Kenneth Kuttler

6.4. SCHUR’S THEOREM 165

Since these two matrices are equal, it follows a = 0. But now it follows that T ∗1 T1 = T1T

∗1

and so by induction T1 is a diagonal matrix D1. Therefore,

T =

(t11 0T

0 D1

)

a diagonal matrix.Now here is a proof which doesn’t involve block multiplication. Since T is normal,

T ∗T = TT ∗. Writing this in terms of components and using the description of the adjointas the transpose of the conjugate, yields the following for the ikth entry of T ∗T = TT ∗.

TT∗︷︸︸︷∑j

tijt∗jk =

∑j

tijtkj =

T∗T︷︸︸︷∑j

t∗ijtjk =∑j

tjitjk.

Now use the fact that T is upper triangular and let i = k = 1 to obtain the following fromthe above. ∑

|t1j |2 =∑j

|tj1|2 = |t11|2

You see, tj1 = 0 unless j = 1 due to the assumption that T is upper triangular. This showsT is of the form 

∗ 0 · · · 0

0 ∗ · · · ∗...

. . .. . .

...

0 · · · 0 ∗

 .

Now do the same thing only this time take i = k = 2 and use the result just established.Thus, from the above, ∑

|t2j |2 =∑j

|tj2|2 = |t22|2 ,

showing that t2j = 0 if j > 2 which means T has the form

∗ 0 0 · · · 0

0 ∗ 0 · · · 0

0 0 ∗ · · · ∗...

.... . .

. . ....

0 0 0 0 ∗

 .

Next let i = k = 3 and obtain that T looks like a diagonal matrix in so far as the first 3rows and columns are concerned. Continuing in this way, it follows T is a diagonal matrix.■

Theorem 6.4.11 Let A be a normal matrix. Then there exists a unitary matrix U suchthat U∗AU is a diagonal matrix. Also if A is normal and U is unitary, then U∗AU is alsonormal.

Proof: From Theorem 6.4.4 there exists a unitary matrix U such that U∗AU equalsan upper triangular matrix. The theorem is now proved if it is shown that the property of

6.4. SCHUR’S THEOREM 165Since these two matrices are equal, it follows a = 0. But now it follows that T*7, = T, 77and so by induction Tj is a diagonal matrix D,. Therefore,t TTH u O0 Da diagonal matrix.Now here is a proof which doesn’t involve block multiplication. Since T is normal,T*T = TT*. Writing this in terms of components and using the description of the adjointas the transpose of the conjugate, yields the following for the ik’” entry of T*T = TT™.TT* TTDo titin =) tates = Do tiytin = Dd Tpitie.J J J JNow use the fact that T is upper triangular and let i = k = 1 to obtain the following fromthe above.2 2 2So tag!” = So Ital? = feaj jYou see, tj; = 0 unless 7 = 1 due to the assumption that Tis upper triangular. This showsT is of the formOe we xNow do the same thing only this time take i = k = 2 and use the result just established.Thus, from the above,So ltogl? = So |tjal? = |tool*,J Jshowing that to; = 0 if 7 > 2 which means T has the form* 0 0 = 000 0 O xNext let 1 = k = 3 and obtain that T looks like a diagonal matrix in so far as the first 3rows and columns are concerned. Continuing in this way, it follows T is a diagonal matrix.aTheorem 6.4.11 Let A be a normal matriz. Then there exists a unitary matrix U suchthat U* AU is a diagonal matrix. Also if A is normal and U is unitary, then U* AU is alsonormal.Proof: From Theorem 6.4.4 there exists a unitary matrix U such that U* AU equalsan upper triangular matrix. The theorem is now proved if it is shown that the property of