Diagonalisation & Powers of Matrices (DP IB Applications & Interpretation (AI): HL): Revision Note

Diagonalisation

What is a diagonal matrix

• A non-zero square matrix is considered to be diagonal if all elements are zero except the elements along its leading diagonal

  • e.g. $\left(\begin{array}{ccc}1& 0& 0\\ 0& 3& 0\\ 0& 0& -1\end{array}\right)$ is diagonal but $\left(\begin{array}{ccc}1& 2& 0\\ 0& 3& 0\\ 0& 0& -1\end{array}\right)$ is not

What is a diagonalisable matrix?

• Matrix $\mathit{M}$ is diagonalisable if there exists a matrix $\mathit{P}$ such that

  • $\mathit{D}={\mathit{P}}^{-1}\mathit{M}\mathit{P}$ is diagonal

• This can also be written as $\mathit{M}=\mathit{P}\mathit{D}{\mathit{P}}^{-1}$

How can I diagonalise a matrix?

• Consider the matrix $\mathit{M}$ which has

  • real distinct eigenvalues ${\lambda }_1$ and ${\lambda }_2$ 

  • with corresponding eigenvectors ${\mathit{x}}_1=\left(\begin{array}{c}{x}_1\\ y_1\end{array}\right)$ and ${\mathit{x}}_2=\left(\begin{array}{c}{x}_2\\ y_2\end{array}\right)$

• You can diagonalise matrix $\mathit{M}$ using

  • $\mathit{P}=\left(\begin{array}{cc}{x}_1& x_2\\ y_1& y_2\end{array}\right)$

  • $\mathit{D}=\left(\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right)$

• For example, consider $\left(\begin{array}{cc}4& 3\\ -1& 0\end{array}\right)$

  • the eigenvalues are 1 and 3

    • $\mathit{D}=\left(\begin{array}{cc}1& 0\\ 0& 3\end{array}\right)$

  • corresponding eigenvectors are $\left(\begin{array}{c}1\\ -1\end{array}\right)$ and $\left(\begin{array}{c}-3\\ 1\end{array}\right)$

    • $\mathit{P}=\left(\begin{array}{cc}1& -3\\ -1& 1\end{array}\right)$

    • ${\mathit{P}}^{-1}=-\frac{1}{2}\left(\begin{array}{cc}1& 3\\ 1& 1\end{array}\right)$

  • $\left(\begin{array}{cc}1& 0\\ 0& 3\end{array}\right)=-\frac{1}{2}\left(\begin{array}{cc}1& 3\\ 1& 1\end{array}\right)\left(\begin{array}{cc}4& 3\\ -1& 0\end{array}\right)\left(\begin{array}{cc}1& -3\\ -1& 1\end{array}\right)$

Matrix powers

How can I find powers of a diagonalisable matrix?

• Write the matrix in diagonalised form

  • $\mathit{M}=\mathit{P}\mathit{D}{\mathit{P}}^{-1}$

• Squaring this gives:

  • ${\mathit{M}}^2=\mathit{P}\mathit{D}{\mathit{P}}^{-1}\mathit{P}\mathit{D}{\mathit{P}}^{-1}$

  • which simplifies to ${\mathit{M}}^2=\mathit{P}{\mathit{D}}^2{\mathit{P}}^{-1}$

• Powers can be found as the product of three matrices

  • ${\mathit{M}}^n=\mathit{P}{\mathit{D}}^n{\mathit{P}}^{-1}$

• Finding higher powers of a diagonal matrix is straight forward

  • ${\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)}^n=\left(\begin{array}{cc}{a}^n& 0\\ 0& b^n\end{array}\right)$

• For example, $\left(\begin{array}{cc}1& 3\\ 1& 1\end{array}\right)=-\frac{1}{2}\left(\begin{array}{cc}1& -3\\ -1& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 3\end{array}\right)\left(\begin{array}{cc}4& 3\\ -1& 0\end{array}\right)$

  • ${\left(\begin{array}{cc}4& 3\\ -1& 0\end{array}\right)}^4=-\frac{1}{2}\left(\begin{array}{cc}1& -3\\ -1& 1\end{array}\right){\left(\begin{array}{cc}1& 0\\ 0& 3\end{array}\right)}^4\left(\begin{array}{cc}1& 3\\ 1& 1\end{array}\right)$

  • ${\left(\begin{array}{cc}4& 3\\ -1& 0\end{array}\right)}^4=-\frac{1}{2}\left(\begin{array}{cc}1& -3\\ -1& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 81\end{array}\right)\left(\begin{array}{cc}1& 3\\ 1& 1\end{array}\right)$

  • ${\left(\begin{array}{cc}4& 3\\ -1& 0\end{array}\right)}^4=\left(\begin{array}{cc}121& 120\\ -40& -39\end{array}\right)$