Question

We say the n × n matrix A is similar to the n × n matrix B if there is an invertible
n × n matrix N so that B = N^(−1)AN. Another way to think about this is that B is
simply the matrix A after the change of basis given by the column vectors of N.
(a) Show that if A is similar to B, then B is similar to A. (i.e., if there is an invertible
matrix N so that B = N^(−1)AN, which is what “A is similar to B” means, then
there must also be some invertible matrix M so that A = M^(−1)BM, which is what
“B is similar to A” means).

Answer 1

M=N^{-1}

Answer 2

I never would have thought of that lol...

Answer 3

why does that make sense? how did you figure that out? if you could ... thanks (:

Answer 4

\[B = N^{-1}AN\]

\[NBN^{-1} =N N^{-1}ANN^{-1} \]
\[NBN^{-1} =IAI=A \]

Answer 5

Functional inverses of the equations. DERP.

Answer 6

just to clarify Zarkon's already clear answer, calling N^(-1)=M is just a labeling trick, and doing his proof with that notation gives the exact result you want
let
M=N^(-1)
then
B=MAM^(-1)
M^(-1)BM=M^(-1)MAM^(-1)M
M^(-1)BM=IAI=A

Answer 7

Show that if A is similar to B, and B is similar to C, then A is similar to C. (i.e., if there is some invertible matrix N1 with B = N1^(−1) AN1, and some invertible matrix N2 with C = N2^(−1)BN2 then there is some invertible matrix N3 with
A = N3^(−1)CN3 .) .... anyone?

Answer 8

Just taking a glance at it I think this proof can be done a similar way...

Answer 9

im just not sure how to work between 3 different bases.

Answer 10

\[B=N_1^{-1}AN_1\to N_1BN_1^{-1}=A\]\[C=N_2^{-1}BN_2\to N_2CN_2^{-1}=B\]sub expression for B from the second line into the first and let\[N_3=N_1N_2\iff N_3^{-1}=N_1^{-1}N_2^{-1}\]

Answer 11

sorry, rather let\[N_3^{-1}=N_1N_2\iff N_3=N_1^{-1}N_2^{-1}\]

Answer 12

ohhhh ok, thank you!!

Answer 13

welcome :D