Linear Algebra help please

Yes that works.

Can u help me with q6c in the above post?

I need help with 6)c)

For 6)a) I showed the 4 conditions hold for inner product.

For 6)b) I use the Gram-Schimdt process to construct an orthonormal set.

Notice that for 6 (b) you don't need to use Gram-Schmidt, as $\{F_1, F_2, F_3\}$ is already an orthogonal set. All you need to do here is to normalize each of the $F_i$ so that $<F_i, F_i> = 1$.

Then you use Gram-Schmidt in part (c) - but note here that the process only needs to be applied to $G$ as the others are already orthogonal.

N.B. I think there's a typo in the part (iii) of the question: $\{F_1, F_2, F_3, F_4\}$ can't be orthonormal - you've been asked to make $\{F_1, F_2, F_3\}$ orthonormal in part (ii) of the question! So looks like the question should read "orthogonal" or should be asking for an orthonormal $\{F'_1, F'_2, F'_3, F'_4\}$, where $\{F'_1, F'_2, F'_3\}$ was the orthonormal basis constructed in part (ii).

Oh so for part ii, I should use the formula F1/||F1||, F2/||F2|| and F3/||F3|| to get an orthonormal set?

And for part iii i should use the formula: W4= G - <G,F3>*F3 - <G,F2>*F2 - <G,F1>*F1

Then obtain F4 by doing W4/||W4||?

That sounds about right to me.

Thanks a lot.

Also is it always the case that if S and T bases for a vector space V then so is SnT?

Thanks a lot.

Also is it always the case that if S and T bases for a vector space V then so is SnT?

Do you mean $S \cap T$ ? Think about $\mathbb{R}^2$ with the usual orthonormal basis (call it S). Rotate that basis by a small amount to get a new basis T. Then $S \cap T = \emptyset$. What do you think now?

Definitely not, consider R. R has a basis {1} and {2}, S intersect T = the empty set

Definitely not, consider R. R has a basis {1} and {2}, S intersect T = the empty set

Out at uni atm and won't get back until after 7pm, so I'll have a proper look at it then.

Can u check this q for me?


I got the kernel as 1 and the images as 3.

Can u check this q for me?


I got the kernel as 1 and the images as 3.

Can u check this q for me?


I got the kernel as 1 and the images as 3.

Ok, your answer is correct, but you need to have some reasoning why the basis for your image is as small as it can be. By rank-nullity, the basis for the image has 3 elements, and as yours does it is fine. But I'm pretty sure thats why the question asks you to state all the theorems you use