Prove that the trace is a similarity invariant.
In other words, if two matrices are similar, then they must have the same trace.
Got the answer from Wikipedia:
tr(AB) = tr(BA)
tr(ABC) = tr(CAB)
tr(P^-1 * A * P) = tr(P^-1 * P * A) = tr(A)
To prove that the trace is a similarity invariant, we need to show that if two matrices are similar, then they must have the same trace.
Let's assume that A and B are similar matrices, which means there exists an invertible matrix P such that A = P^-1 * B * P.
We know that the trace of a matrix, denoted as tr(M), is the sum of its diagonal elements.
Now, let's calculate the trace of A and B separately:
tr(A) = sum of diagonal elements of A
= sum of diagonal elements of (P^-1 * B * P)
= sum of diagonal elements of (P^-1 * (B * P))
Using the associativity property of matrix multiplication, we can write the above expression as:
tr(A) = sum of diagonal elements of ((P^-1 * B) * P)
= sum of diagonal elements of (B * P * P^-1) [since matrix multiplication is associative]
Now, since P*P^-1 = I (the identity matrix), we can simplify the expression further to:
tr(A) = sum of diagonal elements of (B * I)
= sum of diagonal elements of B [since B * I = B, where I is the identity matrix]
Hence, we have shown that tr(A) = tr(B), which means the trace of similar matrices is the same.
In summary, the trace is a similarity invariant, as the trace of two similar matrices is equal.