Let
\[A=\begin{bmatrix}
-4 & -6 & -12 \\
-2 &-1 &-4 \\
2 & 3 & 6
\end{bmatrix}, \quad \mathbf{u}=\begin{bmatrix}
6 \\
5 \\
-3
\end{bmatrix}, \quad \mathbf{v}=\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}, \quad \text{ and } \mathbf{w}=\begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}.\]

(a) Express the vector $\mathbf{u}$ as a linear combination of $\mathbf{v}$ and $\mathbf{w}$.

(b) Compute $A^5\mathbf{v}$.

(c) Compute $A^5\mathbf{w}$.

(d) Compute $A^5\mathbf{u}$.

Solution.

(a) Express the vector $\mathbf{u}$ as a linear combination of $\mathbf{v}$ and $\mathbf{w}$.

Our goal here is to find scalars $c_1, c_2$ such that
\[\mathbf{u}=c_1\mathbf{v}+c_2\mathbf{w}.\] This is the same as the matrix equation
\[\begin{bmatrix}
-2 & -2 \\
0 & -1 \\
1 &1
\end{bmatrix}
\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}=\begin{bmatrix}
6 \\
5 \\
-3
\end{bmatrix}.\] To solve this, we reduced the augmented matrix as follows:
\begin{align*}
\left[\begin{array}{rr|r}
-2 & -2 & 6 \\
0 &-1 &5 \\
1 & 1 & -3
\end{array}\right] \xrightarrow{R_1 \leftrightarrow R_3}
\left[\begin{array}{rr|r}
1 & 1 & -3 \\
0 &-1 &5 \\
-2 & -2 & 6
\end{array}\right]\\[6pt] \xrightarrow[-R_2]{R_3+2R_1}
\left[\begin{array}{rr|r}
1 & 1 & -3 \\
0 &1 &-5 \\
0 & 0 & 0
\end{array}\right] \xrightarrow{R_1-R_2}
\left[\begin{array}{rr|r}
1 & 0 & 2 \\
0 &1 &-5 \\
0 & 0 & 0
\end{array}\right].
\end{align*}
This yields the solution $c_1=2$ and $c_2=-5$.
Hence, we have the linear combination
\[\mathbf{u}=2\mathbf{v}-5\mathbf{w}.\]

(b) Compute $A^5\mathbf{v}$.

We first compute $A\mathbf{v}$. We have
\[A\mathbf{v}= \begin{bmatrix}
-4 & -6 & -12 \\
-2 &-1 &-4 \\
2 & 3 & 6
\end{bmatrix}
\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}
=\begin{bmatrix}
-4 \\
0 \\
2
\end{bmatrix}=2\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}=2\mathbf{v}.
\] Using this relation $A\mathbf{v}=2\mathbf{v}$, we obtain
\[A^2\mathbf{v}=AA\mathbf{v}=A(2\mathbf{v})=2A\mathbf{v}=2(2\mathbf{v})=2^2\mathbf{v}.\] Next, we have
\[A^3\mathbf{v}=AA^2\mathbf{v}=A(2^2\mathbf{v})=2^2A\mathbf{v}=2^2(a\mathbf{v})=2^3\mathbf{v}.\] Repeating this process, we see that $A^5\mathbf{v}=2^5\mathbf{v}$.
Or, we can find this by computing as follows:
\begin{align*}
A^5\mathbf{v}=A^2A^3\mathbf{v}=A^2(2^3\mathbf{v})=2^3A^2\mathbf{v}=2^3(2^2\mathbf{v})=2^5\mathbf{v}.
\end{align*}
In summary, we have
\[A^5\mathbf{v}=2^5\mathbf{v}=32\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}=\begin{bmatrix}
-64 \\
0 \\
32
\end{bmatrix}.\]

(c) Compute $A^5\mathbf{w}$.

First, we note that
\[A\mathbf{w}= \begin{bmatrix}
-4 & -6 & -12 \\
-2 &-1 &-4 \\
2 & 3 & 6
\end{bmatrix} \begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}=\begin{bmatrix}
2 \\
1 \\
-1
\end{bmatrix}=-\begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}=-\mathbf{w}.\] Using this relation $A\mathbf{w}=-\mathbf{w}$ as in part (a), we obtain
\[A^5\mathbf{w}=(-1)^5\mathbf{w}=-\begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}=\begin{bmatrix}
2 \\
1 \\
-1
\end{bmatrix}.\]

(d) Compute $A^5\mathbf{u}$.

Using the linear combination $\mathbf{u}=2\mathbf{v}-5\mathbf{w}$ obtained part (a), we compute
\begin{align*}
A^5\mathbf{w}&=A^5(2\mathbf{v}-5\mathbf{w})\\
&=2A^5\mathbf{v}-5A^5\mathbf{w}\\[6pt] &=2\begin{bmatrix}
-64 \\
0 \\
32
\end{bmatrix}-5\begin{bmatrix}
2 \\
1 \\
-1
\end{bmatrix} &&\text{by (b), (c)}\\[6pt] &=\begin{bmatrix}
-138 \\
-5 \\
69
\end{bmatrix}.
\end{align*}

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

One common pitfall is to compute $A^5$, but this is time consuming and it is very likely to make a mistake by hand computaiton.

The post Compute $A^5\mathbf{u}$ Using Linear Combination first appeared on Problems in Mathematics.

Consider the $2\times 2$ complex matrix
\[A=\begin{bmatrix}
a & b-a\\
0& b
\end{bmatrix}.\]

(a) Find the eigenvalues of $A$.

(b) For each eigenvalue of $A$, determine the eigenvectors.

(c) Diagonalize the matrix $A$.

(d) Using the result of the diagonalization, compute and simplify $A^k$ for each positive integer $k$.

Solution.

(a) Find the eigenvalues of $A$.

Since $A$ is an upper triangular matrix, eigenvalues are diagonal entries.
Hence $a, b$ are eigenvalues of $A$.

(b) For each eigenvalue of $A$, determine the eigenvectors.

Suppose now that $a\neq b$.
Let us find eigenvectors corresponding to the eigenvalue $a$.
We have
\begin{align*}
A-aI=\begin{bmatrix}
0 & b-a\\
0& b-a
\end{bmatrix}
\xrightarrow{R_2-R_1}
\begin{bmatrix}
0 & b-a\\
0& 0
\end{bmatrix}
\xrightarrow{\frac{1}{b-a}R_1}
\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}.
\end{align*}
It follows that the eigenvectors corresponding to $a$ are
\[x_1\begin{bmatrix}
1 \\
0
\end{bmatrix},\] where $x_1$ is any nonzero complex number.

Next, we find the eigenvectors corresponding to the eigenvalue $b$.
We have
\begin{align*}
A-bI=\begin{bmatrix}
a-b & b-a\\
0& 0
\end{bmatrix}
\xrightarrow{\frac{1}{a-b}R_1}
\begin{bmatrix}
1 & -1\\
0& 0
\end{bmatrix}.
\end{align*}
Hence the eigenvectors corresponding to $b$ are
\[x_1\begin{bmatrix}
1 \\
1
\end{bmatrix},\] where $x_1$ is any nonzero complex number.

(c) Diagonalize the matrix $A$.

When $a=b$, then $A$ is already diagonal matrix. So let us consider the case $a\neq b$.
In the previous parts, we obtained the eigenvalues $a, b$, and corresponding eigenvectors
\[\begin{bmatrix}
1 \\
0
\end{bmatrix} \text{ and } \begin{bmatrix}
1 \\
1
\end{bmatrix}.\] Let $S=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix}$ be a matrix whose column vectors are the eigenvectors.
Then $S$ is invertible and we have
\[S^{-1}AS=\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}\] by the diagonalization process.

Remark that this formula is also true even when $a=b$.

(d) Using the result of the diagonalization, compute and simplify $A^k$ for each positive integer $k$.

Using the result of the diagonalization in part (c), we have
\[A=S\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}S^{-1}.\] For each positive integer $k$, we have
\begin{align*}
A^k&=\left(\, S\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}S^{-1} \,\right)^k\\[6pt] &=S\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}^k S^{-1}=S\begin{bmatrix}
a^k & 0\\
0& b^k
\end{bmatrix}S^{-1}\\[6pt] &=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix}
\begin{bmatrix}
a^k & 0\\
0& b^k
\end{bmatrix}
\begin{bmatrix}
1 & -1\\
0& 1
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
a^k & b^k-a^k\\
0& b^k
\end{bmatrix}.
\end{align*}

In summary, we have the formula
\[A^k=\begin{bmatrix}
a^k & b^k-a^k\\
0& b^k
\end{bmatrix}.\] Click here if solved 62

The post Diagonalize the Upper Triangular Matrix and Find the Power of the Matrix first appeared on Problems in Mathematics.

Let $A$ be a $3\times 3$ real orthogonal matrix with $\det(A)=1$.

(a) If $\frac{-1+\sqrt{3}i}{2}$ is one of the eigenvalues of $A$, then find the all the eigenvalues of $A$.

(b) Let
\[A^{100}=aA^2+bA+cI,\] where $I$ is the $3\times 3$ identity matrix.
Using the Cayley-Hamilton theorem, determine $a, b, c$.

(Kyushu University, Linear Algebra Exam Problem)
 

Solution.

(a) Find the all the eigenvalues of $A$.

Since $A$ is a real matrix and $\frac{-1+\sqrt{3}i}{2}$ is a complex eigenvalue, its conjugate $\frac{-1-\sqrt{3}i}{2}$ is also an eigenvalue of $A$.
As $A$ is a $3\times 3$ matrix, it has one more eigenvalue $\lambda$.

Note that the product of all eigenvalues of $A$ is the determinant of $A$.
Thus, we have
\[\frac{-1+\sqrt{3}i}{2} \cdot \frac{-1-\sqrt{3}i}{2}\cdot \lambda =\det(A)=1.\] Solving this, we obtain $\lambda=1$.
Therefore, the eigenvalues of $A$ are
\[\frac{-1+\sqrt{3}i}{2}, \frac{-1-\sqrt{3}i}{2}, 1.\]

(a) Using the Cayley-Hamilton theorem, determine $a, b, c$.

To use the Cayley-Hamilton theorem, we first need to determine the characteristic polynomial $p(t)=\det(A-tI)$ of $A$.
Since we found all the eigenvalues of $A$ in part (a) and the roots of characteristic polynomials are the eigenvalues, we know that
\begin{align*}
p(t)&=-\left(\, t-\frac{-1+\sqrt{3}i}{2} \,\right)\left(\, t-\frac{-1-\sqrt{3}i}{2} \,\right)(t-1) \tag{*}\\
&=-(t^2+t+1)(t-1)\\
&=-t^3+1.
\end{align*}
(Remark that if your definition of the characteristic polynomial is $\det(tI-A)$, then the first negative sign in (*) should be omitted.)

Then the Cayley-Hamilton theorem yields that
\[P(A)=-A^3+I=O,\] where $O$ is the $3\times 3$ zero matrix.

Hence we have $A^3=I$.
We compute
\begin{align*}
A^{100}=(A^3)^{33}A=I^{33}A=IA=A.
\end{align*}

Thus, we conclude that $a=0, b=1, c=0$.

Comment.

Observe that we did not use the assumption that $A$ is orthogonal.

The post Use the Cayley-Hamilton Theorem to Compute the Power $A^{100}$ first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
1 & 2\\
4& 3
\end{bmatrix}.\]

(a) Find eigenvalues of the matrix $A$.

(b) Find eigenvectors for each eigenvalue of $A$.

(c) Diagonalize the matrix $A$. That is, find an invertible matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

(d) Diagonalize the matrix $A^3-5A^2+3A+I$, where $I$ is the $2\times 2$ identity matrix.

(e) Calculate $A^{100}$. (You do not have to compute $5^{100}$.)

(f) Calculate
\[(A^3-5A^2+3A+I)^{100}.\] Let $w=2^{100}$. Express the solution in terms of $w$.

Solution.

(a) Find eigenvalues of the matrix $A$.

To find the eigenvalues of $A$, we calculate the characteristic polynomial $p(t)$ as follows.
We have
\begin{align*}
p(t)&=\det(A-tI)=\begin{vmatrix}
1-t & 2\\
4& 3-t
\end{vmatrix}\\
&=(1-t)(3-t)-8=t^2-4t-5=(t+1)(t-5).
\end{align*}

The eigenvalues of $A$ are roots of its characteristic polynomial $p(t)$.
Hence the eigenvalues of $A$ are $-1$ and $5$.

(b) Find eigenvectors for each eigenvalue of $A$.

We first determine the eigenvectors of the eigenvalue $-1$ by solving the system $(A+I)\mathbf{x}=\mathbf{0}$.
We have
\begin{align*}
A+I=\begin{bmatrix}
2 & 2\\
4& 4
\end{bmatrix}
\xrightarrow[\text{then } \frac{1}{2}R_1]{R_2-2R_1}
\begin{bmatrix}
1 & 1\\
0& 0
\end{bmatrix}.
\end{align*}
This yields that the eigenvectors corresponding to $-1$ are
\[a\begin{bmatrix}
1 \\
-1
\end{bmatrix}\] for any nonzero scalar $a$.

Next, we find the eigenvectors corresponding to the eigenvalue $5$ by solving $(A-5I)\mathbf{x}=\mathbf{0}$.
We have
\begin{align*}
A-5I=\begin{bmatrix}
-4 & 2\\
4& -2
\end{bmatrix}
\xrightarrow[\text{then } \frac{-1}{4}R_1]{R_2+R_1}
\begin{bmatrix}
1 & -1/2\\
0& 0
\end{bmatrix}.
\end{align*}
It follows that the eigenvectors corresponding to $5$ are
\[a\begin{bmatrix}
1 \\
2
\end{bmatrix}\] for any nonzero scalar $a$.

(c) Diagonalize the matrix $A$.

From part (a) and part (b), we have seen that $A$ has eigenvalues $-1$ and $5$ with corresponding eigenvectors
\[\mathbf{u}=\begin{bmatrix}
1 \\
-1
\end{bmatrix} \text{ and } \mathbf{v}=\begin{bmatrix}
1 \\
2
\end{bmatrix}.\] (Here we chose the scalars $a$ to be $1$ but you could use any nonzero values for the scalars $a$.)

Let
\[S=\begin{bmatrix}
\mathbf{u} & \mathbf{v}
\end{bmatrix}
=
\begin{bmatrix}
1 & 1\\
-1& 2
\end{bmatrix}.\] Then the general procedure of the diagonalization yields that the matrix $S$ is invertible and
\[S^{-1}AS=D,\] where $D$ is the diagonal matrix given by
\[D=\begin{bmatrix}
-1 & 0\\
0& 5
\end{bmatrix}.\]

(d) Diagonalize the matrix $A^3-5A^2+3A+I$.

In part (c), we obtained
\[S^{-1}AS=D,\] where
\[S=\begin{bmatrix}
1 & 1\\
-1& 2
\end{bmatrix} \text{ and } D=\begin{bmatrix}
-1 & 0\\
0& 5
\end{bmatrix}.\]

Note that we have $A=SDS^{-1}$ and
\begin{align*}
A^2&=AA=SDS^{-1}\cdot SDS^{-1}=SD^2S^{-1}\\
A^3&=A^2A=SD^2S^{-1}\cdot SDS^{-1}=SD^3S^{-1}.
\end{align*}

These relations gives
\begin{align*}
A^3-5A^2+3A+I&=SD^3S^{-1}-5SD^2S^{-1}+3SDS^{-1}+I\\
&=S(D^3-5D^2+3D+I)S^{-1}.
\end{align*}

Hence we obtain
\begin{align*}
&S^{-1}(A^3-5A^2+3A+I)S\\
&=D^3-5D^2+3D+I\\
&=\begin{bmatrix}
(-1)^3 & 0\\
0& 5^3
\end{bmatrix}
-5\begin{bmatrix}
(-1)^2 & 0\\
0& 5^2
\end{bmatrix}
+3\begin{bmatrix}
-1 & 0\\
0& 5
\end{bmatrix}
+\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
-8 & 0\\
0& 16
\end{bmatrix}.
\end{align*}
This completes the diagonalization of the matrix $A^3-5A^2+3A+I$.

(e) Calculate $A^{100}$.

In part (d), we have seen that $A=SDS^{-1}$, $A^2=SD^2S^{-1}$, $A^3=SD^3S^{-1}$.
Repeating the same argument (or using mathematical induction), we also have
\[A^{100}=SD^{100}S^{-1}.\]

Thus, we have
\begin{align*}
A^{100}&=SD^{100}S^{-1}\\
&=\begin{bmatrix}
1 & 1\\
-1& 2
\end{bmatrix}
\begin{bmatrix}
-1 & 0\\
0& 5
\end{bmatrix}^{100}
\frac{1}{3}\begin{bmatrix}
2 & -1\\
1& 1
\end{bmatrix}\\[6pt] &=\frac{1}{3}\begin{bmatrix}
1 & 1\\
-1& 2
\end{bmatrix}
\begin{bmatrix}
(-1)^{100} & 0\\
0& 5^{100}
\end{bmatrix}
\begin{bmatrix}
2 & -1\\
1& 1
\end{bmatrix}\\[6pt] &=\frac{1}{3}\begin{bmatrix}
2+5^{100} & -1+5^{100}\\
-2+2\cdot 5^{100}& 1+2\cdot 5^{100}
\end{bmatrix}.
\end{align*}

(f) Calculate $(A^3-5A^2+3A+I)^{100}$.

Let
\[B:=A^3-5A^2+3A+I.\]

In part (d), we obtained
\[S^{-1}BS=\begin{bmatrix}
-8 & 0\\
0& 16
\end{bmatrix}.\] Hence we have $B=S\begin{bmatrix}
-8 & 0\\
0& 16
\end{bmatrix} S^{-1}$, and
\begin{align*}
B^{100}&=S\begin{bmatrix}
-8 & 0\\
0& 16
\end{bmatrix}^{100} S^{-1}\\[6pt] &=S\begin{bmatrix}
(-8)^{100} & 0\\
0& 16^{100}
\end{bmatrix} S^{-1}\\[6pt] &=S\begin{bmatrix}
2^{300} & 0\\
0& 2^{400}
\end{bmatrix} S^{-1}
\\[6pt] &=S\begin{bmatrix}
w^3 & 0\\
0& w^4
\end{bmatrix} S^{-1},
\end{align*}
where we put $w=2^{100}$.

Hence we have
\begin{align*}
B^{100}&=\begin{bmatrix}
1 & 1\\
-1& 2
\end{bmatrix}
\begin{bmatrix}
w^3 & 0\\
0& w^4
\end{bmatrix}
\frac{1}{3}\begin{bmatrix}
2 & -1\\
1& 1
\end{bmatrix}\\[6pt] &=\frac{w^3}{3}\begin{bmatrix}
2+w & -1+w\\
-2+2w& 1-2w
\end{bmatrix}.
\end{align*}
Therefore, the result is
\[(A^3-5A^2+3A+I)^{100}=\frac{w^3}{3}\begin{bmatrix}
2+w & -1+w\\
-2+2w& 1-2w
\end{bmatrix}.\]

More diagonalization problems

More Problems related to the diagonalization of a matrix are gathered in the following page:

Diagonalization of Matrices

The post Diagonalize a 2 by 2 Matrix $A$ and Calculate the Power $A^{100}$ first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}\] be a $3\times 3$ matrix. Then find the formula for $A^n$ for any positive integer $n$.

Proof.

We first compute several powers of $A$ and guess the general formula.
We have
\begin{align*}
A^2=\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
=\begin{bmatrix}
1 & 1 & 3 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix},
\end{align*}
\begin{align*}
A^3=A^2A=\begin{bmatrix}
1 & 1 & 3 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
=\begin{bmatrix}
1 & 1 & 5 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix},
\end{align*}
\begin{align*}
A^4=A^3A=\begin{bmatrix}
1 & 1 & 5 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}=\begin{bmatrix}
1 & 1 & 7 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}.
\end{align*}

From these computations, we guess the general formula of $A^n$ is
\[A^n=\begin{bmatrix}
1 & 1 & 2n-1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}.\]

We prove this formula by mathematical induction on $n$.
The base case $n=1$ follows from the definition of $A$.

Suppose that the formula is true for $n=k$.
We prove the formula for $n=k+1$.
We have
\begin{align*}
A^{k+1}&=A^{k}A\\
&=\begin{bmatrix}
1 & 1 & 2k-1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}
&& \text{by the induction hypothesis}\\
&=\begin{bmatrix}
1 & 1 & 2k+1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}\\
&=\begin{bmatrix}
1 & 1 & 2(k+1)-1 \\
0 &0 &1 \\
0 & 0 & 1
\end{bmatrix}.
\end{align*}

Thus the formula holds for $n=k+1$.
Hence the formula is true for any positive integer $n$ by induction.

The post Find the Formula for the Power of a Matrix first appeared on Problems in Mathematics.

Let $A$ be a $3\times 3$ matrix. Suppose that $A$ has eigenvalues $2$ and $-1$, and suppose that $\mathbf{u}$ and $\mathbf{v}$ are eigenvectors corresponding to $2$ and $-1$, respectively, where
\[\mathbf{u}=\begin{bmatrix}
1 \\
0 \\
-1
\end{bmatrix} \text{ and } \mathbf{v}=\begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix}.\] Then compute $A^5\mathbf{w}$, where
\[\mathbf{w}=\begin{bmatrix}
7 \\
2 \\
-3
\end{bmatrix}.\]

Solution.

Since $\mathbf{u}$ is an eigenvector corresponding to the eigenvalue $2$, we have
\[A\mathbf{u}=2\mathbf{u}.\] Similarly, we have
\[A\mathbf{v}=-\mathbf{v}.\] From these, we have
\[A^5\mathbf{u}=2^5\mathbf{u} \text{ and } A\mathbf{v}=(-1)^5\mathbf{v}.\]

To compute $A^5\mathbf{w}$, we first need to express $\mathbf{w}$ as a linear combination of $\mathbf{u}$ and $\mathbf{v}$. Thus, we need to find scalars $c_1, c_2$ such that
\[\mathbf{w}=c_1\mathbf{u}+c_2\mathbf{v}.\] By inspection, we have
\[\begin{bmatrix}
7 \\
2 \\
-3
\end{bmatrix}=3\begin{bmatrix}
1 \\
0 \\
-1
\end{bmatrix}+2\begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix},\] and thus we obtain $c_1=3$ and $c_2=2$,

We compute $A^5\mathbf{w}$ as follows:
\begin{align*}
A^5\mathbf{w}&=A^5(3\mathbf{u}+2\mathbf{v})\\
&=3A^5\mathbf{u}+2A^5\mathbf{v}\\
&=3\cdot 2^5\mathbf{u}+2\cdot (-1)^5\mathbf{v}\\
&=96\mathbf{u}-2\mathbf{v}\\[6pt] &=96\begin{bmatrix}
1 \\
0 \\
-1
\end{bmatrix}-2\begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
92 \\
-2 \\
-96
\end{bmatrix}.
\end{align*}

Therefore, the result is
\[A^5\mathbf{w}=\begin{bmatrix}
92 \\
-2 \\
-96
\end{bmatrix}.\] Click here if solved 37

The post Compute Power of Matrix If Eigenvalues and Eigenvectors Are Given first appeared on Problems in Mathematics.

Let $A= \begin{bmatrix}
1 & 2\\
2& 1
\end{bmatrix}$.
Compute $A^n$ for any $n \in \N$.

Plan.

We diagonalize the matrix $A$ and use this Problem.

Steps.

1. Find eigenvalues and eigenvectors of the matrix $A$.
2. Diagonalize the matrix $A$.
3. Use the result of  this Problem.

Proof.

We first diagonalize the matrix $A$. We solve
\begin{align*}
\det(A-\lambda I) & =\begin{vmatrix}
1-\lambda & 2\\
2& 1-\lambda
\end{vmatrix} \\
&=(1-\lambda)^2-4=\lambda^2-2\lambda-3 =(\lambda+1)(\lambda-3)=0
\end{align*}
and obtain the eigenvalues $\lambda=-1, 3$.

To find an eigenvector $\mathbf{x}$ corresponding to $\lambda=-1$, we solve $(A+I)\mathbf{x}=\mathbf{0}$ or
\[ \begin{bmatrix}
2 & 2\\
2& 2
\end{bmatrix}
\begin{bmatrix} x_1 \\
x_2
\end{bmatrix}=
\begin{bmatrix} 0 \\
0
\end{bmatrix}.\] We obtain an eigenvector $\mathbf{x}=\begin{bmatrix}
1 \\
-1
\end{bmatrix}$ corresponding to $\lambda=-1$.

Similarly, solving $(A-3I)\mathbf{y}=\mathbf{0}$, we obtain an eigenvector $\mathbf{y}=\begin{bmatrix}
1 \\
1
\end{bmatrix}$
corresponding to $\lambda=3$.

Thus the invertible matrix $S=[\mathbf{x} \, \mathbf{y}]=\begin{bmatrix}
1 & 1 \\
-1& 1
\end{bmatrix}$ diagonalizes the matrix $A$, that is,
\[ S^{-1}AS =\begin{bmatrix}
-1 & 0\\
0& 3
\end{bmatrix} \text{ or equivalently }
A=S\begin{bmatrix}
-1 & 0\\
0& 3
\end{bmatrix} S^{-1}.\] Then for each $n \in \N$, we have
\begin{align*}
A^n &= \left (S\begin{bmatrix}
-1 & 0\\
0& 3
\end{bmatrix} S^{-1} \right)^n
=S \begin{bmatrix}
-1 & 0\\
0& 3
\end{bmatrix}^n S^{-1}
=S \begin{bmatrix}
(-1)^n & 0\\
0& 3^n
\end{bmatrix} S^{-1} \\[6pt] &=
\begin{bmatrix}
1 & 1 \\
-1& 1
\end{bmatrix}
\begin{bmatrix}
(-1)^n & 0\\
0& 3^n
\end{bmatrix}
\frac{1}{2}
\begin{bmatrix}
1 & -1 \\
1& 1
\end{bmatrix}
=\frac{1}{2} \begin{bmatrix}
(-1)^n+3^n & (-1)^{n+1}+3^n\\
(-1)^{n+1}+3^n& (-1)^n+3^n
\end{bmatrix}.
\end{align*}

(See this problem for the details of these computations.)

Therefore we obtained the formula
\[A^n=\frac{1}{2} \begin{bmatrix}
(-1)^n+3^n & (-1)^{n+1}+3^n\\
(-1)^{n+1}+3^n& (-1)^n+3^n
\end{bmatrix}.\]

Comment.

Another typical method to compute a power of a square matrix is mathematical induction. To use it, we need to first compute several small powers like $A^2$ and $A^3$ and guess the formula for $A^n$.

If you can guess the formula, then the mathematical induction part is not difficult. But for this specific problem, the formula is a bit complicated to guess as you can see from the solution above. Thus we used diagonalization trick.

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