I think the proof is wrong! v must be a k-component column vector since it can multiply A, and vector 0 must have n components. Accordingly, the transpose of v, which must be a k-component row vector, just cannot be multiplied by vector 0 if k is different from n.
I think the proof is wrong! v must be a k-component column vector since it can multiply A, and vector 0 must has n components. Accordingly, the transpose of v, which must be a k-component row vector, just cannot be multiplied by vector 0 if k is different from n.
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I think the proof is wrong! v must be a k-component column vector since it can multiply A, and vector 0 must have n components. Accordingly, the transpose of v, which must be a k-component row vector, just cannot be multiplied by vector 0 if k is different from n.
chanofkln 1 week ago
I think the proof is wrong! v must be a k-component column vector since it can multiply A, and vector 0 must has n components. Accordingly, the transpose of v, which must be a k-component row vector, just cannot be multiplied by vector 0 if k is different from n.
chanofkln 1 week ago
success
mosle123 10 months ago
I'm from the UK and I love what you're doing here. Keep up the great work!
AlexanderJRLewis 1 year ago
Comment removed
zuki6208 1 year ago
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.
wonderpope 2 years ago
I fucking hate linear algebra
Beefstew2011 2 years ago
Never learned matrixes.
azndude3600 2 years ago
good video as always
RobDaRob 2 years ago