1. ## [SOLVED] 2 proofs

I was wondering if anyone could help me get started on these 2 proofs. I'm having trouble getting through them. They're to help me prep for my upcoming final.

1) Let u and v be vectors in an inner product space. Prove that ||u|| = ||v|| if and only if (u + v) and (u - v) are orthogonal.

2) Let A be an (n × n) matrix, and let λ be an eigenvalue of A. Prove that if k is any scalar, then λ+k is an eigenvalue of A+kI.

Any help is greatly appreciated.

2. Originally Posted by leverin4
I was wondering if anyone could help me get started on these 2 proofs. I'm having trouble getting through them. They're to help me prep for my upcoming final.

1) Let u and v be vectors in an inner product space. Prove that ||u|| = ||v|| if and only if (u + v) and (u - v) are orthogonal.
\begin{aligned}\left|\left|u\right|\right|=\left|\ left|v\right|\right|&\iff \left|\left|u\right|\right|^2=\left|\left|v\right| \right|^2\\&\iff\left=\left\ \ &\iff\left-\left=0\\ &\iff\left+\left-\left-\left=0\\ &\iff\left+\left=0\\&\iff\left=0\qquad\square\end{aligned}

Does this make sense? Some steps are skipped -- see if you can fill them in.

2) Let A be an (n × n) matrix, and let λ be an eigenvalue of A. Prove that if k is any scalar, then λ+k is an eigenvalue of A+kI.

Any help is greatly appreciated.
If $\lambda$ is an eigenvalue of $A$, then $Ax=\lambda x$. But

\begin{aligned}Ax=\lambda x&\implies Ax+kx=\lambda+kx\\ &\implies\ldots\end{aligned}

Can you finish this off?

3. Originally Posted by Chris L T521
\begin{aligned}\left|\left|u\right|\right|=...\\ &\iff\left=\left\\ &\iff\left-\left=0\\ &\iff\left+\left-\left-\left=0\\ &\iff\left+\left+\left+\left=0\\ &\iff\left+\left=0\\&\iff\left=0\qquad\square\end{aligned}

Does this make sense? Some steps are skipped -- see if you can fill them in.
It makes a lot of sense, thanks for your help. I added in what I think you meant to be the missing step; I only saw one, I think. I'm not very good at latex, sorry. Since the problem statement has "if and only if", I also have to prove it backwards, right? Which is essentially the same steps backwards?

4. Originally Posted by leverin4
It makes a lot of sense, thanks for your help. I added in what I think you meant to be the missing step; I only saw one, I think. I'm not very good at latex, sorry. Since the problem statement has "if and only if", I also have to prove it backwards, right? Which is essentially the same steps backwards?
You filled in the missing step correctly. The nice thing about this question is that we can take care of both proof directions with each step!

5. Originally Posted by Chris L T521
If $\lambda$ is an eigenvalue of $A$, then $Ax=\lambda x$. But

\begin{aligned}Ax=\lambda x&\implies Ax+kx=\lambda x+kx\\ &\implies\ \left(A+k\right)x=\left(\lambda+k\right)x\end{alig ned}

Can you finish this off?
Still stuck on this one though, I added the only logical thing I can see. Don't see how it helps me all that much though.

6. Originally Posted by leverin4
Still stuck on this one though, I added the only logical thing I can see. Don't see how it helps me all that much though.
Close! The LHS contains matricies! So when you factor out $x$ from $Ax+kx$, you don't get $(A+k)x$ because you can't take a constant and add it to a matrix. Instead you get...(try to finish the statement)

7. Originally Posted by Chris L T521
Close! The LHS contains matricies! So when you factor out $x$ from $Ax+kx$, you don't get $(A+k)x$ because you can't take a constant and add it to a matrix. Instead you get...(try to finish the statement)
$(A+kI)x$?

8. Originally Posted by leverin4
$(A+kI)x$?
Correct. Now what conclusion can be drawn from $(A+kI)x=(\lambda+k)x$?

9. Originally Posted by Chris L T521
Correct. Now what conclusion can we draw from $(A+kI)x=(\lambda+k)x$?
We can conclude that $(\lambda+k)$ is a scalar that has the same effect on $x$ as $(A+kI)$ and therefore $(\lambda+k)$ is, by definition, an eigenvalue for $(A+kI)$.

10. Originally Posted by leverin4
We can conclude that $(\lambda+k)$ is a scalar that has the same effect on $x$ as $(A+kI)$ and therefore $(\lambda+k)$ is, by definition, an eigenvalue for $(A+kI)$.
Correct!

Does this make better sense now?

11. Originally Posted by Chris L T521
Correct!

Does this make better sense now?
A lot, I was actually stuck at the point where I factor out the $x$ for awhile before I decided to post it here.. You're help has been great. I thanked you in each post.