Action of a linear operator on vectors

[DrClaude]

Not really a homework problem, just doing some self-studying.

Homework Statement

Let $| a \rangle$ by any vector in an $N$-dimensional vector space $\mathcal{V}$, and $\mathbf{A}$ a linear operator on $\mathcal{V}$. The vectors
$$
| a \rangle, \mathbf{A} | a \rangle, \mathbf{A}^2 | a \rangle, \ldots, \mathbf{A}^N | a \rangle
$$
are linearly dependent. Let $\mathcal{M} \equiv \text{Span}\{ \mathbf{A}^k | a \rangle\}_{k=0}^N$. Show that $\mathcal{M}$ has the property that for any vector $| x \rangle \in \mathcal{M}$, the vector $\mathbf{A} | x \rangle$ also belongs to $\mathcal{M}$.

Homework Equations

The Attempt at a Solution

Since the above vectors span $\mathcal{M}$, I can always find $\{c_k\}$ such that I can write
$$
| x \rangle = \sum_{k=0}^N c_k \mathbf{A}^k | a \rangle
$$
Now, if $c_N = 0$, and since $\mathbf{A}$ is a linear operator, I have
$$
\begin{array}{rl}
\mathbf{A} | x \rangle & = \mathbf{A} \sum_{k=0}^{N-1} c_k \mathbf{A}^k | a \rangle \\
& = \sum_{k=0}^{N-1} c_{k} \mathbf{A}^{k+1} | a \rangle
\end{array}
$$
which, by definition, is in $\mathcal{M}$.

But what if $c_N \neq 0$? I need to show that $\mathbf{A}^{N+1} | a \rangle$ is in $\mathcal{M}$. I'm tempted to use the fact that the vectors are linearly dependent to state that I can write
$$
\mathbf{A}^{N} | a \rangle = \sum_{k=0}^{N-1} \alpha_k \mathbf{A}^k | a \rangle
$$
but nothing guarantees me that $\mathbf{A}^{N} | a \rangle$ is not orthogonal to the other vectors, or does it?

Any help appreciated.

 

[micromass]

But what if $c_N \neq 0$? I need to show that $\mathbf{A}^{N+1} | a \rangle$ is in $\mathcal{M}$. I'm tempted to use the fact that the vectors are linearly dependent to state that I can write
$$
\mathbf{A}^{N} | a \rangle = \sum_{k=0}^{N-1} \alpha_k \mathbf{A}^k | a \rangle
$$
but nothing guarantees me that $\mathbf{A}^{N} | a \rangle$ is not orthogonal to the other vectors, or does it?

You can't really write this. Linear dependent means (by definition) that there are scalars $\alpha_k$ which are not all zero such that

$$\alpha_0 | a \rangle + \alpha_1 \mathbf{A}| a \rangle + ... + \alpha_N \mathbf{A}^N| a \rangle=\mathbf{0}$$

Now, you have two situations. Either $\alpha_N\neq 0$, and in that case, you can write something like you wrote in the quote. In the case that $\alpha_n=0$, you need to do something differently. In this case, they key fact is that at least one of the $\alpha_k$ is nonzero.

You can handle both cases simultaniously by remarking that there exists an $M\leq N$ and $\alpha_k$ scalars such that

$$\mathbf{A}^M | a \rangle = \sum_{k=0}^{M-1} \alpha_k \mathbf{A}^k| a \rangle$$

 

[Fredrik]

This isn't precalculus (although at my university, we studied it before calculus. ) I'm moving it to calculus & beyond.

 

[DrClaude]

This isn't precalculus (although at my university, we studied it before calculus. ) I'm moving it to calculus & beyond.

Sorry about that. I hesitated for a while as to which forum to post to.

You can't really write this. Linear dependent means (by definition) that there are scalars $\alpha_k$ which are not all zero such that

$$\alpha_0 | a \rangle + \alpha_1 \mathbf{A}| a \rangle + ... + \alpha_N \mathbf{A}^N| a \rangle=\mathbf{0}$$

Now, you have two situations. Either $\alpha_N\neq 0$, and in that case, you can write something like you wrote in the quote. In the case that $\alpha_n=0$, you need to do something differently. In this case, they key fact is that at least one of the $\alpha_k$ is nonzero.

That sums up my problem quite well. If indeed $\alpha_N=0$, then $\mathbf{A}^N | a \rangle$ is linearly independent of the other vectors, so what does that say about $\mathbf{A}^{N+1} | a \rangle$?

You can handle both cases simultaniously by remarking that there exists an $M\leq N$ and $\alpha_k$ scalars such that

$$\mathbf{A}^M | a \rangle = \sum_{k=0}^{M-1} \alpha_k \mathbf{A}^k| a \rangle$$

Again, I don't see how to treat the case where $M < N$.

 

[micromass]

Again, I don't see how to treat the case where $M < N$.

OK. Can you tell me how you would dela with the case $M=N$? You don't have to write it out completely, a summary would suffice.

 

[DrClaude]

Assuming that
$$
\mathbf{A}^N |a \rangle = \sum_{k=0}^{N-1} \alpha_{k} \mathbf{A}^k |a \rangle
$$
we have
$$
\begin{align}
\mathbf{A}^{N+1} |a \rangle &= \mathbf{A} \sum_{k=0}^{N-1} \alpha_{k} \mathbf{A}^k |a \rangle \\
&= \sum_{k=0}^{N-1} \alpha_{k} \mathbf{A}^{k+1} |a \rangle \\
&= \sum_{k=1}^{N} \alpha_{k-1} \mathbf{A}^{k} |a \rangle
\end{align}
$$
which, by definition of $\mathcal{M}$ is in $\mathcal{M}$, meaning that $\mathbf{A} \left( \mathbf{A}^{N} |a \rangle \right) \in \mathcal{M}$, completing the "proof" that $\mathbf{A} | x \rangle \in \mathcal{M}$ for $| x \rangle \in \mathcal{M}$.

 

[micromass]

Alright. So the basic idea was to multiply by $\mathbf{A}$.

This doesn't work in the case $M<N$. But maybe multiplying with something else works? Like $\mathbf{A}^2$ or $\mathbf{A}^3$?

 

[DrClaude]

Aside: Since I have joined PF, I've mostly posted answers to questions posed by others. It feels very strange to be on the other side. It's been a long time since I've been a student...

Ok, I think I see the light. From the linear dependence of the vectors spanning $\mathcal{M}$, we have that for some $M \leq N$
$$
\mathbf{A}^M | a \rangle = \sum_{k=0}^{M-1} \alpha_k \mathbf{A}^k | a \rangle
$$
with at least one $\alpha_k \neq 0$. Setting $p = N-M+1$, we find
$$
\begin{align}
\mathbf{A}^p \mathbf{A}^M | a \rangle &= \mathbf{A}^p \sum_{k=0}^{M-1} \alpha_k \mathbf{A}^k | a \rangle \\
&= \sum_{k=0}^{M-1} \alpha_k \mathbf{A}^{p+k} | a \rangle \\
\mathbf{A}^{N+1} | a \rangle &= \sum_{k=p}^{N} \alpha_{k-p} \mathbf{A}^{k} | a \rangle
\end{align}
$$
from which we conclude that $\mathbf{A}^{N+1} | a \rangle \in \mathcal{M}$.

Hope I got it right. Thanks micromass for the help.

 

[micromass]

That's it! Congratz!