Linear algebra 1: cauchy schwarz problem

[BWE38]

Homework Statement

If llull = 4, llvll = 5 and u dot v = 10, find llu+vll. u and v are vectors

Homework Equations

llu+vll = llull + llvll cauchy schwarz

The Attempt at a Solution

(1) llu+vll = llull + llvll

(2) (llu+vll)^2 = (llull + llvll)^2

(3) (llu+vll)^2 = llull^2 + 2uv +llvll^2

(4) (llu+vll)^2 = 4^2 + 2(10) +5^2

(5)(llu+vll)^2 = 16 + 20 + 25

(6) sqr(llu+vll)^2 = sqr61

(7) (llu+vll) = 7.81

I think that I am doing it wrong for some reason. Any feed back would be greatly appreciated.

 

[dirk_mec1]

Line (1) is ofcourse not correct.

Hint: square ||u+v||.

 

[HallsofIvy]

Homework Statement

If llull = 4, llvll = 5 and u dot v = 10, find llu+vll. u and v are vectors

Homework Equations

llu+vll = llull + llvll cauchy schwarz

No, that is NOT correct! The Cauchy-Schwartz in-equality says that
$$||u+ v||\le ||u||+ ||v||$$

The Attempt at a Solution

(1) llu+vll = llull + llvll

again, no.

(2) (llu+vll)^2 = (llull + llvll)^2

No.

(3) (llu+vll)^2 = llull^2 + 2uv +llvll^2

That "2uv" (2 time the dot product of u and v) is NOT 0 tells you immediately that ||u+ v|| is NOT equal to ||u||+ ||v||!

(4) (llu+vll)^2 = 4^2 + 2(10) +5^2

(5)(llu+vll)^2 = 16 + 20 + 25

(6) sqr(llu+vll)^2 = sqr61

(7) (llu+vll) = 7.81

I think that I am doing it wrong for some reason. Any feed back would be greatly appreciated.

It looks right to me! What reason do you have to think this is wrong?

 

[BWE38]

No, that is NOT correct! The Cauchy-Schwartz in-equality says that
$$||u+ v||\le ||u||+ ||v||$$


again, no.


No.


That "2uv" (2 time the dot product of u and v) is NOT 0 tells you immediately that ||u+ v|| is NOT equal to ||u||+ ||v||!


It looks right to me! What reason do you have to think this is wrong?

aha! I see what you are saying now. (llu+vll)^2 = uu + uv +vu + vv, which becomes llu+vll^2 = llull^2 + 2uv + llvll^2

llu+vll = llull +llvll, IFF 2uv = 0

Which in this case it isn't. Because uv is equal to 10.