## prove or show false: X^2 = (identity) only if X=I or X=-I,

Linear spaces and subspaces, linear transformations, bases, etc.
testing
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### prove or show false: X^2 = (identity) only if X=I or X=-I,

Show that the statement is generally true or find specific 2 x 2 matrices for which the statement is not true.
1. The matrix equation X^2 = I is satisfied only when X = I or X = -I
2. (A - B)(A + B) = A^2 - B^2 if and only if AB = BA
3. If B = A^2 - 5A + 2I, then AB = BA
4. AB = 0 if and only if BA = 0
5. If A^3 - 7A^2 + 5I = 0, then A^4 = 49A^2 - 5A - 35I
Note: The " I " is the identity matrix, not the number one or the letter ell.

a) I'm sure this is false, but I can't think of any way, off the top of my head, to come up with a matrix that will work.

b) (A - B)(A + B) = (A - B)(A) + (A - B)(B) = A^2 - BA + AB - B^2. For this to equal A^2 - B^2, you have to have -BA + AB = 0, which means AB = BA, right?

I have no idea what to do for c) and e), and I can't think how to come up with an example for d) which I also think is wrong.

stapel_eliz
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a) The matrix equation X^2 = I is satisfied only when X = I or X = -I
Your instinct is right. Since you only have to find a 2-by-2 matrix, make one up, and solve for what you need. For instance:

. . . . .$\left[\begin{array}{cc}a&b\\0&c\end{array}\right] \left[\begin{array}{cc}a&b\\0&c\end{array}\right]\, =\, \left[\begin{array}{cc}a^2&ab\,+\,bc\\0&c^2\end{array}\right] \, =\,\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

Then what can you say about a, b, and c? What matrix can you come up with?
b) (A - B)(A + B) = A^2 - B^2 if and only if AB = BA
You're right.
c) If B = A^2 - 5A + 2I, then AB = BA
Multiply on the left by A. Then note that, by properties of matrices, A(A2) = ()A, A(5A) = 5(A(A)) = 5(A2) = (5A)A, and A(2I) = 2(A(I)) = 2A = 2(IA) = (2I)A. "Factor out" the A on the right-hand side, and subtitute from the right-hand side of the original equation.
d) AB = 0 if and only if BA = 0
As with (b) above, try making something up, in a general sense, and see where that leads. For instance:

. . . . .$\left[\begin{array}{cc}1&1\\2&2\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\, =\, \left[\begin{array}0&0\\0&0\end{array}\right]$

What can you say about a, b, c, and d? Picking non-zero values, what do you get when you reverse the multiplication?
e) If A^3 - 7A^2 + 5I = 0, then A^4 = 49A^2 - 5A - 35I
If A3 - 7A2 + 5I = 0, then A3 = 7A2 - 5I. Multiply this equation on the left by A. This gives you A4 = 7A3 - 5A. From A3 = 7A2 - 5I, substitute the right-hand side in for A3 in 7A3 - 5A. See where that leads....

Have fun!

testing
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Joined: Sun Dec 07, 2008 12:26 am
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thnx