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Consider the matrix $A$ as follows\begin{pmatrix} -3&2 \\ -1&0 \end{pmatrix}

Find the matrix $A^5$ ?

Instead of doing iterative multiplications, any small approach for this ?

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    Hint : Diagonalize the matrix! – 2017-01-25
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    @Peter I am new to diagonalization. Can you explain the approach please ? – 2017-01-25
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    Look here: https://matrixcalc.org/en/#%7B%7B-3,2%7D,%7B-1,0%7D%7D%2A%7B%7B-3,2%7D,%7B-1,0%7D%7D%2A%7B%7B-3,2%7D,%7B-1,0%7D%7D%2A%7B%7B-3,2%7D,%7B-1,0%7D%7D%2A%7B%7B-3,2%7D,%7B-1,0%7D%7D – 2017-01-25
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    First you calculate the eigenvalues (Here $-2$ and $-1$). If they are distinct, you can always diagonalise the matrix. If the multiplicity of an eigenvalue is $k$, you need at least $k$ linear independent eigenvectors, otherwise the matrix cannot be diagonalized. Here you continue with the eigenvectors and form a matrix $T$ with them (Here, you get $T=\pmatrix{2&1\\1&1}$. Then, you have $T^{-1}AT=\pmatrix{-2&0\\0&-1}$, you have diagonalized $A$ – 2017-01-25
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    It is easy to calculate the power of a diagonalmatrix. You just have to power the diagonal elements. Here, we have $D^5=\pmatrix{-32&0\\0&-1}$. Now, we can calculate $A^5=TD^5T^{-1}$. In this case, this does not give a result very fast compared with the repeated matrix multiplication, but for large matrices or a very large $n$, this method is useful. – 2017-01-25
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    If you want to calculate a square root of $A$ (A matrix $B$ with $B^2=A$), there is no direct way. Here you actually need to diagonalize $A$ – 2017-01-25

2 Answers 2

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use the cayley-hamilton theorem. the char poly is $\lambda^2 + 3 \lambda + 2 = 0$ so the matrix $A$ satisfies $$A^2 + 3A + 2I = 0 \to A^2 = -3A - 2I, A^3 = -3A^2 -2A=9A+6I-2A=7A+6I, A^4 = \ldots, A^5 = \ldots $$

you can cut down the work if you use the division algorithm $\lambda^5 = q(\lambda)(\lambda^2 + 3 \lambda + 2) + a\lambda + b$ then $A^5 = aA + bI.$

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Rather than doing 4 iterated multiplications ($A^5 = A \cdot A \cdot A \cdot A \cdot A$), a bit quicker would be to do $A\cdot A = A^2$, then $A^2 \cdot A^2 = A^4$ and finally $A \cdot A^4 = A^5$ which reduces the number of matrix multiplications to perform from 4 to 3.

This approach generalizes for higher powers, take a look at exponentation by squaring. You will only need to carry out $\lfloor \log_2n\rfloor$ multiplications to find $A^n$; in your case $\lfloor \log_2 5\rfloor = 3$.

In general you would try to diagonalize a matrix to easily calculate higher powers. This works in your case, as the matrix is diagonalizable. However, for a $2 \times 2$-matrix and exponent $5$, you probably won't gain much time/efficiency compared to direct computation via iterated multiplication. This will be more interesting for higher powers and/or larger matrices.

The use of diagonalization to compute powers of a matrix in a nutshell: if a matrix $A$ can be written as $PDP^{-1}$ with $D$ a diagonal matrix, then you have: $$A^n = \left( PDP^{-1} \right)^n = \ldots = PD^nP^{-1}$$ where $D^n$ is easy (diagonal elements to the power $n$).