Talk:Linear Algebra/Polynomials of Maps and Matrices

Example 1.2 is wrong
According to my calculations, taking:



T= \begin{pmatrix} \sqrt{3}/2 &-1/2  \\ 1/2        &\sqrt{3}/2 \end{pmatrix} $$

will yield

$$ T^2 - 2T= \begin{pmatrix} \frac{1-2\sqrt{3}}{2} &\frac{2-\sqrt{3}}{2}  \\ \frac{2-\sqrt{3}}{2}        &\frac{1-2\sqrt{3}}{2} \end{pmatrix} $$

From which:

$$ T^2 - 2T - I= \begin{pmatrix} \frac{-1-2\sqrt{3}}{2} &\frac{2-\sqrt{3}}{2}  \\ \frac{2-\sqrt{3}}{2}        &\frac{-1-2\sqrt{3}}{2} \end{pmatrix} $$

Wisapi (discuss • contribs) 21:19, 3 August 2011 (UTC)

Well I get a different wrong answer from

$$ T^2 - 2T= \begin{pmatrix} \frac{1-2\sqrt{3}}{2} &\frac{2-\sqrt{3}}{2}  \\ \frac{\sqrt{3}-2}{2}        &\frac{1-2\sqrt{3}}{2} \end{pmatrix} $$

From which:

$$ T^2 - 2T - I= \begin{pmatrix} \frac{-1-2\sqrt{3}}{2} &\frac{2-\sqrt{3}}{2}  \\ \frac{\sqrt{3}-2}{2}        &\frac{-1-2\sqrt{3}}{2} \end{pmatrix} $$

Drawing vectors on the back of an envelope suggests maybe

$$ -T^4 + T^2 - I = 0 $$

Recent Runes (discuss • contribs) 22:40, 3 August 2011 (UTC)

or

$$ T^2 -\sqrt{3}T - I = 0 $$

Recent Runes (discuss • contribs) 23:04, 3 August 2011 (UTC)


 * Yup! my first calculation was wrong. Yours seems to be right (actually I'd say that yours is right). I'll use the method explained in example 1.6 to see what polynomial I get. This will take some minutes.
 * But how did you use vectors to make this calculation on matrices? Wisapi (discuss • contribs) 23:21, 3 August 2011 (UTC)


 * Well, we know T produces a rotation, so T squared is just the same rotation repeated. If you draw just one of the basis vectors (1,0) or (0,1) as modified by the rotations, and sum the vectors graphically then the geometry is fairly clear. Recent Runes (discuss • contribs) 23:30, 3 August 2011 (UTC)


 * I did the calculations from the Ansatz in example 1.6 and I arrived at

T^2 -\sqrt{3}T + I = 0 $$
 * I also checked it graphically, as you explained, and I got the same result. $$ -T^4 + T^2 - I = 0 $$ is also possible (I tested it graphically), but as this is not a minimal polynomial (the leading coefficient isn't even 1) we should keep the other equation. I'll modify the article. By the way would you mind taking a look at these other comments of mine in this wikibook?
 * Talk:Linear Algebra/Strings and Talk:Linear Algebra/General = Particular + Homogeneous Wisapi (discuss • contribs) 00:11, 4 August 2011 (UTC)
 * Hmmm... I think you want a proper mathematician for that job - I'm just an engineer meself! Recent Runes (discuss • contribs) 19:17, 4 August 2011 (UTC)