GATE EE 2008

Question

Q.51 A is mx n full rank matrix with m>n and I is an identity matrix. Let matrix A' = (A^T.A)^-1 A^T.  Then,which one of the following statement is TRUE?

 (a) AA'A = A

(b) (AA)²= A

(c) AA'A = I

(d) AA'A = A'

[EE, GATE-2008, 2 marks]

Where, X^T: Transpose of X and X^-1= Inverse of X

Comments

By formula A'= (A^T.A)^-1 A^T

                 => A'= A^-1. (A^T)^-1. (A^T)

                => A'= A^-1 ......(1)

Now by formula we can see that option(a) is coming as answer. 

But my question is from eqn (1) we see that A' is A^-1. But it is given A as non-square matrix so how come A^-1 can exist?

 

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A is not invertible as you said but $A^TA$ is invertible and so, $A'A = (A^TA)^{-1}(A^TA)= I$

The expression $(A^TA)^{-1}A^T$ has a special meaning in Pattern Recognition.

$z^* = (A^TA)^{-1}A^Tb$ gives the "best fit line" for a Regression Problem with system $Az=b.$

The matrix $P= A(A^TA)^{-1}A^T$ is called the "Projection Matrix" and it satisfies $P^2=P$ and you can prove it easily.