Question

Linear Algebra Question- simple matrix

Answer 1

Find a 2x2 matrix A with the given properties. A^2 does not = \[I _{2}^{}\], A^{4} = \[I _{2}^{}\]

Answer 2

A 2 x 2 matrix A with the given conditions

\[A^2 \neq I_2\] but

\[A^4 = I_2\]

Answer 3

so matrix multiplication is being used for an n x n matrix... so we need to find an A such that if we use matrix mulitplication A x A we can't have the identity matrix, but A X A X A X A produces an identity matrix.

Answer 4

Imagine you have a vector in 2D space... Like say, your pencil on top of your desk. Is there some operation you can do to your pencil by sliding it around on the desk 4 times that has the same end effect as just leaving it alone? But then if you were to do that movement of your pencil on the table only twice, it would lead to a noticeable change?

Answer 5

If you rotate the pencil at 45 degrees I think that would work?

Answer 6

I really like your explanation on the pencil on the table. That really helps!

Answer 7

:D You're thinking along the right track!

Answer 8

Whoops I meant 90 degrees!

Answer 9

Awesome, perfect. So now there are a couple ways you can try to handle this next step. How do you actually find out what the entries of A are if you know that it represents a 90 degree rotation?

Answer 10

When you rotate I think the matrix is A= [cos(A) -sin(A)..... sin(A) cos (A)] read at row across

Answer 11

Do you just substitute A for 90 degrees or pi/2 radians?

Answer 12

Both are equivalent, \(\sin(90 \deg) = \sin(\pi/2 rad)\), so you don't have anything to really worry about there :P

Answer 13

Okie dokie! Thank you :D That really cleared a lot of things up~

Answer 14

Yeah actually I think there's a more general and clearer way that might help you solve for A if you're interested rather than trying to remember the sines and cosines.

Answer 15

You can just say, ok I know A is a 90 degree rotation so if I start with:

\[A=\begin{pmatrix}
a &b \\c
&d
\end{pmatrix}\]

I know that the matrix transforms ANY and ALL vectors, so I might as well pick a simple test vector:

\[ \begin{pmatrix}
a &b \\c
&d
\end{pmatrix} \begin{pmatrix}
1 \\0

\end{pmatrix} = \begin{pmatrix}
0 \\1

\end{pmatrix}\]
So if you multiply this out, you'll get:

\[ \begin{pmatrix}
1*a+0*b\\1*c+0*d

\end{pmatrix} = \begin{pmatrix}
0 \\1

\end{pmatrix}\]

\[ \begin{pmatrix}
a\\c

\end{pmatrix} = \begin{pmatrix}
0 \\1

\end{pmatrix}\]

which is the same as saying a=0 and c=1. Then just do another test vector to solve for the other two entries of the matrix by rotating it 90 degrees in the picture. I suggest trying \(\binom{0}{1}\) just be careful it's a little tricky!