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I came across the following problem that says:
Let $A$ be a $3 \times 3$ matrix with real entries.If $A$ commutes with all $3 \times 3$ matrices with real entries,then the number of distinct eigenvalues of $A$ is which of the following:

$ 0,1,2,3$ or none of these.

I do not know how to progress with this particular problem. Can someone point me in the right direction? Thanks in advance for your time.

EDIT: Even though the question has been asked few months ago and I got some nice explanations ,I revisited the problem and I am little confused about the conclusion of the problem. The scalar matrices of the form $\lambda I$ satisfies the property. So what should be the number of distinct eigenvalues? Here all eigenvalues will be $1$. So,what is the conclusion? Is the answer $1$ or $0$? Can someone comment on it?

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If you want to go a little further, you could try to prove that the center of the algebra $L(E)$ (the linear operators on a $K$ vector space of any dimension), that is $\{u\in L(E)\,;\, uv=vu\forall v\in L(E)\}$, consists of the scalar operators $K\cdot \mbox{Id}_E$. The key point is that $u\in L(E)$ is scalar if and only if $u(x)$ and $x$ are linearly dependent for every $x\in E$. –  1015 Jun 28 '13 at 17:09

6 Answers 6

up vote 2 down vote accepted

There is a known fact of group theory that says: $Z(GL_n)=\lambda I_n$, for all $\lambda\in\mathbb{R}$, being $Z(GL_n)$ the center of the group: the matrices that commute with every matrix of $GL_n$. I will set the guidelines for the demonstration:

Knowing the equivalence of these matrices with vector space endomorphisms, we take the homomorphism $f:\mathbb{R}^n\longrightarrow \mathbb{R}^n$, that has the matrix $A$ as the homomorphism matrix, being $A\in Z(GL_n)$.

We now show that $\lbrace x,f(x)\rbrace$ are linearly dependent: if they weren't, we could build a base of $\mathbb{R}^n$: $B=\lbrace v_1=x,v_2=f(x),v_3,v_4,...,v_n\rbrace$, for $n$ vectors, and there's only one homomorphism $g$ that swaps the first two vectors and keeps the rest the same: $$G=\left( \begin{array}{ccccc} 0 & 1 & 0 & \cdots & 0\\ 1 & 0 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & 1 \end{array}\right) $$ It must be true that $AG=GA$, so $fg(v_1)=gf(v_1)\Rightarrow f(v_2)=g(v_2)=v_1=x$ which means that $ff(x)=x.$

Now we consider the homomorphism $H$: $$H=\left( \begin{array}{ccccc} 1 & 0 & 0 & \cdots & 0\\ 1 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & 1 \end{array}\right) $$ This is also an isomorphism and $AH=HA$: $$fh(v_1)=f(v_1+v_2)=h(v_2)=v_2\Rightarrow v_2+f(v_2)=v_2\Rightarrow f(v_2)=0$$ We previously had that $x=ff(x)=f(v_2)$, so $x=0$, which is absurd because it was a vector of a minimal base, so our hypothesis was false, and $x, f(x)$ are linearly dependent. This means that $f(x)=\lambda(x)x$, with $\lambda$ a real function, that in the most general way, depends on $x$.

If $x$ and $y$ are linearly independent vectors, then we have that calling $z=x+y$: $$\lambda(z)x+\lambda(z)y=f(z)=f(x)+f(y)=\lambda(x)x+\lambda(y)y$$ This means that $$(\lambda(z)-\lambda(x))x+(\lambda(z)-\lambda(y))y=0$$ And $x,y$ are linearly independent, so: $$\lambda(z)=\lambda(x).$$ So if $x,y$ are linearly independent, $\lambda$ is not a function of the vector, but a constant.

If they're dependent, then we can do a similar trick: if they're dependent, then $y=xa$, for some real $a$: $$\lambda(y)y=f(y)=af(x)=a\lambda(x)x=\lambda(x)y.$$ If $y\not =0$, then $\lambda$ is constant too.

We have proved that for every vector: $Av=\lambda v$, for every $A\in Z(GL_n)$, so: $$A=\lambda\left( \begin{array}{ccccc} 1 & 0 & 0 & \cdots & 0\\ 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & 1 \end{array}\right) $$

The converse is obvious: for every matrix of the form $\lambda I$, the matrix belongs to $Z(GL_n)$, then we have that if a matrix commutes with every other matrix, it has the form of the above, and those matrix have a unique eigenvalue: $\lambda$, which is a root of $n$ degree of its characteristic polynomial.

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If $\mathrm A$ commutes with all matrices, then it commutes with this kind of matrix $$ \mathrm E_{ij} = \text{ the matrix with only zeros but a 1 at line i, column j }$$

Check what it means for $\mathrm A$ to commute with $\mathrm E_{i,j}$.

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Hint: The scalar matrices are precisely those that commute with all other square matrices of the same size.

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Hint: The eigenvalues of $A$ come from its characteristic polynomial, which is a real polynomial with degree $3$. So it always have a real root and it has at most $3$ roots.

Do you agree?

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Yes,that is true.But how should I use the fact that $A$ commutes with all $3 \times 3$ matrices with real entries? –  learner Feb 12 '13 at 14:11

Let $A, B$ be 3×3 matrices with real entries and A commute with B,that is $AB=BA$. Let $\lambda$ be an eigenvalue of $A$. Then there exists a nonzero vector $v$ such that $$Av=\lambda v$$.

Now, we can have the following: $$BAv=B\lambda v$$ By commutativity of $A$ and $B$, we have $$(AB)v=\lambda Bv$$ $$\Rightarrow A(Bv)=\lambda (Bv)$$ $$\Rightarrow A=\lambda I$$ From this we can say that $A$ is a scalar matrix, which is a triangular matrix whose eigenvalue is $\lambda \in \mathbb{R}$.

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(A-λI)(B v )= 0 does not imply A-λI=0. –  Librecoin Feb 12 '13 at 18:40
If we Know the invertibility of $Bv$, I think it is immediate. Can it be invertible? –  ftolessa Feb 13 '13 at 4:52
The vector B v need not be non-zero, since B may be singular. Even if it were non-zero, (A-λI)(B v )= 0 would still not imply A-λI=0. For example, if λ is an eigenvalue of a matrix A, then (A-λI) v = 0 has a non-zero solution v. –  Librecoin Feb 13 '13 at 21:17

To start with you can just grind out the sum.

With $$A = \left[\begin{array}{ccc} \alpha_{1,1} & \alpha_{1,2} & \alpha_{1,3} \\ \alpha_{2,1} & \alpha_{2,2} & \alpha_{2,3} \\ \alpha_{3,1} & \alpha_{3,2} & \alpha_{3,3} \end{array}\right],\; B = \left[\begin{array}{ccc} \beta_{1,1} & \beta_{1,2} & \beta_{1,3} \\ \beta_{2,1} & \beta_{2,2} & \beta_{2,3} \\ \beta_{3,1} & \beta_{3,2} & \beta_{3,3} \end{array}\right] $$

Then $(AB)_{1,1} = \alpha_{1,1} \beta_{1,1} + \alpha_{1,2} \beta_{2,1} + \alpha_{1,3} \beta_{3,1}$.

But $(AB)_{1,1} = (BA)_{1,1} = \beta_{1,1} \alpha_{1,1} + \beta_{1,2} \alpha_{2,1} + \beta_{1,3} \alpha_{3,1}$.

Since $\beta_{i,j}$ can be anything, $\alpha_{1,2} = \alpha_{1,3} = \alpha_{2,1} = \alpha_{3,1} = 0.$

If you repeat that calulation for each cell of $(AB)_{i,j}$ then you get Tharsis's result that $A$ is a scalar matrix, i.e.: $A$ is already of the form $A = \lambda I$ where $I$ is the identity matrix. Eigenvalues of $A$ are those values $\mu$ which satisfy $$det(A - \mu I) = 0.$$ But $A = \lambda I$. So to get eigenvalues you have to solve $$det((\lambda - \mu) I) = 0,$$ i.e.: $$(\lambda - \mu)^3 det(I) = 0.$$ For any one particular $\lambda$ how many such $\mu$ can there be?

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