4
$\begingroup$

I have to show that a set of matrices that commute with the matrix $$\begin{bmatrix} 0&a_2 &a_3 &\cdots & a_n\\ 0& 0 &0 &\cdots &0 \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ 0&0 &0 &\cdots &0 \end{bmatrix}$$ is a subspace and I have to find its basis.

I know that for proving a subspace I have to show that the zero matrix is in this set and that if $A$ and $B$ is in the set than also $A+B$ is in the set and also for if $r \in \mathbb{R}$ and matrix A is from the set than also $rA$ is from this set. I also understand that two matrices commute if $AB=BA$ but I don't know with which matrix should I multiply the given one.

Ok, so I took an arbitrary matrix from $\mathbb{R}^{n \times n}$ $$ \begin{bmatrix} e_{11} &e_{12} &e_{13} & \cdots & e_{1n} \\ e_{21} & \cdots&\cdots&\cdots&e_{2n} \\ \vdots & \vdots &\vdots &\ddots & \vdots \\ e_{n1} & \cdots & \cdots & \cdots &e_{nn} \end{bmatrix} $$ If I try to multiply it with the given one I get

  1. $$\begin{bmatrix} e_{11} &e_{12} &e_{13} & \cdots & e_{1n} \\ e_{21} & \cdots&\cdots&\cdots&e_{2n} \\ \vdots & \vdots &\vdots &\ddots & \vdots \\ e_{n1} & \cdots & \cdots & \cdots &e_{nn} \end{bmatrix} \begin{bmatrix} 0&a_2 &a_3 &\cdots & a_n\\ 0& 0 &0 &\cdots &0 \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ 0&0 &0 &\cdots &0 \end{bmatrix}= \begin{bmatrix} 0&e_{11}a_2 &e_{11}a_3 &\cdots &e_{11} a_n\\ 0& 0 &0 &\cdots &0 \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ 0&0 &0 &\cdots &0 \end{bmatrix} $$

2. $$ \begin{bmatrix} 0&a_2 &a_3 &\cdots & a_n\\ 0& 0 &0 &\cdots &0 \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ 0&0 &0 &\cdots &0 \end{bmatrix} \begin{bmatrix} e_{11} &e_{12} &e_{13} & \cdots & e_{1n} \\ e_{21} & \cdots&\cdots&\cdots&e_{2n} \\ \vdots & \vdots &\vdots &\ddots & \vdots \\ e_{n1} & \cdots & \cdots & \cdots &e_{nn} \end{bmatrix} = \begin{bmatrix} a_2 e_{21}+a_3 e_{31}+ \cdots + a_n e_{n1} & a_2 e_{22}+\cdots+a_n e_{n2} & \cdots & \cdots & a_2 e_{2n}+ \cdots + a_n e_{nn} \\ 0&0&\cdots&\cdots&0 \\ \vdots & \vdots & \vdots & \ddots &0 \\ 0 &0&0 & \cdots &0 \end{bmatrix}$$

Now I have to compare the elements in the both computed matrices and I get a system of equations which I don't know how to solve.

$\endgroup$
  • 1
    $\begingroup$ Take an arbitrary matrix $A\in\mathbb R^{n\times n}$ and multiply it from the left and from the right with the given matrix. Then, in order both products to be one and the same matrix you will find what conditions have to satisfy each matrix from the subspace you are looking for. $\endgroup$ – Svetoslav Feb 17 '16 at 22:41
  • 1
    $\begingroup$ It is the set of solutions of a homogenous linear equation... $\endgroup$ – user251257 Feb 18 '16 at 2:56
  • $\begingroup$ @Svetoslav I multiplied as you said, but now I don't know how to continue with this equations I got. $\endgroup$ – nashoido Feb 18 '16 at 11:25
3
$\begingroup$

Hint: The map $T:\Bbb R^{n \times n} \to \Bbb R^{n \times n}$ given by $$ T(B) = AB - BA $$ is a linear transformation. As such, the kernel of $T$ is a subspace of $\Bbb R^{n \times n}$.

$\endgroup$

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.