Linear Algebra: Projection onto a Subspace

[@PENDANTturnips]

The diagram at the beginning was very helpful! Thanks for the video

[@iossif1000]

This video is not too clear. In order for this to be true, we must have our subspace W spanned by orthogonal or orthonormal vectors to proceed with using the projection formula. We can compute these dim(W) number vectors given dim(W) spanning vectors using the Gram-Schmidt procedure. In your case, your spanning vectors are orthogonal (specifically orthonormal), which is fine. Hope that clears things up in the comments. Computations on point

[@amatoallahouchen5894]

THANKS YOUR VIDEOS ARE REALLY HELPFUL

[@shivdalsaniya]

Thanks It was HELPFUL.

[@uneebhasan6574]

Thank you !

[@Andrewl9110]

if I am not mistaking - in the formula in the beginning of the video, we are not dividing by the length - we are dividing by the length^2. since the length is sqrt(vdotv). Did I go wrong?

[@AdiJ8]

The length implies sqrt(vdotv)

[@lkdsa3kds]

Thank you! Your video helped greatly with my hw

[@zacht8083]

this was really helpful. thank you

[@geniusmathematics9123]

Thank you so much sir

[@manaskrthk]

THANKYOU.

[@buckenfuzz]

do the vectors in the subspace have to be orthogonal?

[@cf9707]

wondering the same thing...

[@notme3154]

Remember vector minus its projection being orthogonal to each vector of a subspace. In this case, vector X minus the projected v_ B in the video does workout to have a dot product of zero for each element ∈ V

[@virajcz7143]

holy shits, this was so good!

[@anoniem012]

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[@UnifiedaxxS]

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[@iossif1000]

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[@anoniem012]

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[@anoniem012]

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