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Linear Algebra, Theory And Applications, 2012a

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180 SPECTRAL THEORY<br />

Then<br />

=<br />

⎛<br />

⎜<br />

⎝<br />

u T i 1<br />

.<br />

u T i n<br />

U ′T AU ′ = U ′T ( )<br />

Au i1 ··· Au in<br />

⎞<br />

⎛<br />

⎟<br />

⎠ ( ) ⎜<br />

λ i1 u i1 ··· λ in u in = ⎝<br />

⎞<br />

λ i1 0<br />

. ..<br />

⎟<br />

⎠ <br />

0 λ in<br />

7.5 Trace <strong>And</strong> Determinant<br />

The determinant has already been discussed. It is also clear that if A = S −1 BS so that<br />

A, B are similar, then<br />

det (A) = det ( S −1) det (S)det(B) =det ( S −1 S ) det (B)<br />

= det(I)det(B) =det(B)<br />

The trace is defined in the following definition.<br />

Definition 7.5.1 Let A be an n × n matrix whose ij th entry is denoted as a ij .Then<br />

trace (A) ≡ ∑ i<br />

a ii<br />

In other words it is the sum of the entries down the main diagonal.<br />

With this definition, it is easy to see that if A = S −1 BS, then<br />

trace (A) =trace(B) .<br />

Here is why.<br />

trace (A) ≡ ∑ i<br />

= ∑ j,k<br />

A ii = ∑ (<br />

S<br />

−1 ) B ij jkS ki = ∑<br />

i,j,k<br />

j,k<br />

B jk δ kj = ∑ B kk =trace(B) .<br />

k<br />

∑ (<br />

B jk S ki S<br />

−1 ) ij<br />

i<br />

Alternatively,<br />

trace (AB) ≡ ∑ ij<br />

A ij B ji =trace(BA) .<br />

Therefore,<br />

trace ( S −1 AS ) =trace ( ASS −1) =trace(A) .<br />

Theorem 7.5.2 Let A be an n×n matrix. Then trace (A) equals the sum of the eigenvalues<br />

of A and det (A) equals the product of the eigenvalues of A.<br />

This is proved using Schur’s theorem and is in Problem 17 below. Another important<br />

property of the trace is in the following theorem.<br />

Theorem 7.5.3 Let A be an m × n matrix and let B be an n × m matrix. Then<br />

trace (AB) =trace(BA) .<br />

Proof:<br />

trace (AB) ≡ ∑ i<br />

( ∑<br />

k<br />

A ik B ki<br />

)<br />

= ∑ k<br />

∑<br />

B ki A ik =trace(BA) <br />

i

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