Matrix Equivalence is Equivalence Relation

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Proof

Checking in turn each of the critera for equivalence:

Reflexive

$\mathbf A = \mathbf{I_m}^{-1} \mathbf A \mathbf{I_n}$ trivially, for all $m \times n$ matrices $\mathbf A$.

Thus reflexivity holds.

$\Box$

Symmetric

Let $\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$.

As $\mathbf P$ and $\mathbf Q$ are both invertible, we have:

 $\displaystyle \mathbf Q \mathbf B \mathbf P^{-1}$ $=$ $\displaystyle \mathbf Q \mathbf Q^{-1} \mathbf A \mathbf P \mathbf P^{-1}$ $\displaystyle$ $=$ $\displaystyle \mathbf{I_m} \mathbf A \mathbf{I_n}$ $\displaystyle$ $=$ $\displaystyle \mathbf A$

Thus symmetry holds.

$\Box$

Transitive

Let $\mathbf B = \mathbf Q_1^{-1} \mathbf A \mathbf P_1$ and $\mathbf C = \mathbf Q_2^{-1} \mathbf B \mathbf P_2$.

Then $\mathbf C = \mathbf Q_2^{-1} \mathbf Q_1^{-1} \mathbf A \mathbf P_1 \mathbf P_2$.

Transitivity follows from the definition of invertible matrix, that the product of two invertible matrices is itself invertible.

$\Box$

Hence the result by definition of equivalence relation.

$\blacksquare$