Cayley Hamilton Theorem

In summary: W5nLCB3ZSB0ZXJlIG5vdCBiZSBjb25zaXN0ZW50LiBUaGF0J3MgYmVpbmcgZnVsbCB0byB0aGluayBpbiBBIHByb3ZpbmcgYW5kIHN0YXRlbWVudCBwLg==In summary, the Cayley Hamilton theorem states that if p(x) is the characteristic polynomial of a matrix A, then when A is plugged into p(x), the result is always the zero matrix. However, this theorem is not as straightforward as it seems, as simply plugging A into the characteristic polynomial expression is
  • #1
Benn
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In my linear algebra course, we just finished proving the cayley hamilton theorem (if p(x) = det (A - xI), then p(A) = 0).

The theorem seems obvious: if you plug in A into p, you get det (A-AI) = det (0) = 0. But, of course, you can't do that (this is especially clear when you consider what A-xI looks like... you can't subtract matrices from real numbers)

Is there any way to salvage the idea of just plugging in A? or is it just a coincidence that it seems so obvious?
 
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  • #2
Benn said:
In my linear algebra course, we just finished proving the cayley hamilton theorem (if p(x) = det (A - xI), then p(A) = 0).

The theorem seems obvious: if you plug in A into p, you get det (A-AI) = det (0) = 0. But, of course, you can't do that (this is especially clear when you consider what A-xI looks like... you can't subtract matrices from real numbers)

Is there any way to salvage the idea of just plugging in A? or is it just a coincidence that it seems so obvious?


We know p(x) is the characteristic polynomial of A. The meaning of p(A) is to take the input A into the characteristic polynomial, not into the formula d(A - xI). Also, note that the 0 in the expression p(A) = 0 isn't referring to a scalar 0, its referring to the zero matrix.
 
  • #3
bins4wins said:
We know p(x) is the characteristic polynomial of A. The meaning of p(A) is to take the input A into the characteristic polynomial, not into the formula d(A - xI). Also, note that the 0 in the expression p(A) = 0 isn't referring to a scalar 0, its referring to the zero matrix.

Yes, thank you.

I understand that. But I wasn't sure if it was just a coincidence that the theorem seemed so obvious when we considered the characteristic polynomial to be det (A-xI), or if there was some way to make rigorous the idea of plugging in A.
 
  • #4
Personally, I find the idea of the notation p(A) fairly loose in terms of rigor, since the definition of p as a function has a domain of the reals. The most amount of rigor you can put in plugging in A is just by defining what it exactly means to plug in A and that is to plug it into the characteristic polynomial expression.
 
  • #5
bins4wins said:
Personally, I find the idea of the notation p(A) fairly loose in terms of rigor, since the definition of p as a function has a domain of the reals. The most amount of rigor you can put in plugging in A is just by defining what it exactly means to plug in A and that is to plug it into the characteristic polynomial expression.

If ##p(x) = c_{n}x^{n} + ... + c_{1}x + c_{0}## where ##p## is defined from ##\mathbb{R}## to ##\mathbb{R}## and ##c_{i} \in \mathbb{R}##, then ##p(A): \{ \text{m x m matrices} \} \rightarrow \{ \text{m x m matrices} \}## is defined by ##p(A) = c_{n}A^{n} + ... + c_{1}A + c_{0}I##. ... I'm completely happy with that.

I must not have been clear in my question. I'm asking for a proof using the idea of 'plugging A into det (A - xI) or an explanation of why there isn't one, not an clarification of the statement of the theorem.
 
  • #6
Benn said:
If ##p(x) = c_{n}x^{n} + ... + c_{1}x + c_{0}## where ##p## is defined from ##\mathbb{R}## to ##\mathbb{R}## and ##c_{i} \in \mathbb{R}##, then ##p(A): \{ \text{m x m matrices} \} \rightarrow \{ \text{m x m matrices} \}## is defined by ##p(A) = c_{n}A^{n} + ... + c_{1}A + c_{0}I##. ... I'm completely happy with that.

I must not have been clear in my question. I'm asking for a proof using the idea of 'plugging A into det (A - xI) or an explanation of why there isn't one, not an clarification of the statement of the theorem.

Well, for one thing, p(A) is a matrix (that happens to have all entries = 0), while det(I.A - A) is a scalar (that happens to equal zero).

More generally, suppose we have p(x) = det(xB + C) = det(Bx + C) for nxn matrices B and C, and suppose we happen to have p(A) = 0. Does it follow that det(AB + C) = 0 or that det(BA + C) = 0? Conversely, if either det(AB + C) = 0 or det(BA + C) = 0, does it follow that p(A) = 0? I am not 100% sure of the answers, but I have my doubts that any of the answers is "yes". (If this turns out to be right, then it would be a sheer accident that you happen to get the correct result that p(A) = 0 in the special case that B = I and C = A.)

RGV
 
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What is the Cayley Hamilton Theorem?

The Cayley Hamilton Theorem is a fundamental result in linear algebra that states that every square matrix satisfies its own characteristic equation. This means that a matrix can be expressed as a polynomial of itself.

Who discovered the Cayley Hamilton Theorem?

The Cayley Hamilton Theorem was discovered by the mathematician Arthur Cayley in 1854.

How is the Cayley Hamilton Theorem used in real-life applications?

The Cayley Hamilton Theorem has many practical applications in fields such as engineering, physics, and computer science. It is used to solve systems of linear equations, analyze the stability of dynamic systems, and in the development of algorithms for data compression and encryption.

Is the Cayley Hamilton Theorem applicable to all types of matrices?

Yes, the Cayley Hamilton Theorem applies to all square matrices, regardless of their size or elements.

Can the Cayley Hamilton Theorem be extended to non-square matrices?

No, the Cayley Hamilton Theorem only applies to square matrices. However, there are similar theorems that apply to non-square matrices, such as the Faddeev-LeVerrier algorithm.

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