A couple questions for linear algebra final review

[BiGyElLoWhAt]

Homework Statement

2: Some proofs:
a) If $\{ v_1 , v_2...v_n \}$ are linearly independent in a real vector space, so are any subset of them.

b) If any subset of vectors $\{ v_1 , v_2...v_n \}$ in a real vector space are linearly dependent, then the whole set of vectors are linearly dependent.

c) If A is invertible so is A^2

d) if A is not invertible, neither is A^3

I think I either know everything else or I can't really ask without asking to be walked through it so I'll just have to spend some time on google for acouple of them.

Any help is appreciated, my final's in 18 hours.

Homework Equations

The Attempt at a Solution

2:
a)I was thinking I could use closure under scalar multiplication and addition to prove that, but is that solid enough?

If $\{ v_1 , v_2...v_n \}$ are linearly independent then by definition

$c_1v_1 + ... + c_{n-1}v_{n-1} ≠ c_nv_n$

(is subset ≈ subspace?)

then for $\{ v_1 , v_2...v_m \}$ where m≤n (not really sure how to notate that all the vectors of this set are contained within the first set) which are all members of the first set, are linearly independent.

This feels weak though... It seems like common sense that if i have some linearly independent vectors, and I throw some of them out, then what I have is still linearly independent; I'm also sure that I need to start at the definition of linearly independent, which basically involves the demonstration of closure under addition and scalar multiplication. I'm just not really sure how to put it technically. I think this is one of those "prove 2+2=4" things and I'm just not seeing how to put it mathematically.

2:
b) Very similar work, I could copy and paste everything I have from up ther and put it here, but once I figure out a, I'll get b.

2:
c) $A^2 = AA$ multiply by $A^{-1}$
... $AAA^{-1} = AI = A$
Proved.
I only put this in here because I'm not sure what to do about d. I know it's very similar, but is it simply
$A^3 = AAA$ and since A is not invertible, A^3 cannot be reduced? How do I word that?

 

[STEMucator]

Homework Statement

2: Some proofs:
a) If $\{ v_1 , v_2...v_n \}$ are linearly independent in a real vector space, so are any subset of them.

b) If any subset of vectors $\{ v_1 , v_2...v_n \}$ in a real vector space are linearly dependent, then the whole set of vectors are linearly dependent.

c) If A is invertible so is A^2

d) if A is not invertible, neither is A^3

I think I either know everything else or I can't really ask without asking to be walked through it so I'll just have to spend some time on google for acouple of them.

Any help is appreciated, my final's in 18 hours.

Homework Equations

The Attempt at a Solution

2:
a)I was thinking I could use closure under scalar multiplication and addition to prove that, but is that solid enough?

If $\{ v_1 , v_2...v_n \}$ are linearly independent then by definition

$c_1v_1 + ... + c_{n-1}v_{n-1} ≠ c_nv_n$

(is subset ≈ subspace?)

then for $\{ v_1 , v_2...v_m \}$ where m≤n (not really sure how to notate that all the vectors of this set are contained within the first set) which are all members of the first set, are linearly independent.

This feels weak though... It seems like common sense that if i have some linearly independent vectors, and I throw some of them out, then what I have is still linearly independent; I'm also sure that I need to start at the definition of linearly independent, which basically involves the demonstration of closure under addition and scalar multiplication. I'm just not really sure how to put it technically. I think this is one of those "prove 2+2=4" things and I'm just not seeing how to put it mathematically.

2:
b) Very similar work, I could copy and paste everything I have from up ther and put it here, but once I figure out a, I'll get b.

2:
c) $A^2 = AA$ multiply by $A^{-1}$
... $AAA^{-1} = AI = A$
Proved.
I only put this in here because I'm not sure what to do about d. I know it's very similar, but is it simply
$A^3 = AAA$ and since A is not invertible, A^3 cannot be reduced? How do I word that?

For a) $\{ v_1 , v_2...v_n \}$ are linearly independent when the vectors satisfy $c_1v_1 + ... + c_nv_n = 0 \iff c_1=...=c_n=0$.

Suppose you take a smaller subset of the vectors, $\{ v_1 , v_2...v_{n-1} \}$. Then these vectors are linearly independent when $c_1v_1 + ... + c_{n-1}v_{n-1} = 0 \iff c_1 = ... = c_{n-1} = 0$.

Using the fact that $c_1v_1 + ... + c_{n-1}v_{n-1} = c_nv_n$, you can say that the smaller subset of vectors is non-singular if $c_nv_n = 0 \iff c_n = 0$.

It must be the case that $c_n = 0$, otherwise you would have a contradiction. Why?

For part c) and d) what about the determinant?

 

[BiGyElLoWhAt]

Aha! thank you I didn't even think about the determinant. But is the only situation where A is non invertible when det(A)=0? If it is then that makes c and d very simple as det(A^n) = det(A)^n.

 

[STEMucator]

Aha! thank you I didn't even think about the determinant. But is the only situation where A is non invertible when det(A)=0? If it is then that makes c and d very simple as det(A^n) = det(A)^n.

If $A$ is invertible (non-singular), $det(A) ≠ 0$.

If $A$ is not invertible (singular) $det(A) = 0$.

It makes c) and d) quite straightforward.

 

[BiGyElLoWhAt]

Awesome, thanks a million.