How to diagonalize a matrix $A$ and then compute $A^{10}$?

Question

Diagonalize the following matrix $$A = \begin{bmatrix} -2 & -8 & -12\\ 1 & 4 & 4\\ 0 & 0 & 1\end{bmatrix}$$ and compute $A^{10}$.

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I have found that the eigenvalues of $A$ are $\lambda _1 = 0$, $\lambda _2 = 1$, $\lambda _1 = 2$. Then, I found the following eigenvectors who form my matrix $P$.

$$v_1=\left (\begin{matrix}1\\ 1\\ 1\end{matrix} \right ), \qquad v_2=\left (\begin{matrix}-1/4\\ 0\\ -1/2\end{matrix} \right ), \qquad v_3=\left (\begin{matrix}0\\ -1/4\\ 0\end{matrix} \right )$$

At the end my matrix

$$P=\left ( \begin{matrix} 1 & 1 &1 \\ -\frac{1}{4}& 0 &-\frac{1}{2} \\ 0 &-\frac{1}{4} & 0 \end{matrix} \right )$$

In linear algebra, two $n$-by-$n$ matrices $A$ and $B$ are called similar if $B=P^{-1}AP$. I get that

$$B=\left ( \begin{matrix} 0 & 0 & 0\\ 0 &1 &0 \\ 0& 0 & 2 \end{matrix} \right )$$

So is this all good? How can I use that to get value of $A^{10}$?

Answer 1

Notice the following: $B = P^{-1}AP \iff A = PBP^{-1}$.

Now consider $A^2$. We see that $A^2 = (PBP^{-1})(PBP^{-1})$. But since $PP^{-1} = I$, this simplifies to $A^2 = PB^2P^{-1}$. Inductively it should be clear that $A^n = PB^nP^{-1}$.

Answer 2

Apply Cayley Hamilton's theorem. You have that the characteristic polynomial is :

$ P(λ) = λ(λ-1)(λ-2) $

Then, $ λ^{10} = P(λ)π(λ) + υ(λ)$ which is the euclidean diaereses.

$υ(λ)$ will be a second-degree polynomial, written as : $ α_2λ^2 + α_1λ + α_ο$ .

Putting the eigenvalues into $λ^{10} = P(λ)π(λ) + υ(λ)$, will make $P(eigenvalue)π(eigenvalue) = 0$ according to Cayley Hamilton and from then on you will be able to calculate the $a_1, a_2, a_3$ by solving a system of 3 equations.

Finally : $ A^{10} = P(A)π(Α) + υ(Α) \Leftrightarrow A^{10} = v(A) = α_2A^2 + α_1A + α_οI$

And this gives you $A^{10}$.

Answer 3

First, I'd like to point out that all three of the eigenvalues can be guessed:

• $\lambda_1 = 0$, because the matrix is singular: the first two columns are clearly linearly dependent.
• $\lambda_2 = 1$, because the $1$ on the bottom right is on the diagonal and is alone in its row.
• $\lambda_3 = 2$, because the trace is $3$ and the first two eigenvalues add up to $2$.

For these and other similar insights, see https://www.lem.ma/3j and the subsequent lessons.

The first eigenvector is obvious, but the other two need Gaussian elimination to determine: $$ v_1=\begin{bmatrix}4 \\ -1 \\ 0 \end{bmatrix}\ \ \ \ \ v_2=\begin{bmatrix}-4 \\ 0 \\ 1 \end{bmatrix}\ \ \ \ \ v_3=\begin{bmatrix}-2 \\ 1 \\ 0 \end{bmatrix} $$

Thus, by the eigenvalue decomposition $$ \begin{align} A &= \begin{bmatrix}4 & -4 & -2 \\ -1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix}4 & -4 & -2 \\ -1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}^{-1}\\ &= \begin{bmatrix}4 & -4 & -2 \\ -1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix}-\frac{1}{2} & -1 & -2 \\ 0 & 0 & 1 \\ \frac{1}{2} & 2 & 2 \end{bmatrix} \end{align} $$

Therefore, by the argument mentioned by Captain, we have $$ \begin{align} A^{10}&=\begin{bmatrix}4 & -4 & -2 \\ -1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}^{10} \begin{bmatrix}-\frac{1}{2} & -1 & -2 \\ 0 & 0 & 1 \\ \frac{1}{2} & 2 & 2 \end{bmatrix}\\ &= \begin{bmatrix}4 & -4 & -2 \\ -1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1024 \end{bmatrix} \begin{bmatrix}-\frac{1}{2} & -1 & -2 \\ 0 & 0 & 1 \\ \frac{1}{2} & 2 & 2 \end{bmatrix} \end{align} $$ The result is: $$ A^{10}=\begin{bmatrix}-1024&-4096&-4100\\ -512 & 2048 & 2048 \\ 0 & 0 & 1 \end{bmatrix} $$ For a thorough discussion of this application of the Eigenvalue Decomposition, see https://www.lem.ma/hr and the subsequent lessons.

Finally, here are two particularly nice examples that combine all of these ideas together and can be answered by doing all the calculations in your head:

$$ \begin{bmatrix}110&55&-164\\ 42& 21 & -62 \\ 88 & 44 & -131 \end{bmatrix}^{2017} $$ (Solution: https://www.lem.ma/-z) $$ \begin{bmatrix}3&-6&4\\ 1& -2 & 2 \\ 0 & 0 & 1\end{bmatrix}^{2016} $$

Answer 4

We have $x^{10} \equiv 511 x^2 - 510 x \bmod x(x-1)(x-2)$ and so $A^{10} = 511 A^2 - 510 A$ by the Cayley–Hamilton theorem.