Showing that A-transpose x A is invertible | Matrix transformations | Linear Algebra | Khan Academy

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Showing that (transpose of A)(A) is invertible if A has linearly independent columns
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[@wonderpope]

So if n>k, the nxn matrix A*A^t isn't necessarily invertible. right? in any case you couldn't use the argument of the columns of A^t being lin. ind.

[@chilinouillesdepommesdeter819]

So clear,thanks for your great work

[@konstantinoskompothekras881]

Great proof. Thank you for sharing your knowledge

[@user-ci8ce5mb5j]

Thanks for the video. Actually I have a question: "Does it work in the other side -- if A T A is invertible => A has independent columns?". Thanks in advance

[@LAnonHubbard]

Did you resolve this? I think the proof is right as the vector 0 will have k components as it is the product of (At)(A)v, and (At)(A) is k by k.

[@yuanjiepeng4589]

Null space?

[@bingjieshi5578]

subtitle is wrong! column vector is 列向量 not 行向量

[@azndude3600]

Never learned matrixes.