Diagonalization and Matrix similarity

[Bertrandkis]

Question1 :
a) Show that if A is nonsingular and diagonalizable then A^-1 is diagonalizable
b) Show that if A is diagonalizable then A^T is diagonalizable

Question 2
Show that if A is similar to B and B similar to C, then A is similar to C.

 

[morphism]

What have you tried so far? This is a straightforward application of the definitions.

 

[HallsofIvy]

Why are you doing a problem like this? If it is schoolwork (and it is certainly worded as such) then it should be in the "Homework and Schoolwork" section.

Also, you can't simply post a problem like this and expect someone else to give you the solution. Show what work you have done. Start, at the least, by quoting the definitions of "nonsingular", "diagonalizable", and "similar".

 

[Bertrandkis]

THIS IS WHAT I TRIED
Question1 :
a) Show that if A is nonsingular and diagonalizable then A[tex]^{-1}[/tex] is diagonalizable

MY SOLUTION
If A is non singular then AA[tex]^{-1}[/tex]=I
If A is diagonalizable if it's similar to a diagonal matrix => P[tex]^{-1}[/tex]AP=D
which implies that AP=PD (1)
We need to show that A[tex]^{-1}[/tex]P=PD [tex]^{-1}[/tex]

so multiply both sides of (1) by A[tex]^{-1}[/tex] we have PA=PD (2)

then multiply both sides of (2) by D[tex]^{-1}[/tex] we have D[tex]^{-1}[/tex]P=A[tex]^{-1}[/tex]P

Therefore A[tex]^{-1}[/tex] is diagonalizable


Question 2
Show that if A is similar to B and B similar to C, then A is similar to C

MY SOLUTION
if A is similar to B then P[tex]^{-1}[/tex]AP=B we can show that AP=PB(1)
if B is similar to C then P[tex]^{-1}[/tex]BP=C (2)
now replace BP in equation (2)by (1) then P[tex]^{-1}[/tex]AP=C
Therefore A is similar to C

 

[Mystic998]

For 1a, it's more useful to write A = PDP^-1, then it's clear what A^-1 has to be. b is the same idea.

For 2, write B = PAP^-1, C = QBQ^-1 (you can't use the same P for both), then the rest is pretty obvious.

 

[HallsofIvy]

THIS IS WHAT I TRIED
Question1 :
a) Show that if A is nonsingular and diagonalizable then A[tex]^{-1}[/tex] is diagonalizable

MY SOLUTION
If A is non singular then AA[tex]^{-1}[/tex]=I
If A is diagonalizable if it's similar to a diagonal matrix => P[tex]^{-1}[/tex]AP=D
which implies that AP=PD (1)
We need to show that A[tex]^{-1}[/tex]P=PD [tex]^{-1}[/tex]

so multiply both sides of (1) by A[tex]^{-1}[/tex] we have PA=PD (2)

then multiply both sides of (2) by D[tex]^{-1}[/tex] we have D[tex]^{-1}[/tex]P=A[tex]^{-1}[/tex]P

No. If you multiply PA= PD on the LEFT by D^-1 you get D^-1PA= D^-1PD. If you multiply PA= PD on the RIGHT by D^-1 you get PAD^-1= P. Neither of those is, directly, D^-1P= A^-1. By the way, have you already proved that, if A is invertible and diagonlizable, the D is also invertible?

Therefore A[tex]^{-1}[/tex] is diagonalizable


Question 2
Show that if A is similar to B and B similar to C, then A is similar to C

MY SOLUTION
if A is similar to B then P[tex]^{-1}[/tex]AP=B we can show that AP=PB(1)
if B is similar to C then P[tex]^{-1}[/tex]BP=C (2)
now replace BP in equation (2)by (1) then P[tex]^{-1}[/tex]AP=C

You can't replace "BP" in (2) by (1)- (1) doesn't say anything about BP- it says that PB=AP. Also, of course, the "P" in your first two equationsis not necessairily the same!

Therefore A is similar to C