A 1|t is an m × m matrix with k = m terms of the form c x, no more than one per row or column. By the induction hypothesis, det A 1|t will be a polynomial in x of degree m with coefficient ± 1. So this entire term is then a polynomial of degree m + 1 with leading coefficient ± 1.

The remaining terms (which constitute the sum that is the determinant of A) are products of complex numbers from the first row with cofactors built from submatrices that lack the first row of A and lack some column of A, other than column t. As such, these submatrices are m × m matrices with k = m 1 < m entries of the form c x, no more than one per row or column. Applying the induction hypothesis, we see that these terms are polynomials in x of degree m 1 or less. Adding the single term from the entry in column t with all these others, we see that det A is a polynomial in x of degree m + 1 and leading coefficient ± 1.

The second case occurs when k < m + 1. Now there is a row of A that does not contain an entry of the form c x. We consider the determinant of A by expanding about this row (Theorem DER), whose entries are all complex numbers. The cofactors employed are built from submatrices that are m × m matrices with either k or k 1 entries of the form c x, no more than one per row or column. In either case, k m, and we can apply the induction hypothesis to see that the determinants computed for the cofactors are all polynomials of degree k or less. Summing these contributions to the determinant of A yields a polynomial in x of degree k or less, as desired.

Definition CP tells us that the characteristic polynomial of an n × n matrix is the determinant of a matrix having exactly n entries of the form c x, no more than one per row or column. As such we can apply P(n) to see that the characteristic polynomial has degree n.

Theorem NEM
Number of Eigenvalues of a Matrix
Suppose that A is a square matrix of size n with distinct eigenvalues λ1,λ2,λ3,,λk. Then

i=1kα A λi = n

Proof   By the definition of the algebraic multiplicity (Definition AME), we can factor the characteristic polynomial as

pA x = c(x λ1)αAλ1 (x λ2)αAλ2 (x λ3)αAλ3 (x λk)αAλk

where c is a nonzero constant. (We could prove that c = (1)n, but we do not need that specificity right now. See Exercise PEE.T30) The left-hand side is a polynomial of degree n by Theorem DCP and the right-hand side is a polynomial of degree i=1kα A λi. So the equality of the polynomials’ degrees gives the equality i=1kα A λi = n.

Theorem ME
Multiplicities of an Eigenvalue
Suppose that A is a square matrix of size n and λ is an eigenvalue. Then

1 γA λ αA λ n

Proof   Since λ is an eigenvalue of A, there is an eigenvector of A for λ, x. Then x A λ, so γA λ 1, since we can extend x into a basis of A λ (Theorem ELIS).

To show that γA λ αA λ is the most involved portion of this proof. To this end, let g = γA λ and let x1,x2,x3,,xg be a basis for the eigenspace of λ, A λ. Construct another n g vectors, y1,y2,y3,,yng, so that

x1,x2,x3,,xg,y1,y2,y3,,yng

is a basis of n. This can be done by repeated applications of Theorem ELIS. Finally, define a matrix S by

S = [x1|x2|x3||xg|y1|y2|y3||yng] = [x1|x2|x3||xg|R]

where R is an n × (n g) matrix whose columns are y1,y2,y3,,yng. The columns of S are linearly independent by design, so S is nonsingular (Theorem NMLIC) and therefore invertible (Theorem NI). Then,

e1|e2|e3||en = In = S1S = S1[x 1|x2|x3||xg|R] = [S1x 1|S1x 2|S1x 3||S1x g|S1R]

So

S1x i = ei1 i g ()

Preparations in place, we compute the characteristic polynomial of A,

pA x = det A xIn  Definition CP = 1 det A xIn  Property OCN = det In det A xIn  Definition DM = det S1S det A xI n  Definition MI = det S1 det S det A xI n  Theorem DRMM = det S1 det A xI n det S  Property CMCN = det S1 A xI n S  Theorem DRMM = det S1AS S1xI nS  Theorem MMDAA = det S1AS xS1I nS  Theorem MMSMM = det S1AS xS1S  Theorem MMIM = det S1AS xI n  Definition MI = pS1AS x  Definition CP

What can we learn then about the matrix S1AS?

S1AS = S1A[x 1|x2|x3||xg|R] = S1[Ax 1|Ax2|Ax3||Axg|AR]  Definition MM = S1[λx 1|λx2|λx3||λxg|AR]  Definition EEM = [S1λx 1|S1λx 2|S1λx 3||S1λx g|S1AR] Definition MM = [λS1x 1|λS1x 2|λS1x 3||λS1x g|S1AR] Theorem MMSMM = [λe1|λe2|λe3||λeg|S1AR]  S1S = I n, (() above)

Now imagine computing the characteristic polynomial of A by computing the characteristic polynomial of S1AS using the form just obtained. The first g columns of S1AS are all zero, save for a λ on the diagonal. So if we compute the determinant by expanding about the first column, successively, we will get successive factors of (λ x). More precisely, let T be the square matrix of size n g that is formed from the last n g rows and last n g columns of S1AR. Then

pA x = pS1AS x = (λ x)gp T x.

This says that (x λ) is a factor of the characteristic polynomial at least g times, so the algebraic multiplicity of λ as an eigenvalue of A is greater than or equal to g (Definition AME). In other words,

γA λ = g αA λ

as desired.

Theorem NEM says that the sum of the algebraic multiplicities for all the eigenvalues of A is equal to n. Since the algebraic multiplicity is a positive quantity, no single algebraic multiplicity can exceed n without the sum of all of the algebraic multiplicities doing the same.

Theorem MNEM
Maximum Number of Eigenvalues of a Matrix
Suppose that A is a square matrix of size n. Then A cannot have more than n distinct eigenvalues.

Proof   Suppose that A has k distinct eigenvalues, λ1,λ2,λ3,,λk. Then

k = i=1k1 i=1kα A λi  Theorem ME = n  Theorem NEM

Subsection EHM: Eigenvalues of Hermitian Matrices

Recall that a matrix is Hermitian (or self-adjoint) if A = A (Definition HM). In the case where A is a matrix whose entries are all real numbers, being Hermitian is identical to being symmetric (Definition SYM). Keep this in mind as you read the next two theorems. Their hypotheses could be changed to “suppose A is a real symmetric matrix.”

Theorem HMRE
Hermitian Matrices have Real Eigenvalues
Suppose that A is a Hermitian matrix and λ is an eigenvalue of A. Then λ .

Proof   Let x0 be one eigenvector of A for the eigenvalue λ. Then by Theorem PIP we know x,x0. So

λ = 1 x,xλ x,x  Property MICN = 1 x,x x,λx  Theorem IPSM = 1 x,x x,Ax  Definition EEM = 1 x,x Ax,x  Theorem HMIP = 1 x,x λx,x  Definition EEM = 1 x,xλ¯ x,x  Theorem IPSM = λ¯  Property MICN

If a complex number is equal to its conjugate, then it has a complex part equal to zero, and therefore is a real number.

Notice the appealing symmetry to the justifications given for the steps of this proof. In the center is the ability to pitch a Hermitian matrix from one side of the inner product to the other.

Look back and compare Example ESMS4 and Example CEMS6. In Example CEMS6 the matrix has only real entries, yet the characteristic polynomial has roots that are complex numbers, and so the matrix has complex eigenvalues. However, in Example ESMS4, the matrix has only real entries, but is also symmetric, and hence Hermitian. So by Theorem HMRE, we were guaranteed eigenvalues that are real numbers.

In many physical problems, a matrix of interest will be real and symmetric, or Hermitian. Then if the eigenvalues are to represent physical quantities of interest, Theorem HMRE guarantees that these values will not be complex numbers.

The eigenvectors of a Hermitian matrix also enjoy a pleasing property that we will exploit later.

Theorem HMOE
Hermitian Matrices have Orthogonal Eigenvectors
Suppose that A is a Hermitian matrix and x and y are two eigenvectors of A for different eigenvalues. Then x and y are orthogonal vectors.

Proof   Let x be an eigenvector of A for λ and let y be an eigenvector of A for a different eigenvalue ρ. So we have λ ρ0. Then

x,y = 1 λ ρ λ ρ x,y  Property MICN = 1 λ ρ λ x,y ρ x,y  Property MICN = 1 λ ρ λ¯x,y x,ρy  Theorem IPSM = 1 λ ρ λx,y x,ρy  Theorem HMRE = 1 λ ρ Ax,y x,Ay  Definition EEM = 1 λ ρ Ax,y Ax,y  Theorem HMIP = 1 λ ρ 0  Property AICN = 0

This equality says that x and y are orthogonal vectors (Definition OV).

Notice again how the key step in this proof is the fundamental property of a Hermitian matrix (Theorem HMIP) — the ability to swap A across the two arguments of the inner product. We’ll build on these results and continue to see some more interesting properties in Section OD.

Subsection READ: Reading Questions

  1. How can you identify a nonsingular matrix just by looking at its eigenvalues?
  2. How many different eigenvalues may a square matrix of size n have?
  3. What is amazing about the eigenvalues of a Hermitian matrix and why is it amazing?

Subsection EXC: Exercises

T10 Suppose that A is a square matrix. Prove that the constant term of the characteristic polynomial of A is equal to the determinant of A.  
Contributed by Robert Beezer Solution [1326]

T20 Suppose that A is a square matrix. Prove that a single vector may not be an eigenvector of A for two different eigenvalues.  
Contributed by Robert Beezer Solution [1327]

T22 Suppose that U is a unitary matrix with eigenvalue λ. Prove that λ had modulus 1, i.e.  λ = 1. This says that all of the eigenvalues of a unitary matrix lie on the unit circle of the complex plane.  
Contributed by Robert Beezer

T30 Theorem DCP tells us that the characteristic polynomial of a square matrix of size n has degree n. By suitably augmenting the proof of Theorem DCP prove that the coefficient of xn in the characteristic polynomial is (1)n.  
Contributed by Robert Beezer

T50 Theorem EIM says that if λ is an eigenvalue of the nonsingular matrix A, then 1 λ is an eigenvalue of A1. Write an alternate proof of this theorem using the characteristic polynomial and without making reference to an eigenvector of A for λ.  
Contributed by Robert Beezer Solution [1327]

Subsection SOL: Solutions

T10 Contributed by Robert Beezer Statement [1324]
Suppose that the characteristic polynomial of A is

pA x = a0 + a1x + a2x2 + + a nxn

Then

a0 = a0 + a1(0) + a2(0)2 + + a n(0)n = pA 0 = det A 0In  Definition CP = det A

T20 Contributed by Robert Beezer Statement [1324]
Suppose that the vector x0 is an eigenvector of A for the two eigenvalues λ and ρ, where λρ. Then λ ρ0, and we also have

0 = Ax Ax  Property AIC = λx ρx  Definition EEM = (λ ρ)x  Property DSAC

By Theorem SMEZV, either λ ρ = 0 or x = 0, which are both contradictions.

T50 Contributed by Robert Beezer Statement [1324]
Since λ is an eigenvalue of a nonsingular matrix, λ0 (Theorem SMZE). A is invertible (Theorem NI), and so λA is invertible (Theorem MISM). Thus λA is nonsingular (Theorem NI) and det λA0 (Theorem SMZD).

pA1 1 λ = det A1 1 λIn  Definition CP = 1 det A1 1 λIn  Property OCN = 1 det λA det λA det A1 1 λIn  Property MICN = 1 det λA det λA A1 1 λIn  Theorem DRMM = 1 det λA det λAA1 λA 1 λIn  Theorem MMDAA = 1 det λA det λIn λA 1 λIn  Definition MI = 1 det λA det λIn + λ1 λAIn  Theorem MMSMM = 1 det λA det λIn + 1AIn  Property MICN = 1 det λA det λIn + AIn  Property OCN = 1 det λA det λIn + A  Theorem MMIM = 1 det λA det A λIn  Property ACM = 1 det λApA λ  Definition CP = 1 det λA0  Theorem EMRCP = 0  Property ZCN

So 1 λ is a root of the characteristic polynomial of A1 and so is an eigenvalue of A1. This proof is due to Sara Bucht.