Adjoint of a Matrix

In today's blog, I present the proof that an n x n matrix is invertible if and only if its determinant is nonzero.

The content in today's blog is taken from Hans Schneider and George Phillip Barker's Matrices and Linear Algebra.

Lemma 1:

Let A be an n x n matrix.

If i ≠ k, then:

∑ (j=1,n) (a_i,j)cof_k,j(A) = 0

Proof:

(1) Assume that k ≠ i.

(2) Let B be the matrix given by:

if h ≠ k, then Row_h(B) = Row_h(A).

if h = k, then Row_h(B) = Row_i(A) where k ≠ i.

(3) It is clear that B contains two identical rows (Row_i and Row_k). A does not necessarily contain any identical rows.

(4) We can see that (see Definition 1, here for definition of Minor M_i,j):

M_k,j(A) = M_k,j(B)

This is true since M_k,j does not include Row_k or Column_j. A,B are only different at Row_k.

(5) Now, by the definition of Cofactor (see Definition 2, here for definition of cofactor):

cof_k,j of B = (-1)^k+jdet(M_k,j(B)) =

(-1)^k+jdet(M_k,j(A))

(6) Since Row_k(B) = Row_i(A), we have:

∑ (j=1,n) b_k,j = ∑ (j=1,n) a_i,j

(7) Now using row k for the cofactor expansion of B, we have (see Theorem 4, here):

det B = ∑ (j=1,n) b_k,jcof_k,j(B)

(8) Since M_k,j(A) = M_k,j(B) and since cof_k,j(A) = (-1)^k+jdet(M_k,j(A)), we have:

∑ (j=1,n) b_k,jcof_k,j(B) = ∑ (j=1,n) b_k,jcof_k,j(A)

(9) Since Row_k(B) = Row_i(A), we have:

∑ (j=1,n) b_k,jcof_k,j(A) = ∑ (j=1,n) a_i,jcof_k,j(A)

(10) Since B includes two duplicate rows: row_i and row_k, it is clear that det(B) = 0. [See Corollary 1.1R, here]

(11) Thus, we are left with:

∑ (j=1,n) a_i,jcof_k,j(A) = 0.

QED

Lemma 2:

Let A be an n x n matrix.

If j ≠ l, then:

∑ (i=1,n) (a_i,j)cofactor_i,l(A) = 0

Proof:

(1) Assume that j ≠ l.

(2) Let B be the matrix given by:

if h ≠ k, then Column_h(B) = Column_h(A).

if h = k, then Column_h(B) = Column_i(A) where k ≠ i.

(3) It is clear that B contains two identical columns (Column_i and Column_k). A does not necessarily contain any identical columns.

(4) We can see that (see Definition 1, here for definition of Minor M_i,j):

M_i,k(A) = M_i,k(B)

This is true since M_i,k does not include Column_k or Row_i. A,B are only different at Column_k.

(5) Now, by the definition of Cofactor (see Definition 2, here for definition of cofactor):

cof_i,k of B = (-1)^i+kdet(M_i,k(B)) =

(-1)^i+kdet(M_i,k(A))

(6) Since Column_k(B) = Column_i(A), we have:

∑ (l=1,n) b_l,k = ∑ (l=1,n) a_l,i

(7) Now using column k for the cofactor expansion of B, we have (see Corollary 4.1, here):

det B = ∑ (l=1,n) b_l,kcof_l,k(B)

(8) Since M_i,k(A) = M_i,k(B) and since cof_i,k(A) = (-1)^i+kdet(M_i,k(A), we have:

∑ (l=1,n) b_l,kcof_l,k(B) = ∑ (l=1,n) b_l,kcof_l,k(A)

(9) Since Column_k(B) = Column_i(A), we have:

∑ (l=1,n) b_l,kcof_l,k(A) = ∑ (l=1,n) a_l,icof_l,k(A)

(10) Since B includes two duplicate columns: column_i and column_k, it is clear that det(B) = 0. [See Corollary 1.1C, here]

(11) Thus, we are left with:

∑ (l=1,n) a_l,icof_l,k(A) = 0.

QED

Definition 1: Adjoint of a matrix

Let A be an n x n matrix. The adjoint of A, written adj A, is the n x n matrix C, where c_i,j = cofactor_j,i, the cofactor of a_j,i in det A.

Theorem 3:

Let A be an n x n matrix. Then A(adj A) = (det A)I and (adj A)A = (detA)I

Proof:

(1) Let D = A(adj A)

(2) Then,

d_i,k = ∑ (j=1,n) a_i,jcof_k,j(A).

[See Definition 1, here for definition of matrix multiplication; see definition 1 above for definition of adjoint]

(3) Now, if i ≠ k, then from Lemma 1 above:

∑ (j=1,n) (a_i,j)cof_k,j(A) = 0

(4) If i = k, then from Theorem 4, here:

det(A) = ∑(i=1,n) a_i,jcof_i,j(A)

(5) Putting this together, we have:

D = A(adj A) =



= det(A)I

(6) Let D = (adj A)A

(7) Then,

d_i,k = ∑ (j=1,n) cof_k,j(A) a_i,k

[See Definition 1, here for definition of matrix multiplication; see definition 1 above for definition of adjoint]

(8) Now, if i ≠ k, then from Lemma 2 above:

∑ (j=1,n) cof_k,j(A)a_i,k = 0

(9) If i = k, then from Theorem 4, here:

∑ (j=1,n) cof_i,j(A)a_i,i = det(A)

(10) Putting this together, we have:

D = (adj A)A =



= det(A)I


QED

Theorem 4: Criteria for Invertibility

An n x n matrix A is invertible if and only if det(A) ≠ 0

Proof:

(1) Assume that an n x n matrix A is invertible.

[See Definition 1, here for definition of invertible]

(2) Then AA^-1 = I

(3) Thus, det(AA^-1) = det(A)det(A^-1) = det(I)

[See Definition 4, here for definition of the determinant function: det(A)]

(4) Since det(I) = 1 [see Corollary 4.3, here], this gives us that:

det(A)*det(A^-1) = 1.

(5) Then det(A) ≠ 0 since 0*det(A^-1) ≠ 1.

(6) Assume that for an n x n matrix A, det(A) ≠ 0

(7) Let X = (det A)^-1(adj A) [We know that this value exists since det(A) ≠ 0. ]

(8) Multiplying A to the left of both sides of equation gives us:

XA = (det A)^-1(adj A)A

(9) Using Theorem 3 above, we know that (adj A)A = (det A)*I

(10) So that we have:

XA = (det A)^-1(det A)*I = 1*I = I

(11) Multplying A to the right of each side of the equation gives us:

AX = A(det A)^-1(adj A)

(12) Using a property 3 of matrix multiplication (see Property 3, here), we have:

AX = (det A)^-1A(adj A)

(13) Using Theorem 3 above, gives us:

AX = (det A)^-1(det A)I = I

(14) So that we have AX = XA = I and by definition of inverse (see Definition 1, here), we can see that X = A^-1.

QED

Corollary 4.1:

If det(A) ≠ 0, then A^-1 = 1/(det A)*(adj A)

Proof:

(1) If det(A) ≠ 0, then A is invertible from Theorem 4 above.

(2) From Theorem 3 above:

(adj A)A = (detA)I

(3) Multiplying A^-1 to both sides gives:

(adj A)AA^-1 =(det A)I*A^-1

(adj A) = (det A)I*A^-1

1/(det A)(adj A) = 1/(det A)*(det A)*I*A^-1 = 1*I*A^-1 = I*A^-1 = A^-1

QED

References

• Hans Schneider, George Philip Barker, Matrices and Linear Algebra, 1989.
• Charles G. Cullen, Matrices and Linear Transformations, Dover Publications, Inc., 1972.