1
$\begingroup$

Let $$\mathbf A =\begin{pmatrix}1&1&1\\1&2&2\\1&2&3\end{pmatrix} \text{and }\mathbf B=\begin{pmatrix}1&0&0\\1&1&0\\1&1&1\end{pmatrix}$$

Then which of following is true

  1. There exists a matrix $\mathbf C$ such that $\mathbf A=\mathbf B\mathbf C=\mathbf C\mathbf B$

  2. There exists no matrix $\mathbf C$ such that $\mathbf A=\mathbf B\mathbf C$

  3. There exists a matrix $\mathbf C$ such that $\mathbf A=\mathbf B\mathbf C$ but $\mathbf A \ne \mathbf C\mathbf B$

  4. There exists no matrix $\mathbf C$ such that $\mathbf A=\mathbf C\mathbf B$.

My attempt: since $\mathbf B$ is invertible I see option 1 to be correct. But it can't be that easy. Also, $\mathbf A$ is symmetric and $\mathbf B$ is triangular so think there some trick here. So I need suggestions. Thanks.

$\endgroup$
  • $\begingroup$ Hints. First one: $3=1+1+1$ and $2=1+1$. Second one:$A$ is symmetric, what can we say about the lines and columns of $B$ and $C$? $\endgroup$ – Martigan Feb 4 '15 at 10:55
  • $\begingroup$ What property of symmetric matrices are u hinting? $\endgroup$ – BigBang Feb 4 '15 at 11:10
  • $\begingroup$ Statement 1. would be true if $B$ and $C$ commute. According to this article, $B$ and $C$ should be upper triangular matrices, which $B$ is not. $\endgroup$ – mvw Feb 4 '15 at 11:12
  • $\begingroup$ If B, C were uper triangular, then A would have been upper. $\endgroup$ – BigBang Feb 4 '15 at 11:30
  • $\begingroup$ Since $A$ is symmetric, if this comes from the product of two matrices, the lines of one should be the column of the other... $\endgroup$ – Martigan Feb 4 '15 at 12:10
2
$\begingroup$

Yes, it's that easy: both matrices $\;A\,,\,\,B\;$ are invertible. and thus

$$A=BC\implies C=B^{-1}A$$

Now, if also for the same $\;C\;$ we'd have $\;A=CB\;$ , then

$$B^{-1}A=C=AB^{-1}\implies AB=BA$$

which is false as you can easily check, and thus you can already solve all four points.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

I consider $$C=\left(\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i \end{array}\right),$$ then $$BC=\left(\begin{array}{ccc}a&b&c\\a+d&b+e&c+f\\a+d+g&b+e+h&c+f+i \end{array}\right),$$ and $$CB=\left(\begin{array}{ccc}a+b+c&b+c&c\\d+e+f&e+f&f\\a+d+g&h+i&i \end{array}\right).$$

If, we want $A=BC$, we get $$C=\left(\begin{array}{ccc}1&1&1\\0&1&1\\0&0&1 \end{array}\right),$$ in the case $A=CB$ we have $$C=\left(\begin{array}{ccc}0&0&1\\-1&0&2\\-1&-1&3 \end{array}\right).$$

Therefore:

1) False 2) False 3) True 4) False

Regards!!

| cite | improve this answer | |
$\endgroup$

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.