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This seems to be an elementary question, gonna ask it anyway.

Suppose that $A$ is a square matrix, and that $p(x)$ is its characteristic polynomial, we know that

(1) $p(x) = \det(xE - A)$

We also know that

(2) $p(x) = x^n + a_{n-1}x^{n-1}+...+a_0$

Now, from (1), we deduce that $p(A) = zero$, where "zero" is the number 0.

From (2), we will get $p(A) = ZERO$, where "ZERO" is the zero matrix.

I know I am wrong, so...where I misunderstood?

Would be appreciate if you go into the details.

Thanks.

Edit

I thought that what I was asking is obvious: you substitute the $x$ with $A$ in both (1) and (2), you should get equal results, since (1) and (2) is equal, this follows from the substitution property of the equality. But as argued in the question, we now get different results, so something must be wrong.

Not an ID
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    What are you asking? The Cayley Hamilton theorem shows that if $p(x) = \det (xI -A)$, and we write $p$ as $p(x) = \sum_k a_k x^k$, then $\sum_k a_k A^k = 0$. We write this as $p(A) = 0$, where the latter $0$ is the matrix zero. – copper.hat Aug 02 '14 at 00:52
  • Sorry, I meant to write that you can't just stuff $A$ as a replacement for $x$ into $\det(xI-A)$. – copper.hat Aug 02 '14 at 01:01
  • @copper.hat And why is that? – Not an ID Aug 02 '14 at 01:05
  • Well, $x$ is a scalar. When you have a polynomial $p(x) = \sum_k a_k x^k$, I can extend the definition to a matrix by setting $p(A) = \sum_k a_k A^k$. But, without writing the $\det$ expression as a polynomial, I cannot extend the definition by just substituting $A$ instead of $x$ in the formula $\det (xI -A)$. – copper.hat Aug 02 '14 at 01:07
  • @copper.hat That is probably true, but I found it not very convincing. Thanks anyway. – Not an ID Aug 02 '14 at 01:12
  • Well, without expanding the expression $\det(xI-A)$ as a polynomial first, how would you define it when $x$ is replaced by a matrix? – copper.hat Aug 02 '14 at 01:25

3 Answers3

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Why should anybody believe $p(X)=\det(XI-A)$ to be true for matrix $X$? In fact, it's not.

The assertion $p(A)=0$ does not mean that the determinant of $\det(AI-A)$ is zero, because that is not how we obtain the expression $p(A)$; rather, it means

$$``\det\begin{pmatrix}A-a_{11} & -a_{12} & \cdots & -a_{1n} \\ -a_{21} & A-a_{22} & \cdots & -a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ -a_{n1} & -a_{n2} & \cdots & A-a_{nn}\end{pmatrix}" ~:= A^n-(a_{11}+a_{22}+\cdots+a_{nn})A^{n-1}+\cdots$$

is zero, because it's this that faithfully imitates how we actually derive the polynomial expression $p(A)$, and it isn't immediately obvious why this will wind up being zero.

anon
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  • Since the function $det$ can be defined as a funtion of matrix using some axioms, I think nothing prevents me from defining $det(\lambda I -A)$ as a function of a matrix, (where $\lambda$ denotes the variable), as pointed in T.A.E. 's answer. Please see my answer for more details. – Not an ID Aug 07 '14 at 01:26
  • @NotanID It's completely obvious that $\det(\lambda I-A)$ can be interpreted as a function of $A$; what's your point, and how's it relevant to my answer? – anon Aug 07 '14 at 01:29
  • Sorry, I misunderstood the first sentence in your answer. Thought you were saying we cannot substitute $x$ with a matrix. – Not an ID Aug 07 '14 at 01:32
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If you Let $M_{n}$ denote the linear space of $n\times n$ matrices over a field $\mathscr{F}$, and if $A \in M_{n}$, then it does make sense to define a function $p : M_{n}\rightarrow \mathscr{F}$. by $p(X)=\mbox{det}(X-A)$. There's nothing wrong with defining such a function, and clearly $p(A)=0$ for such a function, where $0\in\mathscr{F}$.

However, that's not the same as defining a polynomial $p(x)=\mbox{det}(xI-A)$ for a matrix $A$, say $p(x)=a_{0}+a_{1}x+\cdots a_{n}x^{n}$, and then defining $p(A)=a_{0}I+a_{1}A+\cdots a_{n}A^{n}$. For one thing the value of $p(A)$ is, in this case, an element of $M_{n}$ and not an element of $\mathscr{F}$. What is interesting is that $p(A)=0_{n}$, where $0_{n}\in M_{n}$, which is a statement of the Cayley-Hamilton Theorem.

Disintegrating By Parts
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After thought it for a while, I think I know what's going wrong, it's not on (1), but (2), (ultimately, it's on me, I guess).

When we say a matrix satisfies a polynomial

$b_mx^m + ... + b_o$,

what we mean is that

$b_m A^m + ... + b_0 I = ZERO$,

this is not a straightforward substitution, (there is a $I$ came out of nowhere), hence, if anything goes wrong, it won't affect the validity of the properties of substitution, because it is not a substitute operation.

So, it really doesn't matter what it yields when we substitute $x$ with $A$ in (1), or even whether or not it is legit to do so, it won't lead us to $zero = ZERO$, for the reason I just stated in the above paragraph.

If you rewrite the formula as:

$ \det(xE - A) = x^n + a_{n-1}x^{n-1}+...+a_0$

you will clearly see that substitute $x$ with a matrix is not a legit operation, (you cannot add a matrix to the scalar $a_0$).

For any polynomial $f$, If you want to know what $f(A)$ is, you have to do it through (2), as pointed out by @copper.hat, because that's how we defined $f(A)$, and we haven't define it in a second way yet.

I think $f(A)$ is a bad notation, because it misleads us to think we can get $f(A)$ by substitute $x$ in the polynomial with matrix $A$.

Not an ID
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