Eigenvalue of matrix proof by induction

[TanWu]

Homework Statement

Show that if $\alpha$ is an eigenvalue of matrix $B$ then $\alpha^{n}$ is an eigenvalue of matrix $B^{n}$ for positive integer $n$

Relevant Equations

$\alpha$ is eigenvalue
$B\vec x = \alpha \vec x$

We consider base case ($n = 1$), $B\vec x = \alpha \vec x$, this is true, so base case holds.

Now consider case $n = 2$, then $B^2\vec x = B(B\vec x) = B(\alpha \vec x) = \alpha(B\vec x) = \alpha(\alpha \vec x) = \alpha^2 \vec x$

Now consider $n = m$ case,

$B^m\vec x = B(B^{m - 1} \alpha) = B^m \alpha = \alpha^m \vec x$

Now consider $n = m + 1$ case,
$B^{m + 1}\alpha = B(B^m \alpha) = B(\alpha^m \vec x) = \alpha^m(B\vec x) = \alpha^k(\alpha \vec x) = \alpha^{m + 1}\vec x$

Is that correct use of induction?
gratitude expressed to those who help.

 

[nuuskur]

You can formulate more precisely. Instead of saying "consider $n=m$ case", we can say "assume that $B^nx = \alpha ^nx$ for some $n\geqslant 2$". Then for the case $n+1$ we have the equalities
<br /> B^{n+1}x = BB^nx = B\alpha ^nx = \alpha ^nBx = \alpha ^n\alpha x = \alpha ^{n+1}x.<br />
Right now you haven't made it explicit what your induction assumption is. Anyone familiar with the technique is able to fill in the gaps.

 

[docnet]

Homework Statement: Show that if $\alpha$ is an eigenvalue of matrix $B$ then $\alpha^{n}$ is an eigenvalue of matrix $B^{n}$ for positive integer $n$
Relevant Equations: $\alpha$ is eigenvalue
$B\vec x = \alpha \vec x$

We consider base case ($n = 1$), $B\vec x = \alpha \vec x$, this is true, so base case holds.

Now consider case $n = 2$, then $B^2\vec x = B(B\vec x) = B(\alpha \vec x) = \alpha(B\vec x) = \alpha(\alpha \vec x) = \alpha^2 \vec x$

You shouldn't need to show it for $n=2$. After the base case, state the induction hypothesis:
$$B^m\vec x = \alpha^m \vec x$$
is true for some integer $m>1$.

Apply the induction hypothesis to show its true for $m+1$, as you have done so. Then you can conclude it's true for all integers.

 

[nuuskur]

It is not wrong to check the claim up to $m$ manually. Sometimes that's even helpful. The induction hypothesis then is that the claim is true

1. for some $n\geqslant m$ (weak induction);
2. for all $k\leqslant n$, where $n\geqslant m$ (strong induction).

In both cases the task is to prove the claim is true for $n+1$. It doesn't matter which form of induction we use, they are equivalent.

 

[docnet]

Sounds good, though I feel it's more concise/elegant to use weak induction here. But just to clarify, the claim being true for $n+1$, or proving the claim is true for $n+1$, is not any part of the induction hypothesis. Rather, it's the consequence that would follow if the induction hypothesis is true.

 

[nuuskur]

Fair play on your last point. As for the first one, optimisation can be done later. First and foremost, it is important the argument is correct.

 

[docnet]

@PeroK I have a nagging feeling that my statement in #5 about the induction hypothesis is not correct. If you have a moment, can you please point out what's wrong about it? Thank you.

 

[Mark44]

But just to clarify, the claim being true for n+1, or proving the claim is true for n+1, is not any part of the induction hypothesis.

Right. The induction hypothesis in this case is assuming that $B^m\vec x = a^m\vec x$ is true. Using this hypothesis, you show that it follows that $B^{m+1}\vec x = a^{m+1}\vec x$ must also be true. Of course you need to also establish that some base case is true.