Proving Non-Equivalence of Matrices Using Trace in Linear Algebra

[micromass]

Yes, that seems to do it! It looks like you've got it!

 

[Ted123]

Yes, that seems to do it! It looks like you've got it!

Or it may be better to write the second one as

$[A]\neq<b> \iff A \not\sim B \iff A-B \not\in U \iff \text{trace}(A-B)\neq 0 \iff \text{trace}(A)\neq\text{trace}(B)</b>$