How to finds the smallest $n$ such that $I=A^0, A^1, A^2, \dots, A^n$ are linearly dependent?

Question

Let $A$ be the following $2\times 2$ matrix:

$$A=\begin{pmatrix}1&2\\3&4\end{pmatrix}$$

Find the smallest value of $n$ such that matrices $I=A^0, A^1, A^2, \dots, A^n$ are linearly dependent.

I don't quite know how to begin to answer this question since so far we have only defined linear independance for vectors, and these are matrices...

Answer 1

And matrices are vectors, since they are elements of the vector space of all $2\times2$ matrices. And you have$$A_0=\begin{pmatrix}1&0\\0&1\end{pmatrix},A_1=\begin{pmatrix}1 & 2 \\ 3 & 4\end{pmatrix},A_2=\begin{pmatrix}7 & 10 \\ 15 & 22\end{pmatrix}\text{, and }A_3=\begin{pmatrix}37 & 54 \\ 81 & 118\end{pmatrix}.$$Determine their coefficients in the standard basis of $\mathbb{R}^{2\times2}$ and use that information to determine whether or not they are linearly independent.

Answer 2

Linear independence for matrices is defined just like it is defined for vectors. You call matrices $$A_1, A_2,...,A_n$$ linearly independent iff $$C_1A_1 + C_2A_2+...+C_nA_n=0 \implies C_1=C_2=...C_n=0$$

For your matrix of $$A= \begin{bmatrix} 1&2\\3&4\end{bmatrix} $$ you may verify that$ A^2$ is a linear combination of $A$ and $I$.

$$ A^2 = 5A+2I $$

Thus $n=2$ is the smallest number for which $I, A,A^2$ are linearly dependent.

Answer 3

By the Cayley-Hamilton theorem, any matrix is a root of its own characteristic polynomial. The characteristic polynomial of $A$ is $$ p(\lambda)=(1-\lambda)(4-\lambda)-6=\lambda^2-5\lambda-2 $$ This means that $$ 0=A^2-5A-2I $$ and it's quite easy to show that $A$ and $I$ are linearly independent, so the answer is $n=2$.

Answer 4

Instead of straigthforward computation, if you know a bit more linear algebra, one can do the following: What you are actually searching is the degree of the minimal polynomial of $A$. That is, the unitary polynomial $p(x) = x^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ of minimal degree such that $p(A)=A^n+a_{n-1}A^{n-1}+...+a_1A+a_0=0$. Also, it is quite useful to now the not-that-easy-to-prove Cayley theorem, which states that the characterstic polynomial $\chi$ of a matrix is a multiple of the minimal polynomial, i.e. $\chi(A)=0$.

In your example, we thus know that there is a polynomial of degree $2$, the characteristic polynomial $\chi(x)=\det(xI-A)$ which satisfies $\chi(A)=0$, and as $A$ isn't of the form $aI$, we conclude that the answer is $n=2$.

Answer 5

Calculate the powers of $A$ upto $A^2$ and write them in a row:

$$\underbrace{\begin{pmatrix} \fbox{1} & 0 \\ 0 & 1 \end{pmatrix}}_{I}, \underbrace{\begin{pmatrix} \fbox{1} & 2 \\ 3 & 4 \end{pmatrix}}_{A},\underbrace{\begin{pmatrix} \fbox{7} & 10 \\ 15 & 22 \end{pmatrix}}_{A^2}$$

Use the coefficient at the position $(1,1)$ in the matrix $I$ to annihilate the coefficients at $(1,1)$ in $A$ and $A^2$:

$$\underbrace{\begin{pmatrix} 0 & 2 \\ 3 & 3 \end{pmatrix}}_{A-I},\underbrace{\begin{pmatrix} 0 & 10 \\ 15 & 15 \end{pmatrix}}_{A^2-7I}$$

Now notice that $A^2-7I = 5(A-I)$. Therefore:

$$0 = A^2-7I - 5(A-I) = A^2-5A-2I$$

Since $A$ is obviously not a multiple of $I$, we conclude that the smallest $n$ such that $\{I, A, \ldots, A^n\}$ is lineraly dependent is $n = 2$.