Trace and determinant

Trace

Definition

The trace of a square matrix is defined as the sum of the entries on its main diagonal. Let \(A\) be an \(n\times n\) matrix, then

\[\Trace (A) = \sum_{i=1}^n a_{ii}\]

where \(\Trace(A)\) denotes the trace of \(A\).

Lemma

The trace of a square matrix and its transpose are equal.

\[\Trace(A) = \Trace(A^T).\]

Lemma

Trace of sum of two square matrices is equal to the sum of their traces.

\[\Trace(A + B) = \Trace(A) + \Trace(B).\]

Lemma

Let \(A\) be an \(m \times n\) matrix and \(B\) be an \(n \times m\) matrix. Then

\[\Trace(AB) = \Trace(BA).\]

Proof

Let \(AB = C = [c_{ij}]\). Then

\[c_{ij} = \sum_{k=1}^n a_{i k} b_{k j}.\]

Thus

\[c_{ii} = \sum_{k=1}^n a_{i k} b_{k i}.\]

Now

\[\Trace(C) = \sum_{i=1}^m c_{ii} = \sum_{i=1}^m \sum_{k=1}^n a_{i k} b_{k i} = \sum_{k=1}^n \sum_{i=1}^m a_{i k} b_{k i} = \sum_{k=1}^n \sum_{i=1}^m b_{k i} a_{i k}.\]

Let \(BA = D = [d_{ij}]\). Then

\[d_{ij} = \sum_{k=1}^m b_{i k} a_{k j}.\]

Thus

\[d_{ii} = \sum_{k=1}^m b_{i k} a_{k i}.\]

Hence

\[\Trace(D) = \sum_{i=1}^n d_{ii} = \sum_{i=1}^n \sum_{k=1}^m b_{i k} a_{k i} = \sum_{i=1}^m \sum_{k=1}^n b_{k i} a_{i k}.\]

This completes the proof.

Lemma

Let \(A \in \FF^{m \times n}\), \(B \in \FF^{n \times p}\), \(C \in \FF^{p \times m}\) be three matrices. Then

\[\Trace(ABC) = \Trace(BCA) = \Trace(CAB).\]

Proof

Let \(AB = D\). Then

\[\Trace(ABC) = \Trace(DC) = \Trace(CD) = \Trace(CAB).\]

Similarly the other result can be proved.

Lemma

Trace of similar matrices is equal.

Proof

Let \(B\) be similar to \(A\). Thus

\[B = C^{-1} A C\]

for some invertible matrix \(C\). Then

\[\Trace(B) = \Trace(C^{-1} A C ) = \Trace (C C^{-1} A) = \Trace(A).\]

We used this.

Determinants

Following are some results on determinant of a square matrix \(A\).

Lemma

\[\det(\alpha A) = \alpha^n \det(A).\]

Lemma

Determinant of a square matrix and its transpose are equal.

\[\det(A) = \det(A^T).\]

Lemma

Let \(A\) be a complex square matrix. Then

\[\det(A^H) = \overline{\det(A)}.\]

Proof

\[\det(A^H) = \det(\overline{A}^T) = \det(\overline{A}) = \overline{\det(A)}.\]

Lemma

Let \(A\) and \(B\) be two \(n\times n\) matrices. Then

\[\det (A B) = \det(A) \det(B).\]

Lemma

Let \(A\) be an invertible matrix. Then

\[\det(A^{-1}) = \frac{1}{\det(A)}.\]

Lemma

\[\det(A^{p}) = \left(\det(A) \right)^p.\]

Lemma

Determinant of a triangular matrix is the product of its diagonal entries. i.e. if \(A\) is upper or lower triangular matrix then

\[\det(A) = \prod_{i=1}^n a_{i i}.\]

Lemma

Determinant of a diagonal matrix is the product of its diagonal entries. i.e. if \(A\) is a diagonal matrix then

\[\det(A) = \prod_{i=1}^n a_{i i}.\]

Lemma

Determinant of similar matrices is equal.

Proof

Let \(B\) be similar to \(A\). Thus

\[B = C^{-1} A C\]

for some invertible matrix \(C\). Hence

\[\det(B) = \det(C^{-1} A C ) = \det (C^{-1}) \det (A) \det(C).\]

Now

\[\det (C^{-1}) \det (A) \det(C) = \frac{1}{\det(C)} \det (A) \det(C) = \det(A).\]

We used this and this.

Lemma

Let \(u\) and \(v\) be vectors in \(\FF^n\). Then

\[\det(I + u v^T) = 1 + u^T v.\]

Lemma

Let \(A\) be a square matrix and let \(\epsilon \approx 0\). Then

\[\det(I + \epsilon A ) \approx 1 + \epsilon \Trace(A).\]