The Trace of a Matrix
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Dan Nettleton (Iowa State University)
Statistics 611
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The trace of a square matrixn×n
A = [aij ] is
trace(A) = tr(A) =
n
X
aii .
i=1
For example,

5
3 5





tr 
 4 −1 2 = 5 − 1 + 7 = 11.
−3
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Dan Nettleton (Iowa State University)
8 7
Statistics 611
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Some Simple Facts about Trace
Suppose k, k1 , . . . , km ∈ R and A, A1 , . . . , Am are each n × n matrices.
Then
1
tr(A) = tr(A0 )
2
tr(kA) = k · tr(A)
3
tr(A1 + A2 ) = tr(A1 ) + tr(A2 )
4
tr(
Pm
i=1 ki Ai )
=
Pm
i=1 ki
· tr(Ai )
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Dan Nettleton (Iowa State University)
Statistics 611
3 / 11
Result A.17:
(a) tr(AB) = tr(BA). This is known as the cyclic property of the trace.
(b) Ifm×n
A = [aij ], then
0
tr(A A) =
m X
n
X
a2ij .
i=1 j=1
Proof of Result A.17: HW problem.
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Dan Nettleton (Iowa State University)
Statistics 611
4 / 11
Suppose A is an m × n matrix of rank r. Prove that there exist matrices
B and r×n
C such that
m×r
A = BC and rank(B) = rank(C) = r.
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Dan Nettleton (Iowa State University)
Statistics 611
5 / 11
Proof:
Let B = [b1 , . . . , br ] where b1 , . . . , br form a basis for C(A).
Because b1 , . . . , br form a basis, they are LI so that rank(B) = r.
Let cj be the vector of the coefficients of the linear combination of
b1 , . . . , br that gives the jth column of A.
Then A = BC, where C = [c1 , . . . , cn ].
Finally, note that
r = rank(A) = rank(BC) ≤ rank(C) ≤ r ⇒ rank(C) = r.
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Dan Nettleton (Iowa State University)
Statistics 611
6 / 11
Suppose A is an n × n matrix such that AA = kA for some k ∈ R.
Prove that tr(A) = k · rank(A).
(Note that this result implies the trace of an idempotent matrix is equal
to its rank.)
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Dan Nettleton (Iowa State University)
Statistics 611
7 / 11
Proof:
Let r = rank(A). Let n×r
B and r×n
C be matrices of rank r such that A = BC.
Then
BCBC = AA = kA = kBC = B(kr×r
I )C.
Now B of full column rank implies CBC = kr×r
I C, and
C of full row rank implies CB = kr×r
I.
Thus,
tr(A) = tr(BC) = tr(CB) = tr(kr×r
I ) = k · tr(r×r
I ) = k · r = k · rank(A). 2
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Dan Nettleton (Iowa State University)
Statistics 611
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Prove that tr(I − PX ) = n − rank(X).
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Dan Nettleton (Iowa State University)
Statistics 611
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Proof:
We know I − PX is idempotent. Thus, tr(I − PX ) = rank(I − PX ).
We know I − PX is the orthogonal projection matrix onto
C(X)⊥ = N (X0 ).
Thus, C(I − PX ) = N (X0 ), which has dimension n − rank(X).
Thus, rank(I − PX ) = n − rank(X).
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Dan Nettleton (Iowa State University)
Statistics 611
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Alternate Proof:
Because PX is idempotent, tr(PX ) = rank(PX ).
Now note that rank(PX ) = rank(X) because
rank(PX ) = rank(X(X0 X)− X0 ) ≤ rank(X) = rank(PX X) ≤ rank(PX ).
(This also follows from C(X) = C(PX ).)
Thus, tr(I − PX ) = tr(I) − tr(PX ) = n − tr(PX ) = n − rank(X).
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Dan Nettleton (Iowa State University)
Statistics 611
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