Proving and disproving matrix properties?

[cosmopolitanx]

I have no idea how to approach these.. any help would be greatly appreciated.
I can't seem to find this in the book either so if there are any links on the web that relate to these questions please let me know.. Thanks!

Prove or disprove the following statements. I and 0 represent the identity
and zero matrix of the same size as A.

1. If A is a matrix such that A^2 - 3A + 2I = 0, then A - 2I is invertible.
2. If A is a matrix such that A^2 - 5A + 5I = 0, then A - 3I is invertible.
3. If A is invertible, then A^2 + I is invertible.
4. Prove that there is no 5 by 5 matrix such at A^2 + I = 0.

 

[jbunniii]

Hint for (2): note that

A^2 - 5A + 5I = (A - 3I)(A - 2I) - I

 

[jbunniii]

Hint for (1): note that

A^2 - 3A + 2I = (A - 2I)(A - I)

Think about what

(A - 2I)(A - I) = 0 means. Right away you can see two possible solutions for A. What is (A - 2I) in each case?

 

[Dick]

For 2. I would try to prove the contrapositive. If (A-3I) is NOT invertible then A^2 - 3A + 2I is NOT zero. Hint: A-3I not invertible tells you something the existence of a eigenvector of A. For the last two you must mean A to be a real matrix. Otherwise there's a diagonal counterexample for both. Hint: det(A^2)=det(-I).

 

[cosmopolitanx]

hey guys thanks for your help.. so far for one i have
A = I then A-2I is invertible so the statement is true.
for two i have
A = 3I or A = 4I but neither of these work when i plug it back into the original equation?
the last two are real matrices..
for four I'm trying to prove that A^2 can not = -I ?

 

[Mark44]

For 4, don't forget that your working with 5 x 5 matrices. I'm sure that is significant.

 

[Hurkyl]

P.S. keep in mind matrices do not have the cancellation property. While

A=I and A=2I​

are both solutions to

(A-I)(A-2I)=0,​

there is no reason to think those are the only solutions.

 

[Hurkyl]

Really, we cannot help effectively without know what you know about polynomial equations involving matrices. This is part of why our homework guidelines insist that you post what you have done on the problem, and what facts or approaches you believe may be relevant. At the very least, it would help to know what class (and at what level) the question comes from.

 

[Dick]

I agree with Hurkyl. You got 1 wrong by picking half of a hint and then jumping to a wrong conclusion. Why don't you concentrate on one of these problems at a time and fully explain your reasoning? Then we can figure out what you know and what you don't know.

 

[cosmopolitanx]

hey in response to your message hurkyl, I'm taking "elementary linear algebra"
and right now we have covered basic matrix stuff like inverse, determinants, solving systems of linear equations, multiplying/adding matrices etc..
i think these questions relate to chapter two which is about cofactors, adjoint, cramer's rule, eigenvalues/eigenvectors and diagonilization; however, i missed the past two classes so right now I'm trying to self-learn a lot of this stuff from the book...

 

[Dick]

You probably do need chapter 2 for problem 2. It's about eigenvectors. You can handle 1 and 3 without it if you are clever with finding counterexamples. You can also do 4 if you learned determinants.

 

[jbunniii]

hey guys thanks for your help.. so far for one i have
A = I then A-2I is invertible so the statement is true.

A = I is ONE solution. The factorization I gave in my earlier hint shows that there is (at least) one more solution, namely A = 2I. Is A - 2I invertible in this case?

 

[jbunniii]

You probably do need chapter 2 for problem 2. It's about eigenvectors. You can handle 1 and 3 without it if you are clever with finding counterexamples. You can also do 4 if you learned determinants.

Actually problem 2 is trivial with the hint

A^2 - 5A + 5I = (A - 3I)(A - 2I) - I

although this doesn't illustrate any sort of general method for solving similar problems.

Without more context it's hard to know what machinery is expected to be available for these problems.

 

[HallsofIvy]

As jbunniii said originally, "A^2 - 5A + 5I = (A - 3I)(A - 2I) - I = 0" so (A- 3I)(A-2I)= I. Yes, it is trivial!

 

[Dick]

Actually problem 2 is trivial with the hint

A^2 - 5A + 5I = (A - 3I)(A - 2I) - I

although this doesn't illustrate any sort of general method for solving similar problems.

Without more context it's hard to know what machinery is expected to be available for these problems.

That is clever. I didn't quite see where that hint was going.

 

[cosmopolitanx]

thanks everyone for your input it really helped me out..
for one i have A= I or 2I but if A=2I then A-2I is not invertible so the statement is false

for two i don't really understand Jbunniii's hint.. i originally thought A equaled 4I and 3I but when i put these back into the original A^2 - 5A + 5I equation, these matrices do not equal 0.. i decided to use the quadratic formula so now i have A = (5+/- sqroot5)/2 and these answers both give 0 when plugged back into the original equation and they are both invertible for A-3I so for two i have the statement is true?

 

[jbunniii]

thanks everyone for your input it really helped me out..
for one i have A= I or 2I but if A=2I then A-2I is not invertible so the statement is false

for two i don't really understand Jbunniii's hint.. i originally thought A equaled 4I and 3I but when i put these back into the original A^2 - 5A + 5I equation, these matrices do not equal 0.. i decided to use the quadratic formula so now i have A = (5+/- sqroot5)/2 and these answers both give 0 when plugged back into the original equation and they are both invertible for A-3I so for two i have the statement is true?

A^2 - 5A + 5I = (A - 3I)(A - 2I) - I = 0 <==> (A - 3I)(A - 2I) = I

What does the equation

(A - 3I)(A - 2I) = I

mean? What does it say about the matrices (A - 3I) and (A - 2I)?

 

[cosmopolitanx]

.. they are inverses?

 

[jbunniii]

.. they are inverses?

Correct. The inverse of (A-3I) is (A-2I) and vice versa. Therefore you can answer "yes" to the original question (is A - 3I invertible?), and you got the bonus information that (A - 2I) is invertible as well, and the even more bonus information that they are inverses of each other!

 

[FreedayFF]

Anyone has any ideas how to do 3 and 4?

 

[Dick]

For 4 use determinants. For 3, think of a counterexample.

 

[FreedayFF]

can you elaborate the counterexample?

 

[Dick]

Not until you at least TRY. This isn't even your thread. If you believe 4 then you can't do it if n=5. Why not? Why might you be able to do it if n is not 5? Try proving 4 first.

 

[FreedayFF]

by 3 I meant question 3. I tried to find something that when (A^2 + I)x(something) = kI to prove that (A^2 + I) has an inverse. But have not succeeded in a day. By counter example, do you mean pretending that (A^2 + I) has no inverse and prove it wrong?

 

[Dick]

No, I meant find a real matrix A with actual numbers in it such that A is invertible and A^2+I is not invertible. You can find an example where A is 2x2. Hint: can you find a matrix A where A^2+I=0?

 

[FreedayFF]

Hahaha silly me spending the whole day trying to prove that it's right, when in fact, it's wrong. Thanks a lot (:

 

[cosmopolitanx]

i found a counter example by setting
a^2 + bc + 1 = 0
d^2 + bc + 1 = 0
b (a + d) = 0
c (a + d) = 0

where a and d are negatives of each other and bc = neg (a^2) minus one
thanks :)

 

[cosmopolitanx]

any ideas on number four?
i don't know how to prove A^2 cannot = -I for 5x5 matrices.. i know it's possible for 2x2 so why would it be different for a 5x5?
i know dick mentioned something about determinants but finding a determinant of a 5x5 would mean 25 different variables? .. is that the only way to approach this?

 

[Dick]

any ideas on number four?
i don't know how to prove A^2 cannot = -I for 5x5 matrices.. i know it's possible for 2x2 so why would it be different for a 5x5?
i know dick mentioned something about determinants but finding a determinant of a 5x5 would mean 25 different variables? .. is that the only way to approach this?

Finding the determinant of a diagonal matrix like -I is really quite easy. You've covered them, you should know this. And how is det(A^2) related to det(A)?

 

[cosmopolitanx]

ahh i see what you're getting at.. i think i finally got it now :)
thanks again for all your help dick.. and everyone else!