Proving Diagonalizability of an nxn Matrix

[buzzmath]

can anyone help with this problem?

Consider a diagonalizable nxn matrix A with m distinct eigenvalues L1,...,Lm show that (A-L1*I)(A-L2*I)****(A-Lm*I)=0

 

[0rthodontist]

Do you recognize the form (A-LI)? Have you ever solved something like (A-LI)x=0?

 

[buzzmath]

isn't that just used to find the eigenvalues? how would you get rid of x?

 

[0rthodontist]

Don't think in terms of "getting rid of x." Think about the problem conceptually. What ways do you know to show that some matrix is the zero matrix?

 

[buzzmath]

what do you mean by showing that a matrix is the zero matrix? A matrix is a zero matrix if anything multiplied by it is zero? I know that A is similar do a diagonal matrix D with the eigienvalues down the diagonal. I'm not sure if that would help though?

 

[0rthodontist]

Right, a matrix is a zero matrix if any vector multiplied by it is zero. Write the vector as a linear combination of eigenvectors.

 

[buzzmath]

So since A is diagonizable the eigenvectors form an eigenbasis for A. Thus any vector is a linear combination of these eigenvectors. Does that mean that is x is any linear combination of these eigenvectors that (A-L1*I)***(A-Lm*I)*x=0? and if so does it mean since any vector mulitplied by this matrix is zero that this matrix is zero then?

 

[0rthodontist]

Does that mean that is x is any linear combination of these eigenvectors that (A-L1*I)***(A-Lm*I)*x=0?

Can you think of a reason? Try actually writing x as a linear combination rather than just saying that it is.