Sign up ×
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.

Let $A = \begin{pmatrix} 1 & 1& 1\\ 1 & 2 &3 \\ 1 &4 & 5 \end{pmatrix}$ and $D = \begin{pmatrix} 2 & 0& 0\\ 0 & 3 &0 \\ 0 &0 & 5 \end{pmatrix}$.

It is found that right-multiplication by D multiplies each column of A by the corresponding diagonal entry of D, whereas left-multiplication by D multiplies each row of A by the corresponding diagonal entry of D.

Construct a 3 x 3 matrix B, not the identity matrix or zero matrix, such that $AB=BA$.

share|cite|improve this question
What's wrong with $2I$? – Alex Becker Jul 18 '12 at 5:50
$B = A$ or $B = A^2$, etc., will also work. More generally you want to find $B$ which is simultaneously diagonalizable with $A$. – Cocopuffs Jul 18 '12 at 5:54
Hehe, thanks. I was rather trying to construct B using the hint in the middle paragraph and had missed the obvious answers. – Ryan Jul 18 '12 at 5:54
The diagonal hint may be misleading, any diagonal matrix satisfying the requirements is necessarily a multiple of the identity matrix. – Ragib Zaman Jul 18 '12 at 6:11

3 Answers 3

up vote 6 down vote accepted

Note that the determinant of $A$ is $-2$, so your matrix is invertible i.e. there exists some matrix $B=A^{-1}$ such that $AB=BA=I$. If I calculated this correctly $$B = \begin{pmatrix} 1 & 1/2& -1/2\\ 1 & -2 &1 \\ -1 & 3/2 & -1/2 \end{pmatrix}.$$

share|cite|improve this answer

As Alex Becker said, we could use any scalar multiple of $I$ or $A^n$ where $A^n = A\cdot A \dots A$ where $A$ is multiplied by itself $A$ times. We could also use $A^{-1}$ or $(A^{-1})^n$.

Suppose we have any matrices $B$ and $C$ that satisfy $AB=BA$ and $AC=CA$. Then, $A(c_1 B + c_2 C) = c_1 A B + c_2 A C = c_1 BA + c_2 CA = (c_1 B + c_2 C)A$.

From this we conclude that the collection of all matrices that commute with $A$ (under matrix multiplication) forms a vector space, since any linear combinations of matrices that commute with $A$ will also commute with $A$.

share|cite|improve this answer
Does anyone know what is the basis for this vector space? – Andrew Salmon Jul 18 '12 at 6:05
Finding a basis can get quite complicated. If $A$ is diagonalizable it is quite easy, but if $A$ is in Jordan normal form with Jordan blocks of size greater than $1$, determining the matrices which commute with $A$ can get quite tricky, even if $A$ has only the eigenvalue $1$ or $0$. – Geoff Robinson Jul 18 '12 at 8:21

Notice that $A$ is diagonalizable, so any matirx with the same eigenvectors will commute with $A$, for example

$$B = \begin{pmatrix} 0.2481 & 0.31385 & -0.034765 \\ 0.48816 & -0.65821 & 0.76725 \\ -0.20907 & 1.0811 & 0.28335 \end{pmatrix}$$

share|cite|improve this answer

Your Answer


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.