Let AA be a 3x53x5 matrix. Is it possible rank(AAT)=4rank(AA^T)=4? So I know that rank is the amount of leading 11’s in a matrix. So this matrix consists of 33 rows and 55 columns, but how do I prove that the rank of matrix AA multiplied by AA transverse is 44?

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@ Sloan, thankyou for spending a moment on the edit

– user381092

2 days ago

If AA has size 3×53\times 5 (which normally means three rows and five columns), then ATA^T has size 5×35\times 3 and AATAA^T has size 3×33\times 3. So, it is impossible that AATAA^T has 4 columns with leading 1’s since it only has three columns.

– Wore

2 days ago

why would AATAA^T have 4 columns?

– user381092

2 days ago

What I am saying is that AATAA^T has size 3×33\times 3, so it does not even have 4 columns!

– Wore

2 days ago

oh sorry i asked the wrong question. Why would AATAA^T be 3x33x3 how did you calculate its row and column dimensions

– user381092

2 days ago

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1 Answer

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No, it is impossible, AATAA^T is the composition of two linear transformations, the first from R5\mathbb R^5 to R3\mathbb R^3 and the second from R3\mathbb R^3 to R5\mathbb R^5, hence the dimension of the image can be at most 33.

What do you mean the first going from A5A^5 to A3A^3? what is AA for you?

– Wore

2 days ago

Sorry, my brain malfunctioned a bit for a while

– Jorge Fernández Hidalgo

2 days ago

thank you for your answer

– user381092

2 days ago

no problem{}{}{}{}

– Jorge Fernández Hidalgo

2 days ago