It's taking too much time in calculating $A^{30}$ for a $3\times{3}$ non singular matrix. Is there any standard procedure to find such solution?
How to find $A^n $ for a $ 3\times{3 }$ matrix $A$ . $n$ is large say greater than $30$.
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matrices
matrix-equations
matrix-calculus
matrix-decomposition
matrix-rank
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2For which matrix A? Non-singularity is not enough to find $A^{30}$. – 2017-02-01
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1You can reduce the number of multiplications by storing the results and using recursion. $A^{30}=A^{15}*A^{15}$ etc. – 2017-02-01
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0Besides the "binary exponentiation" proposed by @FFF, the other short-cut available is to use a similarity transformation on $A$ to get a diagonal or other [normal form](https://en.wikipedia.org/wiki/Category:Matrix_normal_forms) whose powers are available almost by inspection. – 2017-02-01
3 Answers
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Hint. Diagonalize the matrix or if not possible, try to bring it into Jordan normal form. It depends on the field you are working with. Or else use some symbolic calculator like MAPLE.
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Yes, one can check whether the matrix $A$ is diagonalizable, i.e. $A$ can be written as $P^{-1}DP$ where $P$ is invertible and $D$ is a diagonal matrix, say $\begin{pmatrix} \lambda_1&0&0\\0&\lambda_2&0\\0&0&\lambda_3 \end{pmatrix}$. In that case $A^n=P^{-1}D^nP$ and it's easy to calculate $D^n$.
If this fails there are two general ways to proceed. Either you calculate the Jordan canonical form of $A$ and use that to get higher powers. Or you can use a smart application of the Cayley-Hamilton theorem.
If you do not these things I suggest you first study a basic linear algebra course. If you know some linear algebra but the matrix in question is not diagonalizable, I'm more than willing to demonstrate both methods above.
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Let $P(t)=\det(A-tI)$ the characteristic polynomial of $A$. Then one knows (Cayley) that the matrix $A$ satisfies the relation $P(A)=0$. For a $3\times3$ matrix $A$ this means that there will be scalars $a_0$, $a_1$, $a_2$ such that $$ A^3=a_0+a_1A+a_2A^2. $$ By multiplying both hands of the displayed formula by $A^n$ we get a recursive formula for $A^{n+3}$ in terms of powers of lesser degree. Applying repeatedly this formula we can express any $A^k$ with $k\geq3$ in the form $$ A^k=b_0+b_1A+b_2A^2 $$ for some scalars $b_0$, $b_1$, $b_2$ (depending on $k$).