0
$\begingroup$

I am having troubles finding the pattern here.

So how can I find a matrix with the given requirements of the title?

I note that this occurs with the following:

$$ A = \begin{bmatrix} 1 & -1\\ -1 & 1\\ \end{bmatrix} $$ $$ B = \begin{bmatrix} 1 & 1\\ 1 & 1\\ \end{bmatrix} $$

If I were to multiply $B$ by any scalar, then the resulting matrix would ultimately end up being the zero vector. But how would I do this for matrices of bigger sizes? How could I create other matrices with these requirements?

  • 0
    What are you looking for in dimension $n>2$ ? A triple $(A,B,C)$ such that $AB=AC$ ? Or $B$ and $C$ being given, find $A\neq0$ such that $AB=AC$ ? – 2017-01-28
  • 0
    "If I were to multiply $B$ by any scalar, then the resulting matrix would ultimately end up being the zero vector." What exactly does that mean? – 2017-01-28
  • 0
    It doesn't matter what the dimensions are as long as they have the requirements. – 2017-01-28
  • 0
    I meant to say that $(A)(kB) = 0$. @Thomas Andrews – 2017-01-28
  • 0
    Ah, so you are saying that $AB=AC$ if $C=kB$. That was definitely not clear. Not the least reason is that the $2\times 2$ matrix $A(kB)$ is rarely called a vector in such a context. – 2017-01-28
  • 0
    Why don't you answer my question ? – 2017-01-28
  • 0
    @JeanMarie I said the dimensions doesn't matter - I'm not sure what the other part of your question means though. B and C are not given, I have to construct the two matrices(I think that's what you're asking, clarify me if I'm wrong.) – 2017-01-28

4 Answers 4

0

$$AB=AC \ \iff \ AD=0 \ \ \text{with} \ \ D:=A-B \neq 0 \ \ \text{by hypothesis}$$

$AD=0$ is equivalent to the fact that, for each column vector $D_k$ ($k=1...n$) of $D$, we have:

$\tag{1} \forall k, \ \ AD_k=0 \ \ \iff \ \ D_k \in Ker A.$

2 cases can occur:

  • if rank$(D)=n$: $\iff \ D \ $ invertible$ \iff \ $ (in view of (1)) $ \ \ker(A) \ $ possesses $n$ independent vectors $ \ \iff \ \ker(A)=\mathbb{R^n} \ \iff \ A=0$: impossible.

  • if rank$(D) < n$: A matrix $A$ can be found. Her is the explanation: there exist at least a linear combination :

$$a_1D_1+a_2D_2+\cdots a_nD_n=0$$

In this case, define $A$ as the matrix with all its lines made of entries:

$$a_1,a_2,\cdots a_n$$

Conclusion: Here is a "recipe" for finding $A,B,C$: - take any matrices $B,C$ as long as $B-C$ is not full rank (for example 2 random matrices with two identical last columns).

Pick a vector $\pmatrix{a_1\\a_2\\...\\a_n}$ in $\ker(B-C)$ and take

$$A=\begin{pmatrix}1 \\ 1 \\...\\ 1\end{pmatrix}\pmatrix{a_1 & a_2 &...& a_n}$$

  • 0
    I have corrected a misnaming ($D$ instead of $A$ in a certain line) – 2017-01-29
3

$A(B-C)$ should be $0$, so $B-C$ should be in the null space of $A$.

  • 0
    What do we do when $rank(B-C)=n$ (dimension of the ambient space) ? – 2017-01-28
1

Take the first column of $A$ to be all zeros, and the rows other than the first rows of $B, C$ to be identical.

0

An easier problem would be to find $X,Y\in M_n(\mathbb{R}), X, Y\neq 0$ and $XY=0_n$. Then let, $A = X, B = Y+C,$ and $C$ arbitrary.

Perhaps the easiest examples would be elementary matrices, where $X = E_{ij}, Y = E_{kh}$, then $XY = E_{\delta_{ik} \delta_{jh}} = 0_n$ except when $i=k, j=h$.