Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've heard that for a matrix $A\in M_n(\mathbb{C})$, if $A^3=A$, then $A$ is diagonalizable. Does there happen to be a proof or reference as to why this is true?

Out of curiosity, is it necessary that the entries be from $\mathbb{C}$? Would any field $F$ work just as well, or is this possibly a special fact related to the properties of $\mathbb{C}$?

share|improve this question
1  
If you have such a matrix $\mathbf A$, then it should satisfy the equation $\mathbf A(\mathbf A^2-\mathbf I)=\mathbf 0$. This means $\mathbf A$ could either be the zero matrix or an involutory matrix... –  J. M. Oct 22 '11 at 10:45
6  
@J.M.: No, $\mathbf A$ could for instance be $\pmatrix{1&0\\0&0}$. –  joriki Oct 22 '11 at 10:48
    
Oh, indeed I missed those diagonal matrices with elements from $\{\pm 1,0\}$... –  J. M. Oct 22 '11 at 10:55

4 Answers 4

up vote 13 down vote accepted

You might also want to look at the minimal polynomial $\mu_A$ of $A$. If $A^3-A=0$, then $\mu_A \mid X^3-X=X(X^2-1)$. If $\mathrm{char}(K)\neq 2$ (if $K$ is the underlying field) then this polynomial is equal to $X(X-1)(X+1)$ and a matrix with a minimal polynomial which splits into linear factors with multiplicity 1 is diagonalisable.

share|improve this answer
    
Our answers are nicely complementary in that yours doesn't require the field to be algebraically closed but only works for $A^3$, whereas mine requires the field to be algebraically closed but works for all powers. Your answer can be generalized to other powers, but not all powers and algebraically closed fields, since $X^n-1$ has multiple roots if $\operatorname{char} K\mid n$. –  joriki Oct 22 '11 at 11:53
    
P.S.: I just realized that that's also precisely the case where my proof doesn't work, so your answer actually covers all the same cases and additionally the case $A^3$ without algebraic closure. –  joriki Oct 22 '11 at 12:03

Here is an argument/explanation that does not use Jordan normal form or charateristic polynomials. Instead it uses a more linear transformation-based viewpoint, and properties of projectors. I find this a useful way to think about these kinds of questions, which is why I'm positing it.


Note first that $A^3 = A$ implies that $A^4 = A^2$. This means that $A^2$ is a projection. In general, if $P^2 = P$, for some linear transformation of a vector space $V$ (over any field), then $V$ is the direct sum of the image of $P$ and the image of $I- P$. (Here $I$ denotes the identity.) Furthermore, on the the image of $P$, the transformation $P$ acts by the identity, while on the image of $I - P$, it acts by zero.

So in our case, the $3$-dim'l vector space $V$ on which your matrix $A$ is acting splits as the direct sum of the image of $A^2$, on which $A^2$ acts by the identity, and as the image of $I - A^2$, on which (using the equation $A - A^3 = 0$) we see that $A$ acts by zero.

So we have partially diagonalized $A$; we have decomposed $V$ into a sum of two subspaces, each invariant under $A$, with $A^2 = I$ on the first, and $A = 0$ on the second.

To complete the diagonalization, we make the same kind of argment, but now we may assume that $A^2 = I$.

From $A^2 = I$, we find that $\bigl(\dfrac{I-A}{2}\bigr)^2 = \dfrac{I-A}{2}$. Thus $(I-A)/2$ is again a projector, and so the subspace on which $A^2 = I$ decomposes as a sum of two subspaces, one on which $(I-A)/2 = I,$ which is to say $A = - I$, and one on which $(I-A)/2 = 0$, which is to say $A = I$.

Putting it altogether, we decomposed our original space $V$ as the sum of three $A$-invariant subspaces, on which $A$ acts by $0$, $-1$, and $1$ respectively.


The argument works over any field where $2$ is invertible. If we are in char. $2$, then the decomposition into the sum of spaces on which $A = 0$ and $A^2 = I$ is still possible, but (as Joriki points out in his answer) we can't necessarily diagonalize a matrix satisfying $A^2 = I$.

One way to see this is to note that in char. $2$, $A^2 = I$ is equivalent to $(A-I)^2 = 0$, and so we can construct matrices $A$ such that $A^2 = I$ by choosing nilpotent matrices $N$ such that $N^2 = 0$, and then setting $A = I + N$. If $N \neq 0$ (which is possible for $n\times n$ matrices with $n > 1$), then such an $A$ (identity plus non-zero nilpotent) is not diagonalizable (in any characteristic; but away from char. $2$, matrices of this form can't satisfy $A^2 = 0$).

share|improve this answer

An $n$ by $n$ matrix $A$ with coefficients in a field $K$ is diagonalizable if and only if its minimal polynomial $f$ has no multiple roots and splits over $K$.

Clearly, if $A$ is diagonalizable, $f$ has no multiple roots and splits over $K$.

For the converse, one can use the Chinese Remainder Theorem to reduce the statement to the case $A=0$.

share|improve this answer

This holds because it holds for the Jordan normal form. Every matrix over $\mathbb C$ is similar to a matrix in Jordan normal form. The third power of a Jordan normal block of size at least $2$ is

$$ \pmatrix{\lambda&1\\ &\lambda&\ddots\\ &&\ddots&1\\ &&&\lambda&}^3 = \pmatrix{\lambda^3&3\lambda^2\\ &\lambda^3&\ddots\\ &&\ddots&3\lambda^2\\ &&&\lambda^3&} $$

(with further non-zero entries further above the diagonal). If $A^3=A$, then this also holds for any similar matrix, and thus for the Jordan normal form, which implies $\lambda=\lambda^3$ and $3\lambda^2=1$. These equations don't have a common solution, which shows that all Jordan blocks are of size $1$, that is, the Jordan normal form is diagonal. This clearly also works for any other integer power of $A$.

Regarding arbitrary fields, this exposition proves that every matrix over an algebraically closed field is similar to a matrix in Jordan normal form. Thus the result holds in any algebraically closed field $\mathbb F$ in which there is no common solution for $\lambda=\lambda^3$ and $3\lambda^2=1$. For $\operatorname{char}\mathbb F=3$, the second equation reads $0=1$, which has no solutions. For $\operatorname{char}\mathbb F\ne3$, we can multiply the first equation by $3$ and substitute for $3\lambda^2$ from the second equation to get $3\lambda=\lambda$ and thus $3=1$ (since $\lambda=0$ doesn't solve the second equation). Subtracting $1$ yields $2=0$, so the statement holds if $\operatorname{char}\mathbb F\ne2$.

For $\operatorname{char}\mathbb F=2$, we have $\pmatrix{1&1\\0&1}^3=\pmatrix{1&1\\0&1}$, and this matrix is not diagonalizable.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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