# How can I find the nth exponent of the matrix using the diagonalization algorithm?

Hey guys, simple question in linear algebra.

I want to find the nth exponent of this matrix:

$$\left[ \begin{array}{cc} 1 & 1 \\ 0 & 2 \end{array} \right]$$

I'm trying to use the diagonal algorithm, first by finding the eigenvalues, so I get $t_1=1$, and $t_2=2$, and than I realize that the matrix is not diagonalizable... So how can I still calculate the nth exponent?

Thanks

• Compute the second and third power. Can you see a pattern? – Rasmus Mar 17 '11 at 22:58
• I need to use eigenvalues, is something wrong with what I wrotr? Is there any mistake in the consept? – user6163 Mar 17 '11 at 23:02
• The problem is that the matrix is diagonalizable; see my answer. – InterestedGuest Mar 17 '11 at 23:59

Nir, it is impossible to have less linearly independent eigenvectors than distinct eigenvalues; each distinct eigenvalue corresponds to the span of at least one eigenvector. Eigenvectors for your values of 1 and 2 are $\begin{bmatrix} 1\\ 0 \end{bmatrix}$ and $\begin{bmatrix} 1\\ 1 \end{bmatrix}$ respectively. By the definition of diagonalization, if you have an eigenbasis for $\mathbb{R^{2}}$, your matrix is diagonalizable. Since you have a basis consisting of two eigenvectors (and we know that for any vector space of dimension n, n linearly independent vectors form a basis, in addition to knowing that any eigenvectors that correspond to different eigenvalues are linearly independent), you can diagonalize your matrix. From here on, you can decompose your matrix into the product of an eigenbasis matrix, the eigenvalue matrix, and the inverse of the eigenbasis matrix, and compute the powers.

I am assuming this is not homework, since you did not tag it as such. Here is the full plan of action:

Now that you know the corresponding eigenvectors for the eigenvalues, you can represent your matrix $A$ as $\left[ \begin{array}{cc} 1 & 1 \\ 0 & 2 \end{array} \right]$ $=$ $\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]$ $\left[ \begin{array}{cc} 1 & 0 \\ 0 & 2 \end{array} \right]$ $\left[ \begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array} \right]$

Now that you have $A=SDS^{-1}$ where $S$ is the matrix with your eigenbasis as columns, and $D$ is the diagonal matrix consisting of the eigenvalues, you can find an $A^{k}$ by representing $A^{k}$ in the following way: $A^{k}=(SDS^{-1})^{k}=\underset{k}{\underbrace{(SDS^{-1})(SDS^{-1})...(SDS^{-1})}}$. As you can see, all $S$ and $S^{-1}$ besides the ones on the very left and very right will cancel, leaving you with $A^{k}=SD^{k}S^{-1}$. Now, you are left with an equation $\left[ \begin{array}{cc} 1 & 1 \\ 0 & 2 \end{array} \right]^{k}$ $=$ $\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]$ $\left[ \begin{array}{cc} 1 & 0 \\ 0 & 2 \end{array} \right]^{k}$ $\left[ \begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array} \right]$ and you can simply exponentiate the diagonal values and multiply the matrices out to get the result.

You mentioned that you realized that the matrix is not diagonalizable, so we can try finding eigenvectors by hand, to make sure that the process is clear.

For the eigenvalue 1, you need to find when $(A-1I)\vec{x}$ is equal to zero. So, you compute the kernel of $(A-I)$: $\left[ \begin{array}{cc} 0 & 1 \\ 0 & 1 \end{array} \right]$. Row-reducing, you get $\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]$, so the kernel is represented by the span of $\begin{bmatrix} 1\\ 0 \end{bmatrix}$, meaning that a basis for $\xi_{1}$ is $\begin{bmatrix} 1\\ 0 \end{bmatrix}$.

Same process for the eigenvalue 2: Find the kernel of $(A-2I)$, meaning the kernel of $\left[ \begin{array}{cc} -1 & 1 \\ 0 & 0 \end{array} \right]$. As you can see, the kernel consists of the span of $\begin{bmatrix} 1\\ 1 \end{bmatrix}$, meaning that a basis for $\xi_{2}$ is $\begin{bmatrix} 1\\ 1 \end{bmatrix}$.

Generally, an $n\times n$ matrix is always diagonalizable if it has $n$ distinct eigenvalues, which your matrix does.

This matrix is diagonalizable. As you said the eigenvalues are $\lambda = 1$ and $\lambda = 2$ with corresponding eigenvectors $\begin{bmatrix} 1 & 0\end{bmatrix}^T$ and $\begin{bmatrix} 1 & 1\end{bmatrix}^T$ for example. Hence $$\begin{bmatrix} 1 & 0\\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}.$$ Since $\begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & -1\\ 0 & 1 \end{bmatrix}$ we obtain $$\begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix}^n = \begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & 2^n \end{bmatrix} \begin{bmatrix} 1 & -1\\ 0 & 1 \end{bmatrix}.$$

In general, you can use the Cayley–Hamilton theorem to reduce all powers of a matrix to a low degree polynomial in the matrix.