# Thread: Easy Linear Algebra Proofs

1. ## Easy Linear Algebra Proofs

Hello I'm reading Lang's Intro to Linear Algebra & I've noticed that he uses squaring an equation,
working out the algebra and then square-rooting to prove a theorem.
I'm trying to get used to proofing for analysis &
I'd like to know whether squaring is considered an adequate method of proof?

A quick example is;

||xA|| = |x| ||A||

1/ ||xA||² = {√[(xA) · (xA)]}² = xA · xA

(Using the definition ||A|| = √(A · A) = √(a_1² + a_2² + ... + a_n²)

2/ xA · xA = (xa_1, xa_2, ..., xa_n) · (xa_1, xa_2, ..., xa_n)

(Using the definition of vector A : (a_1, a_2, ..., a_n) )

3/ x²a_1² + x²a_2² + ... + x²a_n² = x²(a_1² + a_2² + ... + a_n²)

(Using the Dot Product property for components)

4/ x² A · A = |x|² ||A||²

(Rewriting the squared components in 3/ as A · A,
x² as |x|² to account for ± x values & A · A as ||A||²
as just representing the definition of ||A|| squared).

5/ Take a square root, and voila!

Here he set out to prove his theorem by squaring out one side,
working out the algebra and achieving the other side i.e. proving an equality.

I'm just concerned as to whether this would constitute a rigid proof
in an analysis book as the ones I've read
(albeit I was lacking the mathematical maturity I have now, which is still in it's infancy!)
would seemingly come out of nowhere

Another question is the proof of orthogonality.

||A + B|| = ||A - B|| iff A · B = 0

1/ ||A + B||² = ||A - B||² <==> {√[(A + B) · (A + B)]}² = {√[(A - B) · (A - B)]}²

2/ (A + B) · (A + B) = (A - B) · (A - B) <==> A² + 2A · B + B² = A² - 2A · B + B²

3/ 2A · B = - 2A · B

4/ 4A · B = 0

5/ A · B = 0

If A · B = 0 then the above is true, but if A · B ≠ 0
how are you supposed to show ||A + B|| ≠ ||A - B|| ?

I'm thinking you're supposed to find a contradiction.
You assume all of the above & do the proof like I did,
then when you get to the end, 5/,
you find A · B = 0 but you know that A & B are not orthogonal
so we see that the assumption cannot be true.

Is that considered a proof or just a small exercise?

Note: I'm supposed to be proving all of this just using 4 properties of the dot product
as this is a way to achieve generality;
1/ A · B = B · A
2/ A · (B + C) = A · B + A · C
3/ (xA) · B = x(A · B)
4/ A · A > 0 iff A ≠ 0

2. Are you French?

3. If French people do this --> , then yes

Otherwise, no I'm Irish

Hello I'm reading Lang's Intro to Linear Algebra & I've noticed that he uses squaring an equation,
working out the algebra and then square-rooting to prove a theorem.
I'm trying to get used to proofing for analysis &
I'd like to know whether squaring is considered an adequate method of proof?

A quick example is;

||xA|| = |x| ||A||

1/ ||xA||² = {√[(xA) · (xA)]}² = xA · xA

(Using the definition ||A|| = √(A · A) = √(a_1² + a_2² + ... + a_n²)

2/ xA · xA = (xa_1, xa_2, ..., xa_n) · (xa_1, xa_2, ..., xa_n)

(Using the definition of vector A : (a_1, a_2, ..., a_n) )

3/ x²a_1² + x²a_2² + ... + x²a_n² = x²(a_1² + a_2² + ... + a_n²)

(Using the Dot Product property for components)

4/ x² A · A = |x|² ||A||²

(Rewriting the squared components in 3/ as A · A,
x² as |x|² to account for ± x values & A · A as ||A||²
as just representing the definition of ||A|| squared).

5/ Take a square root, and voila!

Here he set out to prove his theorem by squaring out one side,
working out the algebra and achieving the other side i.e. proving an equality.

I'm just concerned as to whether this would constitute a rigid proof
in an analysis book as the ones I've read
(albeit I was lacking the mathematical maturity I have now, which is still in it's infancy!)
would seemingly come out of nowhere
They don't "come out of nowhere", they come out of the definition of |v| as $\sqrt{\sum x_i^2}$.

Another question is the proof of orthogonality.

||A + B|| = ||A - B|| iff A · B = 0

1/ ||A + B||² = ||A - B||² <==> {√[(A + B) · (A + B)]}² = {√[(A - B) · (A - B)]}²

2/ (A + B) · (A + B) = (A - B) · (A - B) <==> A² + 2A · B + B² = A² - 2A · B + B²

3/ 2A · B = - 2A · B

4/ 4A · B = 0

5/ A · B = 0

If A · B = 0 then the above is true, but if A · B ≠ 0
how are you supposed to show ||A + B|| ≠ ||A - B|| ?
No, the above is true as long as $A\cdot B$ is NOT equal to 0, other wise we cannot divide by $A\cdot B$.
If $A\cdot B= 0$ then ||A+ B|| and ||A- B|| are both the lengths of the hypotenuse of right triangles with legs of length ||A|| and ||B|| and so are equal.

I'm thinking you're supposed to find a contradiction.
You assume all of the above & do the proof like I did,
then when you get to the end, 5/,
you find A · B = 0 but you know that A & B are not orthogonal
so we see that the assumption cannot be true.

Is that considered a proof or just a small exercise?

Note: I'm supposed to be proving all of this just using 4 properties of the dot product
as this is a way to achieve generality;
1/ A · B = B · A
2/ A · (B + C) = A · B + A · C
3/ (xA) · B = x(A · B)
4/ A · A > 0 iff A ≠ 0

5. Originally Posted by HallsofIvy
They don't "come out of nowhere"
No, I was talking about the proof's I'd seen in my real analysis book. My question was about squaring as an adequate method of proofing because I've never really seen just squaring an equation working out the algebra & then rooting as a method of proof bar some basic algebra stuff. I wonder would a real analysis book contain proofs that used this method of proofing?

Originally Posted by HallsofIvy
No, the above is true as long as A · B is NOT equal to 0
No, you're missing what I'm saying.

The proof I've laid out is when ||A + B|| & ||A - B|| are legs of a hypoteneuse, yes.

My question was, if you were given ||A + B|| = ||A - B|| and asked to do the proof to find out if this equality was true even though A and B were not orthogonal, would what I had written after the correct proof make sense?

if you were given ||A + B|| = ||A - B|| and asked to do the proof to find out if this equality was true even though A and B were not orthogonal, would what I had written after the correct proof make.
Suppose that $U~\&~V$ are any two vectors such that $\left\| {U + V} \right\| = \left\| {U - V} \right\|$.
Then it follows that $2U\cdot V=-2U\cdot V$.
Which implies $U\cdot V=0$.