Let $A$ be a $3\times 3$ matrix over $\mathbb {R}$ such that $AB =BA$ for all matrices $B$ over $\mathbb {R}$ then what can we say about such matrix $A$. or such matrix $A$ must be orthogonal matrix? Can we say anything about its eigen values? I tried by taking random examples also tried to construct such $3 \times 3$ matrices. But i am not able to get any proper conclusion and proof.


marked as duplicate by Marc van Leeuwen linear-algebra Jan 26 '15 at 11:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 7
    $\begingroup$ $A$ must be a scalar multiple of the identity. This is a nice exercise. $\endgroup$ – Qiaochu Yuan May 9 '12 at 7:48
  • $\begingroup$ can i get little hint please $\endgroup$ – srijan May 9 '12 at 7:53
  • $\begingroup$ Choose $B$ to be some particularly simple matrices. $\endgroup$ – Qiaochu Yuan May 9 '12 at 7:53
  • $\begingroup$ Take some trivial $B$'s like a $B$ will only one column non-zero and all the other entries being zero and you will get some constraints on the entries of $A$. $\endgroup$ – user17762 May 9 '12 at 7:54
  • 1
    $\begingroup$ Do you know what $Ae_i$ and $e_i^tA$ look like for the column ($3\times1$) matrices $e_i$ having entry $1$ in the $i$th row and $0$ elsewhere? Now try making a $B$ from this somehow, say $B=(\delta_{ai}\delta_{bj})=e_ae_b^t$. $\endgroup$ – bgins May 9 '12 at 8:22

If we say that $A_i$ is the $i$th column of $A=(a_{ij})$ and $A^{(j)}$ is the $j$th row and $e_i$ is the $3\times1$ unit column vector with $1$ in the $i$th row (and consequently $0$ everywhere else), then for $B=e_ie_j^t$, which is $1$ at $(i,j)$ and $0$ elsewhere, we find that $$AB=Ae_ie_j^t=A_ie_j^t$$ and $$BA=e_ie_j^tA=e_iA^{(j)}$$ are both $3\times3$ matrices, the first with the $i$th column of $A$ in its $j$th column, the second with the $jth$ row of $A$ in its $i$th row. For these to be equal, we must have $a_{ij}=0$ whenever $i\ne j$. Furthermore, the single nonzero entry in the products above is again at $(i,j)$, where we find: $$a_{ii}=(AB)_{ij}=(BA)_{ij}=a_{jj}\,.$$ Thus $A=\lambda I$ must be a scalar (multiple of the identity) matrix.

So how does this look? Well, here goes: $$ e_1=\left[\matrix{1\\0\\0}\right],\quad e_2=\left[\matrix{0\\1\\0}\right],\quad e_3=\left[\matrix{0\\0\\1}\right], $$ $$ e_1^t=\left[\matrix{1&0&0}\right],\quad e_2^t=\left[\matrix{0&1&0}\right],\quad e_3^t=\left[\matrix{0&0&1}\right], $$ $$ e_1e_1^t=\left[\matrix{1&0&0\\0&0&0\\0&0&0}\right],\quad e_1e_2^t=\left[\matrix{0&1&0\\0&0&0\\0&0&0}\right],\quad e_1e_3^t=\left[\matrix{0&0&1\\0&0&0\\0&0&0}\right],\quad $$ $$ e_2e_1^t=\left[\matrix{0&0&0\\1&0&0\\0&0&0}\right],\quad e_2e_2^t=\left[\matrix{0&0&0\\0&1&0\\0&0&0}\right],\quad e_2e_3^t=\left[\matrix{0&0&0\\0&0&1\\0&0&0}\right],\quad $$ $$ e_3e_1^t=\left[\matrix{0&0&0\\0&0&0\\1&0&0}\right],\quad e_3e_2^t=\left[\matrix{0&0&0\\0&0&0\\0&1&0}\right],\quad e_3e_3^t=\left[\matrix{0&0&0\\0&0&0\\0&0&1}\right],\quad $$ $$ A=\left[\matrix{ a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} }\right],\quad A_1=\left[\matrix{a_{11}\\a_{21}\\a_{31}}\right],\quad A_2=\left[\matrix{a_{12}\\a_{22}\\a_{32}}\right],\quad A_3=\left[\matrix{a_{13}\\a_{23}\\a_{33}}\right], $$ and using a superscript for my nonstandard row notation, $$ A^{(1)}=\left[\matrix{a_{11}&a_{12}&a_{13}}\right],\quad A^{(2)}=\left[\matrix{a_{21}&a_{22}&a_{23}}\right],\quad A^{(3)}=\left[\matrix{a_{31}&a_{32}&a_{33}}\right]=(A^t)_3^t. $$ So for example with $i=2$ and $j=3$, we get $B=e_ie_j^t=e_2e_3^t$ as above so that $AB=A(e_2e_3^t)=(Ae_2)e_3^t$ since matrix multiplication is associative. Furthermore, $$ Ae_2= \left[\matrix{ a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} }\right] \cdot \left[\matrix{0\\1\\0}\right]= \left[\matrix{ a_{11}\cdot0 + a_{12}\cdot1 + a_{13}\cdot0\\ a_{21}\cdot0 + a_{22}\cdot1 + a_{23}\cdot0\\ a_{31}\cdot0 + a_{32}\cdot1 + a_{33}\cdot0 }\right] =\left[\matrix{a_{12}\\a_{22}\\a_{32}}\right] =A_2 $$ so that $$ AB=(Ae_2)e_3^t=A_2e_3^t =\left[\matrix{a_{12}\\a_{22}\\a_{32}}\right] \cdot\left[\matrix{0&0&1}\right] =\left[\matrix{ a_{12}\cdot0&a_{12}\cdot0&a_{12}\cdot1\\ a_{22}\cdot0&a_{22}\cdot0&a_{22}\cdot1\\ a_{32}\cdot0&a_{32}\cdot0&a_{32}\cdot1 }\right] =\left[\matrix{0&0&a_{12}\\0&0&a_{22}\\0&0&a_{32}}\right] $$ and, similarly (without all the gory details), $$ e_3^tA=\left[\matrix{0&0&1}\right]\cdot \left[\matrix{ a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} }\right]= \left[\matrix{a_{31}&a_{32}&a_{33}}\right] =A^{(3)} $$ so that $$ BA=(e_2e_3^t)A=e_2(e_3^tA)=e_2A^{(3)}= \left[\matrix{0\\1\\0}\right]\cdot \left[\matrix{ a_{31}&a_{32}&a_{33}} \right]= \left[\matrix{0&0&0\\a_{31}&a_{32}&a_{33}\\0&0&0}\right]. $$ Thus using $B=e_2e_3^t$, we can see from $AB=BA$ that $a_{22}=a_{33}$ and $a_{i2}=0=a_{3j}$ for $i\ne2,j\ne3$.

  • 1
    $\begingroup$ Note that this does not show $a_{ii}=a_{jj}$ for all $i,j$. $\endgroup$ – Alex Becker May 9 '12 at 8:45
  • $\begingroup$ @AlexBecker: Thanks, I fixed it now. $\endgroup$ – bgins May 9 '12 at 9:01
  • $\begingroup$ @bgins Can you explain how matrices $A$ and $B$ are being multiplied to each other. I am sorry i am not able to get you point. How this matrix $B=e_ie_j^t$ has come? $\endgroup$ – srijan May 9 '12 at 9:41
  • 1
    $\begingroup$ I added a second section graphically depicting many of the the matrices, certainly enough to see what's going on. I also changed my nonstandard notation for the rows of $A$ by introducing parentheses around the superscripted row numbers. What you need to know are the definition of the $ij$th entry of a matrix, namely that it is the entry in row $i$ and column $j$, and the rule for matrix multiplication. $\endgroup$ – bgins May 9 '12 at 20:36

Here is one way to prove this, a bit less elementary than what was suggested in the comments (but it avoids any computations). It is a fundamental fact (and easy to prove) that if $A,B$ are commuting matrices, then every eigenspace for $B$ must be $A$-stable (i.e., if $v$ is in the eigenspace of$~B$ for eigenvalue$~\lambda$, then so is $A\cdot v$). Also, a $1$-dimensional subspace $\langle v\rangle$ is $A$-stable precisely if $v$ is an eigenvector for $A$ (immediate from the definition).

On one hand if $A$ is a scalar multiple of identity, then it clearly commutes with every $B$ (by definition any linear operator commutes with scalar multiplications). On the other hand if $A$ is not a scalar multiple of identity, then there exists some vector $v$ that is not an eigenvector of $A$ (if every nonzero vector were eigenvector, one easily shows they all have the same eigenvalue $\lambda$, and $A$ would be $\lambda I$). Then choose $B$ such that $\langle v\rangle$ is an eigenspace for $B$ (easy). By the above, if $B$ would commute with $A$, then $\langle v\rangle$ would be $A$-stable, and therefore $v$ an eigenvector for $A$. But it isn't, so $A$ and $B$ do not commute.

Note that the facts that the base field is $\Bbb R$, that the dimension is$~3$ or even that it is finite are irrelevant to this question.


Here is a simple exercise which I find useful few times.

Exercise Let $D$ be a diagonal matrix, with pairwise distinct diagonal entries. Then $D$ commutes only with diagonal matrices.

Proof: Let $CD=DC$. Equaling the $ij$ entries in these matrices we get:


Thus, if $i \neq j$we have $d_{ii} \neq d_{jj}$ and thus $c_{ij}=0$..

Now back to your problem. Pick $B$ a diagonal matrix with pairwise distinct diagonal entries. Then $AB=BA$ implies that $A$ is diagonal.

Next pick $B$ the matrix with all ones, that is $b_{ij}=1 \forall i,j$. $AB=BA$ and $A$ diagonal implies that all entries of $A$ are equal.


Here's another solution. Since $A$ is $3\times3$ and $3$ is odd, $A$ has a real eigenvalue $\lambda$ (if you're comfortable with complex eigenvalues, we don't even need the dimension to be odd.) We claim that $A=\lambda I$. To prove this, let $w = (w_1,w_2,w_3)$ be an arbitrary vector in $\mathbb R^3$; it suffices to show that $wA =\lambda w$.

Let $v$ be an eigenvector corresponding to $\lambda$. Some coordinate of $v$ must be nonzero, and $v$ can be scaled at will; without loss of generality, say the first coordinate of $v$ equals $1$. Now set $B$ to be the matrix whose columns are $w_1v$, $w_2v$, and $w_3v$. We easily compute that $AB$ is the matrix whose columns are $\lambda w_1v$, $\lambda w_2v$, and $\lambda w_3v$; in particular, the top row of $AB$ equals $(\lambda w_1,\lambda w_2,\lambda w_3) = \lambda w$. This is also the top row of $BA$, which equals the top row of $B$ times $A$, which equals $w$ times $A$. This establishes that $wA =\lambda w$ as desired.


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