Matrix Syllogism

Logic Level 4

Consider a m × n m \times n matrix A A of coefficients, a vector x → \overrightarrow{x} of unknowns and a vector b → \overrightarrow{b} of solutions satisfying A x → = b → A \overrightarrow{x} = \overrightarrow{b}

You're given that there exists a left inverse L L of A A satisfying L A = I n L A = I_n

Consider the following argument

  1. A x → = b → A \overrightarrow{x} = \overrightarrow{b}
  2. Multiplying both sides with L L on the left, L A x → = L b → L A \overrightarrow{x} = L \overrightarrow{b}
  3. Hence, x → = L b → \overrightarrow{x} = L \overrightarrow{b} satisfies 1 1
  4. Substituting x x back in 1 1 , A L b → = b → A L \overrightarrow{b} = \overrightarrow{b}
  5. Hence, L L is a right inverse of A A .

But that was not what we were originally given. Which is the first statement that went wrong?

Inspired by Artin's Algebra
5 4 2 3 1

Agnishom Chattopadhyay by Agnishom Chattopadhyay , 22, India

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1 solution

We could summarize a part of our argument like this:

A x = b ⇒ L A x = L b ⇒ I n x = L b ⇒ x = L b . \begin{aligned}Ax=b&\Rightarrow& LAx=Lb\\ &\Rightarrow& I_nx=Lb\\ &\Rightarrow& x=Lb.\end{aligned}

In statement three, we say that x = L b x = Lb satisfies the original equation, but this is not necessarily true since implication ( ⇒ ) (\Rightarrow) does not necessarily work both ways.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

The solution is trying to say that just because you derive that x = L b x = Lb , it does not necessary satisfy the original equation of A x = b Ax = b . We will supply a counter-example that establishes this.

Let L = [ − 1 − 1 1 5 2 − 5 2 1 ] , A = [ 1 2 3 4 5 6 ] , b = [ 1 1 2 ] . L = \begin{bmatrix} -1 & -1 & 1 \\ \frac{5}{2} & -\frac{5}{2} & 1 \end{bmatrix} , \quad A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix} , \quad b = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} .

It is easy to verify that L A = I 2 LA = I_2 . Also, x = L b = [ 0 2 ] , x = Lb = \begin{bmatrix} 0 \\ 2 \end{bmatrix} , but this vector does not satisfy A x = b Ax = b . (You may want to think about why this is true.)

Jon Haussmann - 4 years, 11 months ago

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Thanks. This is very helpful.

Agnishom Chattopadhyay - 4 years, 11 months ago

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