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Linear Algebra, Theory And Applications, 2012a

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7.4. SCHUR’S THEOREM 179<br />

Proof: From Theorem 7.4.4 there exists a unitary matrix U such that U ∗ AU equals<br />

an upper triangular matrix. The theorem is now proved if it is shown that the property of<br />

being normal is preserved under unitary similarity transformations. That is, verify that if<br />

A is normal and if B = U ∗ AU, then B is also normal. But this is easy.<br />

B ∗ B = U ∗ A ∗ UU ∗ AU = U ∗ A ∗ AU<br />

= U ∗ AA ∗ U = U ∗ AUU ∗ A ∗ U = BB ∗ .<br />

Therefore, U ∗ AU is a normal and upper triangular matrix and by Lemma 7.4.10 it must be<br />

a diagonal matrix. <br />

Corollary 7.4.12 If A is Hermitian, then all the eigenvalues of A are real and there exists<br />

an orthonormal basis of eigenvectors.<br />

Proof: Since A is normal, there exists unitary, U such that U ∗ AU = D, a diagonal<br />

matrix whose diagonal entries are the eigenvalues of A. Therefore, D ∗ = U ∗ A ∗ U = U ∗ AU =<br />

D showing D is real.<br />

Finally, let<br />

U = ( u 1 u 2 ··· u n<br />

)<br />

where the u i denote the columns of U and<br />

⎛<br />

⎞<br />

λ 1 0<br />

⎜<br />

D = ⎝<br />

. ..<br />

⎟<br />

⎠<br />

0 λ n<br />

The equation, U ∗ AU = D implies<br />

AU = ( )<br />

Au 1 Au 2 ··· Au n<br />

= UD = ( )<br />

λ 1 u 1 λ 2 u 2 ··· λ n u n<br />

where the entries denote the columns of AU and UD respectively. Therefore, Au i = λ i u i<br />

and since the matrix is unitary, the ij th entry of U ∗ U equals δ ij and so<br />

δ ij = u ∗ i u j ≡ u j · u i .<br />

This proves the corollary because it shows the vectors {u i } are orthonormal. Therefore,<br />

they form a basis because every orthonormal set of vectors is linearly independent. <br />

Corollary 7.4.13 If A is a real symmetric matrix, then A is Hermitian and there exists a<br />

real unitary matrix U such that U T AU = D where D is a diagonal matrix whose diagonal<br />

entries are the eigenvalues of A. By arranging the columns of U the diagonal entries of D<br />

can be made to appear in any order.<br />

Proof: This follows from Theorem 7.4.6 and Corollary 7.4.12. Let<br />

U = ( u 1 ··· u n<br />

)<br />

Then AU = UD so<br />

AU = ( Au 1 ··· Au n<br />

) (<br />

= u1 ··· u n<br />

) (<br />

D = λ1 u 1 ··· λ n u n<br />

)<br />

Hence each column of U is an eigenvector of A. It follows that by rearranging these columns,<br />

the entries of D on the main diagonal can be made to appear in any order. To see this,<br />

consider such a rearrangement resulting in an orthogonal matrix U ′ given by<br />

U ′ = ( u i1 ··· u in<br />

)

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