1.3. SCHUR’S THEOREM 13

Definition 1.3.1 A complex n× n matrix U is said to be unitary if U∗U = I. HereU∗ is the transpose of the conjugate of U. The matrix is unitary if and only if its columnsform an orthonormal set in Cn. This follows from the way we multiply matrices in whichthe i jth entry of U∗U is obtained by taking the conjugate of the ith row of U times the jth

column of U.

Theorem 1.3.2 (Schur) Let A be a complex n×n matrix. Then there exists a unitarymatrix U such that

U∗AU = T, (1.6)

where T is an upper triangular matrix having the eigenvalues of A on the main diagonal,listed with multiplicity1.

Proof: The theorem is clearly true if A is a 1×1 matrix. Just let U = 1, the 1×1 matrixwhich has entry 1. Suppose it is true for (n−1)× (n−1) matrices and let A be an n× nmatrix. Then let v1 be a unit eigenvector for A. Then there exists λ 1 such that

Av1 = λ 1v1, |v1|= 1.

Extend {v1} to a basis and then use the Gram - Schmidt process to obtain

{v1, · · · ,vn}

an orthonormal basis of Cn. Let U0 be a matrix whose ith column is vi. Then from thedefinition of a unitary matrix Definition 1.3.1, it follows that U0 is unitary. Consider U∗0 AU0.

U∗0 AU0 =

 v∗1...v∗n

( Av1 · · · Avn)=

 v∗1...v∗n

( λ 1v1 · · · Avn)

Thus U∗0 AU0 is of the form (λ 1 a0 A1

)where A1 is an n− 1× n− 1 matrix. Now by induction, there exists an (n−1)× (n−1)unitary matrix Ũ1 such that

Ũ∗1 A1Ũ1 = Tn−1,

an upper triangular matrix. Consider

U1 ≡(

1 0

0 Ũ1

).

An application of block multiplication shows that U1 is a unitary matrix and also that

U∗1 U∗0 AU0U1 =

(1 0

0 Ũ∗1

)(λ 1 ∗0 A1

)(1 0

0 Ũ1

)=

(λ 1 ∗0 Tn−1

)= T

where T is upper triangular. Then let U = U0U1. Since (U0U1)∗ = U∗1 U∗0 , it follows that

A is similar to T and that U0U1 is unitary. Hence A and T have the same characteristic1‘Listed with multiplicity’ means that the diagonal entries are repeated according to their multiplicity as roots

of the characteristic equation.

1.3. SCHUR’S THEOREM 13Definition 1.3.1 A complex n x n matrix U is said to be unitary if U*U =I. HereU* is the transpose of the conjugate of U. The matrix is unitary if and only if its columnsform an orthonormal set in C”. This follows from the way we multiply matrices in whichthe ij" entry of U*U is obtained by taking the conjugate of the i” row of U times the j'"column of U.Theorem 1.3.2 (Schur) Let A be a complex n x n matrix. Then there exists a unitarymatrix U such thatU*AU =T, (1.6)where T is an upper triangular matrix having the eigenvalues of A on the main diagonal,listed with multiplicity!Proof: The theorem is clearly true if A is a 1 x | matrix. Just let U = 1, the 1 x 1 matrixwhich has entry 1. Suppose it is true for (n — 1) x (n—1) matrices and let A be ann xnmatrix. Then let v; be a unit eigenvector for A. Then there exists 2; such thatAv, =A), |v;| = 1.Extend {v,} to a basis and then use the Gram - Schmidt process to obtain{v1 et 5 Un}an orthonormal basis of C”. Let Up be a matrix whose i” column is v;. Then from thedefinition of a unitary matrix Definition 1.3.1, it follows that Up is unitary. Consider Uj AUo.vi viUj AUp = ; ( Av; ss) AUn )= : ( Aiwy sss AUn )A a0 Anwhere A, is an n—1 Xx n—T1 matrix. Now by induction, there exists an (n—1) x (n—1)unitary matrix U; such thatVUThus Uj AU is of the formUSA, =Th-1;an upper triangular matrix. Consider(1 0u=(1 2)An application of block multiplication shows that U; is a unitary matrix and also thateyye _({1 0 A, * 1 0)\) (A * _viva ( yo CO a Coo =O ny Jotwhere T is upper triangular. Then let U = UpU}. Since (UpU1)* = UfU;, it follows thatA is similar to T and that UoU, is unitary. Hence A and T have the same characteristic'Listed with multiplicity’ means that the diagonal entries are repeated according to their multiplicity as rootsof the characteristic equation.