# Linearly independence of various sets of $2\times2$ matrices

The following questions are not T/F questions.

I'm trying to understand this complex subsets independency or dependency.

Let $A$ be an $2\times 2$ matrix over the real numbers.

1. The subset $\{A^2, A^5, A^{11}\}$ is always linearly dependent.
2. The subset $\{I, A, A^2\}$ is always linearly dependent.
3. It is possible that the subset $\{A^2, A^5, A^{11}\}$ is linearly independent.
4. It is possible that the subset $\{I, A^2, A^5, A^{11}\}$ is linearly independent.
5. It is possible that the subset $\{I, A, A^2\}$ is linearly independent.

$I$ refers to the identity matrix.

Examples would be appricated.

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I suppose you meant "linearly (in)dependent in the vector space of all $\,2\times 2\,$ real matrices", right? What have you done? Do you know something about a matrix characteristic polynomial? –  DonAntonio Oct 11 '12 at 10:46
What do you think is the dimension of the space of $2 \times 2$ matrices? –  wj32 Oct 11 '12 at 10:52
@wj32 That is not really apposite, because the Cayley-Hamilton theorem says that powers of an $n\times n$ matrix $A$ generate an $n$-dimensional subspace of the space of all $n\times n$ matrices. –  MJD Oct 11 '12 at 11:56
@MJD: Oops, my bad! –  wj32 Oct 11 '12 at 12:00

Start with the definition of linearly dependent. A set of elements $\{v_1,\ldots,v_n\}$ of a vector space $V$ over field $K$ is linearly dependent if there are some $\lambda_1,\ldots,\lambda_n \in K$ such that $\lambda_1v_1 + \ldots + \lambda_nv_n = 0$. In your case, the vector space is the space $V$ of all linear mappings $\mathbb{R}^2 \rightarrow \mathbb{R}^2$, and the field is $\mathbb{R}$. Note that this space is isomorphic to $\mathbb{R}^4$. It follows that every set with more than 4 elements must be linearly dependent, because $\text{dim}(V) = 4$.
Now, every matrix $A \in \mathbb{R}^{nxn}$ has an associated characteristic polynomial $p_A \in \mathbb{R}[x]$ with $\text{deg }p_A = n$, i.e. $p_A(x) = a_nx^n + \ldots + a_1x + a_0$. And, very importantly, you always have $p_A(A) = 0$! (One says that the characteristic polynomial annihilates the matrix). Thus, for every matrix $A \in \mathbb{R}^{2x2}$ you can find $\lambda_0,\ldots,\lambda_2$ such that $\lambda_0I + \lambda_1A + \lambda_2A^2 = 0$. This answers question (2) and (5).
For the other questions, observe that you can also interpret the above as a way to write $A^2$ as a linear combination of $I$ and $A$, more precisely as $A^2 = \frac{1}{\lambda_2}(\lambda_0 I + \lambda_1 A)$. You can extend that to generate any higher power too - for $A^3$ you get $$\begin{eqnarray} A^3&=&AA^2=A\frac{1}{\lambda_2}(\lambda_0 I + \lambda_1 A) =\frac{1}{\lambda_2}(\lambda_0 A + \lambda_1 A^2) =\frac{1}{\lambda_2}\left(\lambda_0 A + \lambda_1 \frac{1}{\lambda_2}(\lambda_0 I + \lambda_1 A)\right) \\ &=&\left(\frac{\lambda_0}{\lambda_2}+\frac{\lambda_1^2}{\lambda_2^2}\right)A + \frac{\lambda_0\lambda_1}{\lambda_2^2}I \end{eqnarray}$$ The exact coefficients are not important - the important fact is that for every n you can find coefficients $\mu_1,\mu_2 \in \mathbb{R}$ such that $A^n = \mu_1I + \mu_2A$. Thus, the subspace spanned by a set of powers of $A$ has at most dimension 2. It follows that every set with 3 or more members is linearly dependent.
I think towards the end, you are working too hard. All you need to know is that every power of $A$ can be written as $cA+d$ for some numbers $c$ and $d$. –  Gerry Myerson Oct 11 '12 at 12:08
@GerryMyerson Isn't the prett much the same thing? Note that $I$ and $A^1$ isn't always in the set, so even with your approach the answer isn't totally obvious... –  fgp Oct 11 '12 at 12:17
I don't care whether $I$ and/or $A$ is/are in the set; everything in the set can be written as $cA+d$, which is a 2-dimensional vector space, so, if there are three or more elements in the set, end of story. I certainly don't have to go looking for a polynomial $q$ so that some product is some special polynomial of degree 11. –  Gerry Myerson Oct 11 '12 at 12:31