Hey guys,

Can you tell me if this solution looks alright?

http://s708.photobucket.com/albums/w...rent=hwsol.png

Thanks!

James

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- May 2nd 2011, 07:29 PMjames121515Linear Algebra: Linear Independence question
Hey guys,

Can you tell me if this solution looks alright?

http://s708.photobucket.com/albums/w...rent=hwsol.png

Thanks!

James - May 2nd 2011, 09:49 PMjames121515
Sorry the picture didn't come up. Here it is again.

http://i708.photobucket.com/albums/w...1515/hwsol.png - May 3rd 2011, 02:24 AMAckbeet
The row reduction looks good to me. I'm not sure I would your matrix equation the way you did. I would write it this way:

$\displaystyle \left[\begin{array}{rrr}\mathbf{v}_{1} &\mathbf{v}_{2} &\mathbf{v}_{3}\end{array}\right]\left[\begin{array}{r}c_{1}\\ c_{2}\\ c_{3}\end{array}\right]=\left[\begin{array}{rrr}1 &2 &0\\ \lambda^{2} &\lambda &0\\ 1 &4 &1\\ 2 &8 &2\end{array}\right]\left[\begin{array}{r}c_{1}\\ c_{2}\\ c_{3}\end{array}\right].$

Your column matrix

$\displaystyle \left[\begin{array}{r}c_{1}\\ c_{2}\\ c_{3}\end{array}\right]$

disappeared on the RHS for some reason. Other than that, as I say, it looks good to me!

[EDIT] See Deveno's post for a correction. - May 3rd 2011, 05:28 AMDeveno
one must be extremely careful in accounting for each step of the row reduction.

i found that to eliminate λ from the row-reduced matrix, it was necessary to divide by λ. this suggests that λ = 0 is a special case.

and indeed, when λ = 0, we have 2(1,0,1,2) + 2(0,0,1,2) = (2,0,4,8), so that {(1,0,1,2), (2,0,4,8), (0,0,1,2)} is a linearly dependent set.

thus the statement: "regardless of the parameter λ, the only scalar values of c1, c2, c3 such that c1v1 + c2v2 + c3v3 = 0 are c1,c2,c3 all 0" is UNTRUE

i have exhibited a counter-example.