Linear Algebra / Proof

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Homework Statement



Let A be any square matrix and PsubA(lambda) be its characteristic polynomial, show that
PsubA(A) = 0.



Homework Equations





The Attempt at a Solution



I can show this for a general 2x2 matrix case with entries a, b,c,d and understand how it would be true of all square matrices, but I'm just not sure how to show this is true for any square matrix. We are studying Jordan Canonical Form of a matrix so, I'm thinking I should somehow use that. Any help would be appreciated.
 

Answers and Replies

  • #2
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Isn't det(A - [itex]\lambda[/itex]I) the characteristic polynomial?

If [itex]\lambda[/itex] is an eigenvalue of A, then the expression above evaluates to zero. I.e., PA([itex]\lambda[/itex]) = 0.
 
  • #3
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I'm sorry Mark, but I can't agree with your proof. The Cayley-Hamilton theorem seems more difficult than that.

The characteristic polynomial of a 2x2-matrix is

[tex]det\left(\begin{array}{cc} a-\lambda & b\\ c & d-\lambda\end{array}\right)[/tex]

You can't go on substituting A for lambda. Because then you will have a matrix in a matrix...
 
  • #4
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Well, clearly a - A and d - A don't make sense, since you can't subtract a matrix from a scalar, but what's wrong with A - AI?
 
  • #5
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Well, [tex]\lambda I[/tex] is a scalar product, while AI is a matrix product. It's not obvious to me that you can suddenly change the meaning of a product.

For example, consider the polynomial [tex]det(\lambda I)=0[/tex], then this polynomial is actually [tex]\lambda^n=0[/tex] (with n the dimension of I). Thus a root of this polynomial is a nilpotent matrix.
However, if you immediately substitute A for lambda, then you get det(A)=0. And thus a root of this polynomial would be a noninvertible matrix.

Since not every noninvertible matrix is nilpotent, we get two different answers. So the two methods are not equivalent.
 
  • #6
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I retract what I said before. I remembered most of what Cayley-Hamilton says (roughly, square matrices satisfy their own characteristic equations), and misapplied it here.
 
  • #7
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Well, [tex]\lambda I[/tex] is a scalar product, while AI is a matrix product. It's not obvious to me that you can suddenly change the meaning of a product.

For example, consider the polynomial [tex]det(\lambda I)=0[/tex], then this polynomial is actually [tex]\lambda^n=0[/tex] (with n the dimension of I). Thus a root of this polynomial is a nilpotent matrix.
However, if you immediately substitute A for lambda, then you get det(A)=0. And thus a root of this polynomial would be a noninvertible matrix.

Since not every noninvertible matrix is nilpotent, we get two different answers. So the two methods are not equivalent.

There was a hint to use Schur's theorem to show that A may be assumed to be upper triangular, then the characteristic polynomial would be (a11 - λ1)(a22 - λ2) ...(ann-λn) right?
 
  • #8
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Yes, so you only need to prove things for triangular matrices.

Your characteristic polynomial is indeed correct. Now try to fill in A (your triangular matrix) in the polynomial. Do you get 0?
 
  • #9
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Yes, so you only need to prove things for triangular matrices.

Your characteristic polynomial is indeed correct. Now try to fill in A (your triangular matrix) in the polynomial. Do you get 0?
yes, because that will make the a11 entry 0 in the first matrix then the a22 0 in the 2nd matrix and so on which will result in the zero matrix??
 
  • #10
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Well, you still have to multiply all those matrices...
 
  • #11
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Well, you still have to multiply all those matrices...
are you saying I need to show that or are you trying to move me in another direction? I know the det =0 for each (ann-λn). can I use that?
 
  • #12
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No, I'm not trying to push you in another direction. You were going in a great direction!
 
  • #13
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No, I'm not trying to push you in another direction. You were going in a great direction!
thats where I'm having trouble. When I multiply the 1st two together the first two diagonal entries become zero then times the 3rd makes the 3rd diagonal entry zero and so on, but how do I notate that elegantly? Isn't that what I'm supposed to do??
 
  • #14
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That is exactly what you should do!

However, I do not think that there is a clean notation for this. It's going to get messy no matter what...
 
  • #15
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That is exactly what you should do!

However, I do not think that there is a clean notation for this. It's going to get messy no matter what...
In 2 previous problems (not on here) I had to show that if J is any diagonal matrix then
PsubJ(J)=0 also I had to show that any Jordan block J PsubJ(J)=0 where PsubJ(λ) is its characteristic polynomial. With this being proven, can I use the fact that A has a Jordan canonical form and A is similar to J? Looking at my notes, I think this may be what I need to use to prove this problem. So I need to use A=QJQ^-1 replace A with this expression. I'm getting stuck though. Micromass, have you any ideas on this?
 

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