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Simple
Simple
3:19 AM
if A is diagonalizable, how to show A is difference of two squares
can someone give me a hint, many thanks
 
 
1 hour later…
Martin Sleziak
Martin Sleziak
4:43 AM
@Simple Perhaps a reasonable start should be showing that any diagonal matrix can be written as $D=B^2-C^2$ for some matrices $B$, $C$
 
Simple
Simple
5:29 AM
the differences of the diagonal of B,C is A. A =QDQ^-1, B＝QE_1Q^-1 and C=QE_2Q^-1. then E_1E_2=0
 
 
1 hour later…
Martin Sleziak
Martin Sleziak
6:45 AM
I don't really follow what you wanted to say by the previous message.
What I meant was that if $D=B^2-C^2$ then $A=P^{-1}DP=(P^{-1}BP)^2-(P^{-1}CP)^2$
Here is a related question on main: math.stackexchange.com/questions/365810/…
> Let $A$ be a real $n\times n$ matrix. We say that $A$ is a difference of two squares if there exist real $n\times n$ matrices $B$ and $C$ with $BC = CB = 0$ and $A = B^2 − C^2$.
Now If A is a diagonal matrix, then I have to show that that it is a difference of two squares.
 

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