Linear Algebra Multiple Choice

[Ted123]

Homework Statement

[PLAIN]http://img697.imageshack.us/img697/9307/linvd.jpg

The Attempt at a Solution

Last condition = 1) ?

How about the others?

 

[lanedance]

so how about starting by finding the eigenvalues?

 

[Ted123]

so how about starting by finding the eigenvalues?

Eigenvalues are

[itex]x= \frac{1}{2}\left( - \sqrt{a^2 -4b} -a\right)[/itex]

[itex]x= \frac{1}{2}\left( \sqrt{a^2 -4b} -a\right)[/itex]

 

[lanedance]

ok, so you have the general formula, but finding the eigenvalues for each case, should give you good hint...

if you're not sure why try reading
http://en.wikipedia.org/wiki/Diagonalizable_matrix

 

[Ted123]

ok, so you have the general formula, but finding the eigenvalues for each case, should give you good hint...

if you're not sure why try reading
http://en.wikipedia.org/wiki/Diagonalizable_matrix

I went for

1st condition - 2
2nd condition - 1
3rd condition - 3
4th condition - 1

but one of these is wrong. Can you see which one?

 

[lanedance]

why not tell me your reasoning?

 

[lanedance]

for the last one, a^2 - 4b < 0 is a subset, but more importantly consider when a^2 - 4b = 0

 

[Ted123]

for the last one, a^2 - 4b < 0 is a subset, but more importantly consider when a^2 - 4b = 0

So is the last one 4?


If a^ - 4b = 0 then both eigenvalues are the same (not distinct) but this isn't enough to conclude that A is not diagonalisable.

 

[lanedance]

yep, but i would say if a^ - 4b = 0 then both eigenvalues are the same (not distinct), so this isn't enough to conclude that A is diagonalisable (as for this we need 2 linearly independent eigenvectors)