# Matrix proof

1. Dec 7, 2013

### Baharx

Matrix proof :(

1. Let A and B be two similar matrices.

characteristic in the space λ is an eigen value,

show that : sized V_λ^A = sized V_λ^B

2. Let A invertible matrix.

A ∈ ℝ nxn and invertible matrix ⇔ 0, A is not an eigen value.

3. Let A and B be two similar matrices.

Nullity (A) = Nullity (B)

2. Dec 7, 2013

### HallsofIvy

Staff Emeritus
I don't know what this means. What is V_λ? What does "^A" mean? Do you possible mean the "eigenspace" corresponding to eigenvalue λ? That is, the subspace of all v such that Av= λv? And "sized" is the dimension of that space?

A is a matrix, it can't be an eigenvalue! I presume you mean "0 is not an eigenvalue for A". If it where, there would exist a non-zero vector, v, such that Av= 0. What does that tell you about $A^{-1}(0)$?

First, do you know what a "similar matrices" are? Two matrices, A and B, are similar, if and only if there exist an invertible matrix, P, such that $PAP^{-1}= B$ which is the same as saying that $PA= PB[/tex]. So if v is a non-zero vector such that Av= 0, then PAv= 0= PBv. Since P is invertible, if follows that Bv= 0. 3. Dec 7, 2013 ### Baharx thank you so much for your reply. i solved 2 and 3. yes i have to solve that way all problems : [itex]PAP^{-1}= B$ mean is A and B similar matrix.

for 1 question;

how can i write;

Let A and B be two similar matrices. $PAP^{-1}= B$

characteristic in the space λ is an eigen value,

show that : sized VλA = sized VλB

(sized or size V )

still i couldn't solve.

---------------

Question 2:

Let A be a square matrix. Prove that A is invertible if and only if 0 is not an eigenvalue of A by
showing:
A cuter way to do this problem would be to say:
0 is an eigenvalue ⇔ det(A − 0I) = 0 ⇔ det A = 0 ⇔ A is not invertible.
(a) If 0 is not an eigenvalue of A, then A is invertible.
If 0 is not an eigenvalue, then Ax = 0x has only the trivial solution, so A is invertible.
(b) If A is invertible, then 0 is not an eigenvalue of A.
If A is invertible, then A has trivial nullspace, so Ax = 0 has only the trivial solution. But this equation is the same as Ax = 0x, so we see that there are no eigenvectors corresponding to λ = 0

Question 3:

B is row equivalent to A if B = CA, where C is invertible. We can now establish two important results:
Theorem: If B ∼ A, then Null(B) =Null(A).

Suppose x ∈ Null(A). Then Ax = 0. Since B ∼ A, then for some invertible matrix C B=CA, and it follows that Bx=CAx=C0=0, so x∈Null(B). Therefore Null(A) ⊆ Null(B).
Conversely, if x ∈ Null(B), then Bx = 0. But B = CA, where C is invertible, being the product of elementary matrices. Thus Bx = CAx = 0.
Multiplying by C^−1 gives Ax = (C^−1).0 = 0,
so x ∈ Null(A), and Null(B) ⊆ Null(A). So the two sets are equal, as advertised.

QUESTION 1..??? :/

4. Dec 7, 2013

### Baharx

Btw question 3 is maybe we can solve like this; (if its not true my solve, pls tell me.)

Suppose x ∈ Nullity(A) then Ax=0
$PAP^{-1}= B$

Bx = P^-1AP x= I. Ax = 0 → x ∈ Nullity(B)

Suppose x ∈ Nullity(B) then Bx=0

Ax = P^-1BP x= I. Bx = 0 → x ∈ Nullity(A)

Nullity(B) ⊆ Nullity(A). So the two sets are equal,

Nullity(B) = Nullity(A)

5. Dec 7, 2013

### Ray Vickson

What does "sized" or "size" mean? I have read many books on linear algebra over the past 50 years and have NEVER seen that term. If you want help you really do need to answer questions that people ask you.

6. Dec 7, 2013

7. Dec 7, 2013

### Baharx

i am sorry i was trying to translate.. Size hmmm length or a scalar "norm".

8. Dec 7, 2013

### Baharx

ok, sorry for that.

9. Dec 8, 2013

### Baharx

ok i will write again.

Let A and B be two similar matrices. $PAP^{-1}= B$

characteristic in the space λ is an eigen value,

show that :

length $V^{A}_{λ}$ = length $V^{B}_{λ}$

dimension $V^{A}_{λ}$ = dimension $V^{B}_{λ}$

Thanks.. probably dimension.

Last edited: Dec 8, 2013