[Linear Algebra] Construct an n x 3 matrix D such that AD=I3

[bornofflame]

Homework Statement

Suppose that A is a 3 x n matrix whose columns span R³. Explain how to construct an n x 3 matrix D such that AD = I₃.

"Theorem 4"
For a matrix A of size m x n, the following statements are equivalent, that is either all true or all false:
a. For each b in R^m, Ax = b has a solution
b. Each b in R^m is a linear combination of A.
c. The columns of A span R^m.
d. A has a pivot position in every row.

Homework Equations

AD = I₃
Ax = b
I₃ = [e₁ e₂ e₃]
AB = [Ab₁ ... b_n]

The Attempt at a Solution

So far what I've managed to put together is the following:
Since the columns of A span R³, it follows that, per Theorem 4:
For each b in R³, Ax = b has a solution
Each b in R³ is a linear combination of A.
A has a pivot position in every row.

Because we can write AB = [Ab₁ ... b_n], D as [d₁ d₂ d₃] and I₃ as [e₁ e₂ e₃], we can also write
AD = [Ad₁ Ad₂ Ad₃] = [e₁ e₂ e₃].
We can substitute Ad₁ = e₁ for Ax = b in this case and we know that, since the columns of A span R³ that this equation has a solution. This logic can be applied to the columns [d₁ d₂ d₃] to solve for D.

Is this correct? If not, am I at least on the right track? Thanks.

 

[Dick]

Homework Statement

Suppose that A is a 3 x n matrix whose columns span R³. Explain how to construct an n x 3 matrix D such that AD = I₃.

"Theorem 4"
For a matrix A of size m x n, the following statements are equivalent, that is either all true or all false:
a. For each b in R^m, Ax = b has a solution
b. Each b in R^m is a linear combination of A.
c. The columns of A span R^m.
d. A has a pivot position in every row.

Homework Equations

AD = I₃
Ax = b
I₃ = [e₁ e₂ e₃]
AB = [Ab₁ ... b_n]

The Attempt at a Solution

So far what I've managed to put together is the following:
Since the columns of A span R³, it follows that, per Theorem 4:
For each b in R³, Ax = b has a solution
Each b in R³ is a linear combination of A.
A has a pivot position in every row.

Because we can write AB = [Ab₁ ... b_n], D as [d₁ d₂ d₃] and I₃ as [e₁ e₂ e₃], we can also write
AD = [Ad₁ Ad₂ Ad₃] = [e₁ e₂ e₃].
We can substitute Ad₁ = e₁ for Ax = b in this case and we know that, since the columns of A span R³ that this equation has a solution. This logic can be applied to the columns [d₁ d₂ d₃] to solve for D.

Is this correct? If not, am I at least on the right track? Thanks.

Looks fine to me.

 

[StoneTemplePython]

I agree with the above.

extension:
Since you are in working in reals, you have an inner product, and may may want to consider pushing the problem a bit to find the solution with the minimum length --i.e. getting the minimum length (squared 2 norm) solution for each vector $\mathbf d_1$, $\mathbf d_2$ and $\mathbf d_3$ (equivalently, minimum squared Frobenius norm for $D$). Specifically said solution would be given by

$D = A^T\big(AA^T\big)^{-1}$
or
$\mathbf d_j = A^T\big(AA^T\big)^{-1} \mathbf e_j$
- - - -
There's a way to derive this via Lagrange Multipliers, and another way to derive it via Singular Value Decomposition. Both are instructive. This is a very nice result, that not a lot of people are familiar with, for some reason.
- - - -
it also has a nice common sense feel to it, because

$I_3 = AD = A\Big(D\Big) = A\Big( A^T\big(AA^T\big)^{-1}\Big) = A A^T\big(AA^T\big)^{-1}= \big(A A^T\big)\big(AA^T\big)^{-1} = I_3$

so to get comfortable with this solution, you'd just need to convince yourself that $\big(AA^T\big)$ is in fact non-singular.

 

[bornofflame]

Thank you, both. I'm still not comfortable with Linear Alebra yet and so feel quite shaky in what I've learned. Unfortunately there are no tutors for it on campus.

Thanks for the extension, StoneTemplePython. We haven't touched Lagrange Multipliers or Singular Value Decomposition yet, but that's something to look forward to.