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enter image description here

I can figure out the first part of the question just fine, by using the Gram-Schmidt process, except I am having trouble with finding an orthonormal basis for the orthogonal complement of U, or "U perp".

Would I need to find the orthonormal basis for the space U and then use that basis to find the orthonormal basis for its orthogonal complement? Or is it possible to find the orthonormal basis for U-perp just by finding the orthogonal complement of the set of vectors in the picture then using Gram-Schmidt process on those?

Again, I can find the orthonormal basis for the space U just fine, except I have no idea where to go from here.

EDIT: Here is the orthonormal basis for the space U generated by the three vectors pictured. enter image description here

EDIT2: I solved for the null-space of the three vectors and came up with a fourth vector, then I applied Gram-Schmidt to the fourth vector with respect to the first three and obtained this as a result, would this one vector here be the orthonormal basis of the orthogonal complement of U? V4 is the vector obtained from solving for the null-space of the first three.

enter image description here

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    $\begingroup$ Orthogonalize the basis for $U$, then complete this to a basis for $\mathbb{R}^4$, orthogonalize the added vectors with respect to your original set. $\endgroup$ – Mnifldz Nov 20 '19 at 4:17
  • $\begingroup$ I used the Gram-Schmidt process on the three vectors in the question, except I'm not entirely sure how to complete this to a basis for R^4? $\endgroup$ – Alex Charron Nov 20 '19 at 4:39
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    $\begingroup$ You need to find a fourth vector which isn't in the span of your original 3. This last vector need not be orthogonal, but you will need to orthogonalize it with one additional application of Gram-Schmidt. If you've orthogonalized your vectors already, I'd recommended adding your work to this post so people can see what you've done. $\endgroup$ – Mnifldz Nov 20 '19 at 4:41
  • $\begingroup$ Alright I have found a fourth vector, then applied Gram-Schmidt on the fourth vector with respect to the original set, would the Orthonormalized V4 in the picture i added be correct? $\endgroup$ – Alex Charron Nov 20 '19 at 5:19
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Extend the given basis for $U$ to a basis for $\mathbb{R}^4$ before applying Gram-Schmidt to the entire thing. Then the first three vectors of the result give you a basis for $U$ and the last, being orthogonal to all three, gives you a basis for $U^\perp$.

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  • $\begingroup$ Ok, so I solved for the null-space of the original three vectors, and found a vector V4 that is orthogonal to all three original vectors, then applied Gram-Schmidt to the entire thing again, so is V4 orthonormalized with respect to the first three vectors the basis for 𝑈⊥? $\endgroup$ – Alex Charron Nov 20 '19 at 5:11
  • $\begingroup$ The fourth vector is a basis for $U^\perp$. (I don't know what is meant by "V4 orthonormalized with respect to the first three vectors") $\endgroup$ – Daniel McLaury Nov 20 '19 at 5:27
  • $\begingroup$ I meant that I applied Gram-Schmidt to all four vectors (the three given vectors and the fourth vector) $\endgroup$ – Alex Charron Nov 20 '19 at 5:30
  • $\begingroup$ Yes: it's orthogonal to everything in V and together they span the entire space. $\endgroup$ – Daniel McLaury Nov 20 '19 at 22:42

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