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

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In summary, using the method of reasoning with the kernel, it can be shown that the matrix C does not exist, as the first three columns of A cannot be mapped from the dependent columns of B. This is independent of the invertibility of C.
  • #1
songoku
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Homework Statement
Please see below
Relevant Equations
Matrix Multiplication
1681226190098.png


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
 
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  • #2
songoku said:
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.
 
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  • #3
fresh_42 said:
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
 
  • #5
Thank you very much fresh_42
 
  • #6
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.
 
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  • #7
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.
 
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