Pseudo Inverse & QR Decomposition

[ahamdiheme]

Homework Statement

Let A$$\in$$R^mxn be factorized as A=QR where Q$$\in$$R^mxn has orthonormal columns. Prove that A⁺=R⁺Q^T

Homework Equations

R is an upper triangular matrix

The Attempt at a Solution

I tried to apply the definition A⁺=(A^TA)⁺A⁺
I ended up here: R⁺(RR⁺)^TQ^T
I'm not sure if i am going in the right direction
thanks for your assistance

 

[mighty2000]

Dear student,

Your attempt is a good start. Let's continue from where you left off:

We know that R is an upper triangular matrix, which means that R+ is also upper triangular. Therefore, we can write R+ as a matrix with its elements above the main diagonal being zero. Let's say that R+ is a matrix with elements r_ij, where i>j.

Now, let's look at the expression RR+QT. Since R+ is upper triangular, we know that RR+ will also be upper triangular. This means that the elements below the main diagonal of RR+ will be zero. So, we can write RR+ as a matrix with elements r_ij, where i<=j.

Therefore, when we multiply RR+ with QT, we will only get non-zero elements for the elements r_ij where i<=j. This means that the resulting matrix will be upper triangular.

Now, let's look at the expression (RR+)T. Since R+ is upper triangular, (RR+)T will also be upper triangular. This means that the elements below the main diagonal of (RR+)T will be zero. So, we can write (RR+)T as a matrix with elements r_ij, where i<=j.

Therefore, when we multiply (RR+)T with QT, we will only get non-zero elements for the elements r_ij where i<=j. This means that the resulting matrix will be upper triangular.

Now, let's combine the above results:

R+(RR+)TQT = R+[(RR+)TQT] = R+[(RR+)T][QT] = [(RR+)T][QT]

= [R(R+T)][QT] = [R(RT+)][QT] = R(RT+QT) = R(RT+QT) = R(RT+QT)

= RR+QT = RR+QT = A+

Therefore, A+=R+QT.

I hope this helps. Let me know if you need any further clarification.