Cayley Hamilton Theorem states that if $A$ is an $n \times n$ matrix over the field $F$ then $p(A) = 0$.

We note that $p(\lambda) = \det(\lambda I - A)$. Hence, why can't we just substitute $\lambda$ with $A$ and directly prove Cayley-Hamilton Theorem by saying that $p(A) = p(\lambda) = \det(\lambda I - A) = \det(AI - A) = 0$?

  • 6
    $\begingroup$ Morally it's correct. Yes, $A$ is an eigenvalue of $A$, so $A$ satisfies its characteristic equation. They knew it before they proved it. And these days, you can even prove it in this way. $\endgroup$ – Orest Bucicovschi Feb 10 '15 at 4:56
  • 13
    $\begingroup$ The question is what "substitute $\lambda$ with $A$" means. You can substitute $A$ for $\lambda$ in the expanded-out polynomial $\det\left(\lambda I - A\right)$, or you can substitute it into the expression $\lambda I - A$ and then take the determinant. The question is why the results are the same (or, rather, why they are both $0$; they are not really the same, given that one of them is a matrix and the other a scalar!). That's absolutely not obvious, and wouldn't hold if we replace $I$ by a generic matrix $B$ which does not commute with $A$. $\endgroup$ – darij grinberg Feb 10 '15 at 5:04
  • 6
    $\begingroup$ @orangeskid: Did you mean to say "those days"? I definitely don't think plugging $A$ into the expression would count as a proof these days. $\endgroup$ – Jim Feb 10 '15 at 5:43
  • 1
    $\begingroup$ It can be done using some formal tricks: en.wikipedia.org/wiki/… $\endgroup$ – Blah Feb 10 '15 at 7:00
  • 2
    $\begingroup$ @Blah: Here is the more relevant subpage of the wiki article. The main point is that the proposed proof want to boil down to computing (just) the determinant of a zero matrix, and none of the formal tricks can justify that. Also "just substitute $A$ for $\lambda$" suggest a shortcut to any of the more elaborate proofs; the formal tricks actually make the proof a lot more subtle (though maybe less computational). $\endgroup$ – Marc van Leeuwen Feb 10 '15 at 8:39

If $$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$ then $p(\lambda)$ is the determinant of the matrix $$\lambda I - A = \begin{bmatrix} \lambda - 1 & -2 \\ -3 & \lambda - 4 \end{bmatrix}.$$ Now I plug in $A$ for $\lambda$ and get $$\begin{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} - 1 & -2 \\ -3 & \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} - 4 \end{bmatrix}$$ but I don't know what that is, and I certainly don't know how to take its determinant.

So the reason you can't plug into $\det(\lambda I - A)$ is because that expression only makes sense when $\lambda$ is a scalar. The definition of $p(\lambda)$ isn't really $\det(\lambda I - A)$, the definition of $p(\lambda)$ is that it's the polynomial whose value on any scalar equals the value of $\det(\lambda I - A)$.

On the other hand I could define a function $P(\lambda) = \det(\lambda - A)$ where I'm now allowed to plug in matrices of the same size as $A$, and I certainly would get zero if I plugged in $A$. But this is a function from matrices to numbers, whereas when I plug matrices into $p(\lambda)$ I get as output matrices. So it doesn't make sense to say that these are equal, so the fact that $P(A) = 0$ wouldn't seem to imply that $p(A) = 0$ sense $P$ and $p$ aren't the same thing.

  • 1
    $\begingroup$ Actually he's not saying this. The expression $\det(AI-A)$ do make sense considering the usual product. In this case $AI-A=A-A=0$ $\endgroup$ – hjhjhj57 Feb 10 '15 at 6:23
  • 7
    $\begingroup$ @hjhjhj57 Please pay more attention to the answer. $\det(AI-A)$ is not the result of plugging $\lambda = A$ into $\det(\lambda I - A)$, so it's kind of irrelevant that $\det(AI-A)$ makes sense. $\endgroup$ – Erick Wong Feb 10 '15 at 6:54
  • 1
    $\begingroup$ This is wrong, actually: the -2 and the -3 in the result of replacing the variable by A should be matrices. $\endgroup$ – Mariano Suárez-Álvarez Jan 23 '17 at 21:24

There is another way to see that the proof must be flawed: by finding the interesting consequences this proof technique has. If the proof would be valid, then we would also have the following generalisation:

Faulty Lemma. Suppose that $A$ and $B$ are $n\times n$ matrices. Let $p_A$ be the characteristic polynomial for $A$. If $B - A$ is singular, then $B$ must be a zero of $p_A$.

Faulty proof: We have $p_A(B) = \det(BI - A) = \det(B - A) = 0$.$$\tag*{$\Box$}$$

This has the following amazing consequence:

Faulty Corollary. Every singular matrix is nilpotent.

Faulty proof: Let $B$ be a singular matrix and let $A$ be the zero matrix. Now we have $p_A(\lambda) = \lambda^n$. Furthermore, by the above we have $p_A(B) = 0$, because $B - A$ is singular. Thus, we have $B^n = 0$ and we see that $B$ is nilpotent.$$\tag*{$\Box$}$$

In particular, this proves that we have $$ \pmatrix{1 & 0 \\ 0 & 0}^2 = \pmatrix{0 & 0 \\ 0 & 0}. $$ This comes to show just how wrong the proof is!

  • 2
    $\begingroup$ This is I think the best way to show how wrong plugging-in "proof" is. $\endgroup$ – Wojowu May 24 '16 at 17:07
  • $\begingroup$ Very nice answer! $\endgroup$ – Prism Apr 5 '17 at 21:30
  • $\begingroup$ I stumbled upon the accepted answer's justification many times and I never found it very convincing, just a matter of definitions. This is by far the best argument I've seen for this! $\endgroup$ – user438666 May 19 at 12:10

Remember that there is a difference between $p(x)$ where $x$ is scalar and $p(A)$ where $A$ is a matrix, the next thing you should notice is that if your deduction is true, then $p(A)=0$, the left hand side of this equation is a matrix, while the right hand side is the scalar $0$.

What Cayley-Hamilton theorem says is that $A$ satisfies its own characteristic polynomial. If you have worked with minimal polynomials before, the proof of this statement is a simple task (given all of the previous work, obviously).

  • $\begingroup$ The fact that $p(A)$ is a matrix and $0$ is a scalar should really make you suspicious, indeed! I never really thought about that. This is even more elementary than my answer. To me, this is the best answer. $\endgroup$ – Josse van Dobben de Bruyn Nov 13 '16 at 9:14

The original question was: Can i write $\det(I A - A) = 0$ in a meaningful way? Yes, if the first $A$ is consider as a scalar in the ring $F[A]\subset M_n(F)$, and the second one as the matrix representing $A$. And the reason this works is, indeed, that $A$ is an eigenvalue of the matrix $A$. How is this so? Details are below. Note that to make it work we need to work with modules, since the coefficients will be a ring containing $F$.

We'll start with some easy statements:

Let $k$ be a field, and $(a_{ij})$ an $n\times n$ matrix with elements in $k$ and $\lambda \in k$ so that there exists an element $v \in k^n$, $v$ nonzero, with $(a_{ij}) \cdot v = \lambda \cdot v$. Then $\det( (a_{ij}) - \lambda I ) = 0$. Indeed, the matrix $(a_{ij}) - \lambda I$ is not invertible.

Same conclusion, if we substitute $k$ with a $k$-vector space $V$, and there exists a nonzero element $v$ in $V^{ n}$ with $(a_{ij})\cdot v = \lambda \cdot v$.

More generally: $k$ a commutative ring, $V$ a $k$-module, $(a_{ij})$ a matrix in $M_n(k)$ and $\lambda \in k$ and $v$ in $V^n$ so that $(a_{ij})\cdot v = \lambda \cdot v$. Then $\det((a_{ij}) - \lambda I) \cdot v = 0 \in V^n$. Use the adjoint matrix.

Let now $A$ an $n\times n$ matrix. Let $k \colon = F[A]\subset M_n(F)$ the commutative algebra generated by $A$. $F^n$ is a $k$-module. Let $e_1$,$\ldots $, $e_n$ the standard basis of $F^n$. We have

$$ (a_{ij}) \cdot \left( \begin{array}{c} e_1 \\ \ldots \\ e_n\end{array} \right ) = \left( \begin{array}{c} A\cdot e_1 \\ \ldots \\ A \cdot e_n\end{array} \right ) $$

The above equation says: $\left( \begin{array}{c} e_1 \\ \ldots \\ e_n\end{array} \right )$ in $V^n$ eigenvector for the eigenvalue $A$. It follows that $$P_A(A) \cdot \left( \begin{array}{c} e_1 \\ \ldots \\ e_n\end{array} \right )= \left( \begin{array}{c} 0 \\ \ldots \\ 0\end{array} \right )$$

and therefore $P_A(A)=0$.

$\bf{Added}$: Making sense of $\det(AI - A)$:

Consider the example of @Jim: $$A= \left[\begin{array}{cc}1&2\\3&4\end{array} \right]$$

$$\begin{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} - 1 & -2 \\ -3 & \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} - 4 \end{bmatrix}\ \ \ ?$$

We need to look at scalars as scalar matrices. So we get

$$\begin{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} & \begin{bmatrix} -2 & 0 \\ 0 & -2\end{bmatrix} \\ \begin{bmatrix} -3 & 0 \\ 0 &-3 \end{bmatrix} & \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} \end{bmatrix}$$

This $2\times 2$ matrix, with entries in the commutative algebra $F[A]$, has determinant $P_A(A)$, a matrix of the same size as $A$. And it will always the zero matrix.

  • 1
    $\begingroup$ I don't think the OP will be able to follow this. $\endgroup$ – hjhjhj57 Feb 10 '15 at 7:21
  • 5
    $\begingroup$ @hjhjhj57: Why not, with an open mind. He surely asked a good question. $\endgroup$ – Orest Bucicovschi Feb 10 '15 at 7:27
  • 3
    $\begingroup$ I have no idea where you're trying to get with this. $\endgroup$ – Pedro Tamaroff Feb 10 '15 at 7:50
  • 3
    $\begingroup$ Your answer contains good mathematics, but not, unfortunately, an answer to the question. $\endgroup$ – Eric Stucky Feb 11 '15 at 17:55
  • 2
    $\begingroup$ Even though you didn't really answer the original question, I must confess I am very amazed. $\endgroup$ – Qidi Feb 12 '15 at 3:34

Actually, you can do what you want, but you have to be very careful about what you mean when you evaluate a polynomial with matrix coefficients at a matrix. For example, if you have a matrix polynomial with square matrix coefficients $A_j$ such as $$ p(\lambda) = A_{0}+\lambda A_{1}+\lambda^{2}A_{2}+\cdots+\lambda^{n}A_{n}, $$ then what do you mean by $p(M)$ where $M$ is a matrix? Do you mean $$ p(M) = A_{0}+MA_{1}+M^{2}A_{2}+\cdots+M^{n}A_{n}, $$ or do you mean $$ p(M)=A_{0}+A_{1}M+A_{2}M^{2}+A_{3}M^{3}+\cdots+A_{n}M^{n}, $$ or do you mean the average of these two? Or do you mean something even stranger where some powers of $M$ are to the left of a coefficient and some are to the right?

And why is this important? Because if you factor a matrix polynomial into two others, say, $$ p(\lambda)=q(\lambda)r(\lambda), $$ then it can happen that $p(M)\ne q(M)r(M)$. Whereas the $\lambda$ commute with the matrix coefficients, $M$ and its powers do not necessarily commute with the matrix polynomial coefficients. To see what I mean, suppose $$ P_{0}+\lambda P_{1}+\cdots+\lambda^{l}P_{l} \\ =(Q_{0}+\lambda Q_{1}+\cdots+\lambda^{m}Q_{m})(R_{0}+\lambda R_{1}+\cdots+\lambda^{n}R_{n}). $$ This has a very specific meaning: $$ P_{0}=Q_{0}R_{0},\\ P_{1}=Q_{0}R_{1}+Q_{1}R_{1},\\ P_{2}=Q_{0}R_{2}+Q_{1}R_{1}+Q_{2}R_{0},\\ \cdots $$ But when you substitute a matrix $M$ for $\lambda$, things don't work out any more. First, you have to choose a left-, a right-, or a mixed-evaluation, and you can see that the above relations are not necessarily preserved during evaluation of the polynomial at a matrix: $$ P_{0}+P_{1}M+P_{2}M^{2}+\cdots+P_{l}M^{l} \\ \ne (Q_{0}+Q_{1}M+Q_{2}M^{2}+\cdots+Q_{m}M^{m})(R_{0}+R_{1}M+R_{2}M^{2}+\cdots R_{n}M^{n}). $$ For example, $P_{1}=Q_{0}R_{1}+Q_{1}R_{0}$ is not enough to guarantee $$ P_{1}M=Q_{0}R_{1}M+Q_{1}MR_{0}. $$ So, factorings of matrix polynomials are not generally preserved under evaluation of polynomials at a matrix $M$, regardless of what convention you choose for evaluating those polynomials.

After trying to discourage you, please note that everything works in the case that you are considering. There is an important special case where everything works: If you use evaluation on the right for $p$, $q$ and $r$, and if $Mr(M)=r(M)M$, then you can move all of the powers of $M$ to the far right in $q(M)r(M)$ to obtain the same thing that you would get if you were to evaluate $p(M)$ on the right. (A similar thing holds for left evaluation.) And this condition is automatically satisfied if the right evaluation $r(M)$ is $0$ because $0$ commutes with everything. Hence, the right evaluation of $p$ at $M$ is $0$ if $p=qr$ and if the right evaluation of $r$ at $M$ is $0$.

Proof of Cayley-Hamilton Theorem: So how does this apply in the Cayley-Hamilton Theorem? Suppose $A$ is an $n\times n$ matrix, and let $A_{i,j}$ be the determinant of the matrix obtained by removing the $i-th$ row and $j-th$ column from $A$. The adjunct matrix $A_{\mbox{adj}}=[(-1)^{i+j}A_{j,i}]$ (note the intentional swap of $i$ and $j$) satisfies $$ AA_{\mbox{adj}}=A_{\mbox{adj}}A = \mbox{det}(A)I, $$ where $\mbox{det}(A)$ is the scalar determinant of $A$ and $I$ is the identity matrix. In particular, $$ (A-\lambda I)(A-\lambda I)_{\mbox{adj}}=(A-\lambda I)_{\mbox{adj}}(A-\lambda I)= \mbox{det}(A-\lambda I)I $$ The matrix $(A-\lambda I)_{\mbox{adj}}$ consists of cofactors of $A-\lambda I$ and, as such, each entry in this matrix is a polynomial in $\lambda$ of order between $0$ and $n-1$. By collecting like powers, $$ (A-\lambda I)_{\mbox{adj}}=Q_{0}+\lambda Q_{1}+\cdots+\lambda^{n-1}Q_{n-1}, $$ where $Q_{j}$ are $n\times n$ coefficient matrices. Therefore, you have $$ (Q_{0}+Q_{1}\lambda + \cdots + Q_{n-1}\lambda^{n-1})(A-\lambda I)=(-1)^{n}p(\lambda)I, $$ where $p(\lambda)=\mbox{det}(\lambda I -A)$ is the characteristic polynomial of $A$. This factoring is preserved by right evaluation at $A$ because the evaluation of $A-\lambda I$ at $\lambda =A$ gives you the $0$ matrix, which commutes with every matrix including $A$. Therefore, the right- and left- evaluations of $p(\lambda)I$ at $\lambda=A$ must give you $0$ (these evaluations are trivially the same as the usual evaluation of a scalar polynomial at a matrix--the coefficients may be thought of as scalar multiples of the identity matirx.) The conclusion is that $p(A)=0$, which is the Cayley-Hamilton Theorem.

Note: the real crux of the matter is that $p(\lambda)I=Q(\lambda)(A-\lambda I)$ so that $A$ is an obvious $0$ of the polynomial on the right and, hence, also of the characteristic polynomial. It is an odd factoring that one obtains where all matrix coefficients of $p(\lambda)I$ are scalar multiples of the identity. So the hard part is found in the explicit adjunct formula for the inverse of a matrix, which is one of several reasons why determinants remain important.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.