Prove there does not exist invertible matrix C satisfying A = CB

[songoku]

Homework Statement

Please see below

Relevant Equations

Matrix Multiplication

My attempt:

Let C =
$$\begin{pmatrix}
c_{11} & c_{12} & c_{13} \\
c_{21} & c_{22} & c_{23} \\
c_{31} & c_{32} & c_{33}
\end{pmatrix}$$

If C is multiplied by B, then:

1)
a₂₁ = c₂₁ . b₁₁
0 = c₂₁ . b₁₁ $\rightarrow c_{21}=0$

2)
a₃₁ = c₃₁ . b₁₁
0 = c₃₁ . b₁₁ $\rightarrow c_{31}=0$

3)
a₃₂ = c₃₂ . b₂₂
0 = c₃₂ . b₂₂ $\rightarrow c_{32}=0$

But a₃₃ = c₃₁ . b₁₃ + c₃₂ . b₂₃ + c₃₃ . b₃₃ = 0, which contradicts the restriction from the question

So actually matrix C does not exist, not only invertible matrix C does not exist but also non - invertible matrix C can not exist.

Is this what the question wants? Or I am missing something?

Thanks

 

[fresh_42]

Homework Statement: Please see below
Relevant Equations: Matrix Multiplication

View attachment 324741

My attempt:

Let C =
$$\begin{pmatrix}
c_{11} & c_{12} & c_{13} \\
c_{21} & c_{22} & c_{23} \\
c_{31} & c_{32} & c_{33}
\end{pmatrix}$$

If C is multiplied by B, then:

1)
a₂₁ = c₂₁ . b₁₁
0 = c₂₁ . b₁₁ $\rightarrow c_{21}=0$

2)
a₃₁ = c₃₁ . b₁₁
0 = c₃₁ . b₁₁ $\rightarrow c_{31}=0$

3)
a₃₂ = c₃₂ . b₂₂
0 = c₃₂ . b₂₂ $\rightarrow c_{32}=0$

But a₃₃ = c₃₁ . b₁₃ + c₃₂ . b₂₃ + c₃₃ . b₃₃ = 0, which contradicts the restriction from the question

So actually matrix C does not exist, not only invertible matrix C does not exist but also non - invertible matrix C can not exist.

Is this what the question wants? Or I am missing something?

Thanks

Looks ok. Whether you should calculate it, or reason by a vector in the kernel cannot be said. That depends on the context that you didn't provide.

 

[songoku]

Looks ok. Whether you should calculate it, or reason by a vector in the kernel cannot be said. That depends on the context that you didn't provide.

If the context you mean is related to "Relevant Equations", I want to clarify that what I wrote in there was actually the method I could think of to solve this question, not the method I must use so I really want to learn another approach to solve the question.

I have no idea how to reason by a vector in the kernel. Kernel is null space so what I have in mind is something like this:

Let C be the transformation matrix T that transforms B to A so T(B) = A.
Kernel of T is the set of all B such that T(B) = 0 but not all elements in A is zero so I don't really know how to use kernel to solve the question.

Thanks

 

[fresh_42]

Your solution is probably the shortest.

 

[songoku]

Thank you very much fresh_42

 

[Infrared]

To say how you can approach this thinking about kernels: If $C$ is invertible, then the kernel (I'm more used to using the word 'nullspace' when describing matrices and kernel when describing linear maps, but this is just terminology) of $B$ and the kernel of $A=CB$ are the same. However, the first three columns of $A$ are independent, so there is no nonzero element of the kernel of the form $\begin{pmatrix} * \\ * \\ * \\0\end{pmatrix}$ whereas for $B$, the first three columns are dependent, so there is such an element in the kernel.

 

[mathwonk]

to rephrase this nice comment, the matrix C maps the first three columns of B to the first three columns of A, but that is impossible, since dependent columns cannot map to independent ones. I.e. the fact that A and B have 4 columns is a smoke screen, and one can ask the question about their 3x3 left parts, where it is clear. Note also that this does not use invertibility of C either. In terms of the kernel, it uses only that the kernel of CB contains the kernel of B, whether C is invertible or not.