Linear Algebra - Lecture 25 - The Invertible Matrix Theorem

[@Rand3mAnclrew]

Thank you for making videos. There are so little resources online for learning Linear Algebra its astounding. Could you please make sure to put emphasis on proofs? My textbook (Larson's Elementary Linear Algebra) seems prove every proof by saying "We'll leave it to you"

[@lover-of-Shuhada]

😂😂..

[@flossy3432]

you saved my life

[@Brahma2012]

Thank you James

[@huipingyang4096]

Statement f and I says the same thing is onto and one-to-one? is this true?

[@chenlecong9938]

Also in 9.31 statements I and F together are implying that the transformation t is both onto and one to one.Are you implying that one to one is in a sense,onto?since by definition,onto is when one output is given by AT LEAST one input?

[@HamblinMath]

No, one-to-one and onto are separate concepts. A generic *function* can be one-to-one but not onto, or onto but not one-to-one. However, a *linear transformation* from R^n to R^n cannot be one without being the other. That's one of the special properties of linear transformations.

[@chenlecong9938]

James Hamblin thanks for kindly replying.

[@ghsjgsjg53chjdkhjydhdkhfmh74]

Wow😮 thank you!

[@Cizzo8]

I'm confused which video it actually is that g,h,i and k,l are explained by you. I don't think it's lecture 18?

[@chenlecong9938]

7.21 it seems like you are implying that e sub i is one of the column of the identity matrix.then x must’ve been the column of the inverse of A.then you’re already assuming A has an inverse matrix and is thus invertible?sorry mate,really don’t get it.

[@HamblinMath]

That's what "e_i" means: the i-th column of the n x n identity matrix. Statement (g) says that Ax=b has a solution for *every* vector b, so I'm applying that to the vector b = e_i.

[@ghsjgsjg53chjdkhjydhdkhfmh74]

I don't understand proving g to a😓😓 can anyone please explain??

[@liaodaniel1684]

One can continue to use the same D and show that DA=I_n (by similar approach, but with (e_j)^T), so A is invertible because it has inverse D.

[@user-xn4yu5rn9q]

Thank you so much

[@Ricky-gc7cj]

Thank!

[@EDROCKSWOO]

LEBRON JAMES is a god.