# Matrix $\mathbf C$ such that $\mathbf A = \mathbf B \mathbf C=\mathbf C\mathbf B$

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.

• 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 $C$? – Martigan Feb 4 '15 at 10:55
• What property of symmetric matrices are u hinting? – BigBang Feb 4 '15 at 11:10
• 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. – mvw Feb 4 '15 at 11:12
• If B, C were uper triangular, then A would have been upper. – BigBang Feb 4 '15 at 11:30
• Since $A$ is symmetric, if this comes from the product of two matrices, the lines of one should be the column of the other... – Martigan Feb 4 '15 at 12:10

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.

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!!