Linear Algebra Proof: c=0 Eigenvalue for Non-Invertible Matrix

[evilpostingmong]

Homework Statement

Show that A has c=0 as an eigenvalue if and only if A is non invertible.

Homework Equations

The Attempt at a Solution

Let A be a square matrix (mxm). Then Av=(a1,1v1+...+am,1vm)+...+(a1,mv1+...+am,mvm).
Since an identity matrix is square, and Icv is an mxm matrix,
we have (a1,1v1+...+a1,mv1)+..+(am,1vm+...+am,mvm)
For (a1,1+...+a1,m)v1+...+(am,1+...+am,m)vm=Icv,
(ak,1+...+ak,m) must=c. But if the matrix is mxn, n>m
then we have (ak,1+...+ak,m+...+ak,n) which is>(ak,1+...+ak,m).
Unless all rows after m are 0 (making the matrix back to mxm)
all (ak,1+...+ak,n) must be zero along with c so that
Av=0=Icv=I(0)v=0.

 

[Dick]

You don't need to write out any components of anything. Define what it means for a matrix to be noninvertible in terms of the kernel.

 

[evilpostingmong]

You don't need to write out any components of anything. Define what it means for a matrix to be noninvertible in terms of the kernel.

Hey Dick, hows it going? Anyway, in terms of the kernel, a matrix is noninvertible
if the transformation is not injective, so the kernel has elements besides zero.
So it is not just 0 that gets mapped to 0. Within the kernel are those vectors
in rows that become 0 after row reducing. So letting v be a 1xn vector,
and A having more rows than v but same amount of columns as v has rows,
we multiply and row reduce to see that the excessive rows become 0.

 

[Dick]

Great, how about with you? Ok, so if the matrix is noninvertible there is a nonzero vector v such that Av=0. What does that tell you about eigenvalues of A?

 

[evilpostingmong]

Great, how about with you? Ok, so if the matrix is noninvertible there is a nonzero vector v such that Av=0. What does that tell you about eigenvalues of A?

If Av=0 and v=/=0 then A is the zero transformation of v if v is in the kernel. So for Av to =0
knowing v is not 0, v must be multiplied by 0 to get to 0.
Since Iv=v=/=0, and Av=0*v,
0*v=Icv=cIv=0*Iv=0*v

 

[Dick]

If Av=0 and v=/=0 then A is the zero transformation of v if v is in the kernel. So for Av to =0
knowing v is not 0, v must be multiplied by 0 to get to 0.
Since Iv=v=/=0, and Av=0*v,
0*v=Icv=cIv=0*Iv=0*v

It's all in there, but it doesn't clearly answer my question. I think you mean to say simply, if Av=0 and v is nonzero then 0 is an eigenvalue because _______. Try and make a clear statement of that 'because'. Start with the definition of an eigenvalue and say why 0 is one.

 

[evilpostingmong]

It's all in there, but it doesn't clearly answer my question. I think you mean to say simply, if Av=0 and v is nonzero then 0 is an eigenvalue because _______. Try and make a clear statement of that 'because'.

v is not zero, so the eigenvalue must=0 Av=cv=0*v=0

 

[Dick]

Ok, so you've got "if A is noninvertible then 0 is an eigenvalue". Now do the converse. "if 0 is an eigenvalue then A is noninvertible".

 

[evilpostingmong]

Ok, so you've got "if A is noninvertible then 0 is an eigenvalue". Now do the converse. "if 0 is an eigenvalue then A is noninvertible".

If 0 is an eigenvalue, then there is some nonzero v in the kernel of A such that Av=0.

 

[Dick]

If 0 is an eigenvalue, then there is some nonzero v in the kernel of A such that Av=0.

I would say "If 0 is an eigenvalue then there is a nonzero vector v such that Av=0*v=0. So v is in the kernel of A. Since the kernel is nontrivial A is noninvertible." Compare that with your version, which doesn't even contain the word 'noninvertible'. A proof something is noninvertible should probably contain the word noninvertible.

 

[HallsofIvy]

A proof something is noninvertible should probably contain the word noninvertible.

Good advice!

 

[evilpostingmong]

I would say "If 0 is an eigenvalue then there is a nonzero vector v such that Av=0*v=0. So v is in the kernel of A. Since the kernel is nontrivial A is noninvertible." Compare that with your version, which doesn't even contain the word 'noninvertible'. A proof something is noninvertible should probably contain the word noninvertible.

That makes sense