If we let AAA be a m×nm\times n matrix and let BB be a n×pn\times p matrix. Suppose that both AA and BB have a left inverse. Show that ABAB has a left inverse.

i know that matrix ATA^TA is an invertible nn by nn symmetric matrix. Hence. (AT)−1ATA=I(A^T)^{-1}A^TA = I. So then A−1left=(ATA)−1ATA_{\text{left}}^{-1} = (A^TA)^{-1}A^T

But how do I go about proving that ABAB has a left inverse?

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Directly use the definition of left inverse: a matrix LL such that LA=InLA=I_n. Take MB=IpMB=I_p and try what happens with (ML)(AB)(ML)(AB).

– egreg

2 days ago

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1 Answer

1

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Hint: Try computing (B−1leftA−1left)(AB)(B_{\text{left}}^{-1}A_{\text{left}}^{-1})(AB). What do you get?

@ parsiad how do you compute that?

– geohelp

2 days ago

@mtphys Just with the definition of left inverse, and associativity. e.g. A−1leftA=IA_{\text{left}}^{-1} A = I

– arkeet

2 days ago

@ arkeet(B−1leftA−1left)(AB)=I2(B_{\text{left}}^{-1}A_{\text{left}}^{-1})(AB) = I^2??

– geohelp

2 days ago

@mtphys: I2=II^2=I

– parsiad

2 days ago

@ parsiad. So since it equals I is that enough for the proof?

– geohelp

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