If an $n\times n$ matrix has $n$ independent eigenvectors, then $A=PDP^{-1}$: in the basis of eigenvectors it is just a diagonal stretch, and its powers are $A^k=PD^kP^{-1}$.
Many questions in science and economics ask what happens when the same linear change is applied again and again: a population updated every year, a Markov chain run for a hundred steps, a recurrence such as the Fibonacci numbers. Each time the answer involves a power $A^k$ — and multiplying a matrix by itself fifty times is slow, error-prone, and tells you nothing about the pattern.
For a diagonal matrix, though, powers are effortless: you just raise each diagonal entry to the power $k$. This section shows that many matrices are diagonal “in disguise”. If we measure vectors using a basis of eigenvectors instead of the standard basis, the matrix only stretches each coordinate by its eigenvalue. Writing this down gives the factorisation $A=PDP^{-1}$, one of the most useful ideas in linear algebra.
Similar Matrices and Diagonalisable Matrices
If $D=\begin{bmatrix}d_1&0\\0&d_2\end{bmatrix}$ then $D^2=\begin{bmatrix}d_1^2&0\\0&d_2^2\end{bmatrix}$, and in general $D^k=\begin{bmatrix}d_1^k&0\\0&d_2^k\end{bmatrix}$ — the same is true for $n\times n$ diagonal matrices. We would like to transfer this convenience to other matrices.
📘 Definition 7.4 — Similar matrices (recall)
As in Section 6.3, two $n\times n$ matrices $A$ and $B$ are similar if there is an invertible $n\times n$ matrix $P$ such that $$B=P^{-1}AP\qquad(\text{equivalently } A=PBP^{-1}).$$
📐 Theorem 7.5 — Similar matrices share their characteristic polynomial
If $A$ and $B$ are similar $n\times n$ matrices, then they have the same characteristic polynomial, and therefore the same eigenvalues (with the same algebraic multiplicities), the same determinant and the same trace.
Why. If $B=P^{-1}AP$, then $B-\lambda I=P^{-1}AP-\lambda P^{-1}P=P^{-1}(A-\lambda I)P$, so $$\det(B-\lambda I)=\det(P^{-1})\det(A-\lambda I)\det P=\det(A-\lambda I),$$ because $\det(P^{-1})\det P=1$. The converse is false: $\begin{bmatrix}2&1\\0&2\end{bmatrix}$ and $\begin{bmatrix}2&0\\0&2\end{bmatrix}$ have the same characteristic polynomial $(2-\lambda)^2$, but $P^{-1}(2I)P=2I$ for every $P$, so nothing except $2I$ itself is similar to $2I$.
📘 Definition 7.5 — Diagonalisable matrix
A square matrix $A$ is diagonalisable if it is similar to a diagonal matrix, that is, if $$A=PDP^{-1}$$ for some invertible matrix $P$ and some diagonal matrix $D$. Finding such $P$ and $D$ is called diagonalisation.
The Diagonalisation Theorem
Which matrices are diagonalisable, and how do we find $P$ and $D$? Write $P=[\mathbf{v}_1\ \cdots\ \mathbf{v}_n]$ by columns and $D=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$. Then
$$AP=[A\mathbf{v}_1\ \cdots\ A\mathbf{v}_n]\qquad\text{and}\qquad PD=[\lambda_1\mathbf{v}_1\ \cdots\ \lambda_n\mathbf{v}_n].$$
So $AP=PD$ says exactly that $A\mathbf{v}_i=\lambda_i\mathbf{v}_i$ for every column. And $P$ is invertible exactly when its columns are linearly independent. This proves the central result.
📐 Theorem 7.6 — The diagonalisation theorem
An $n\times n$ matrix $A$ is diagonalisable if and only if $A$ has $n$ linearly independent eigenvectors.
In that case $A=PDP^{-1}$, where the columns of $P$ are $n$ linearly independent eigenvectors $\mathbf{v}_1,\dots,\mathbf{v}_n$ and $D$ is the diagonal matrix whose diagonal entries $\lambda_1,\dots,\lambda_n$ are the corresponding eigenvalues, in the same order. Such a set of eigenvectors is a basis of $\mathbb{R}^n$, called an eigenvector basis (or eigenbasis).
When are there enough independent eigenvectors? Eigenvectors from different eigenvalues never get in each other's way.
📐 Theorem 7.7 — Distinct eigenvalues give independent eigenvectors
If $\mathbf{v}_1,\dots,\mathbf{v}_k$ are eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda_1,\dots,\lambda_k$, then $\{\mathbf{v}_1,\dots,\mathbf{v}_k\}$ is linearly independent. Consequently, an $n\times n$ matrix with $n$ distinct real eigenvalues is diagonalisable.
Why (for two vectors). Suppose $c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{0}$. Multiplying by $A$ gives $c_1\lambda_1\mathbf{v}_1+c_2\lambda_2\mathbf{v}_2=\mathbf{0}$; multiplying the original equation by $\lambda_2$ gives $c_1\lambda_2\mathbf{v}_1+c_2\lambda_2\mathbf{v}_2=\mathbf{0}$. Subtracting, $c_1(\lambda_1-\lambda_2)\mathbf{v}_1=\mathbf{0}$. Since $\lambda_1\ne\lambda_2$ and $\mathbf{v}_1\ne\mathbf{0}$, $c_1=0$, and then $c_2=0$. The general case repeats this argument by induction. Note that “$n$ distinct eigenvalues” is sufficient but not necessary: $I$ has only the eigenvalue $1$, yet it is diagonal already.
With repeated eigenvalues, everything depends on the size of the eigenspaces. Putting together bases of all the eigenspaces always gives an independent set, so we get the following test.
📐 Theorem 7.8 — Test for diagonalisability
Let $A$ be an $n\times n$ matrix whose distinct eigenvalues are $\lambda_1,\dots,\lambda_p$. The following are equivalent:
the characteristic polynomial factors completely into real linear factors, and for every eigenvalue the geometric multiplicity equals the algebraic multiplicity.
When these hold, the bases of $E_{\lambda_1},\dots,E_{\lambda_p}$ together form an eigenvector basis of $\mathbb{R}^n$.
🧭 Problem-solving strategy — diagonalising a matrix
Eigenvalues. Solve $\det(A-\lambda I)=0$. If some roots are not real, $A$ is not diagonalisable over $\mathbb{R}$. If the $n$ roots are real and distinct, you already know $A$ is diagonalisable.
Eigenspaces. For each eigenvalue, find a basis of $E_\lambda=\operatorname{Nul}(A-\lambda I)$.
Count. If the bases contain fewer than $n$ vectors in total, stop: $A$ is not diagonalisable.
Build $P$ and $D$. Use the $n$ eigenvectors as the columns of $P$, and put each eigenvalue in the matching column of $D$.
Check $AP=PD$ — this avoids computing $P^{-1}$ and catches almost every arithmetic slip.
✏️ Example 1 — Diagonalising a $2\times2$ matrix
Diagonalise $A=\begin{bmatrix}4&-2\\1&1\end{bmatrix}$ (the matrix of Section 7.1, Example 1).
💡 Show solution▶
Eigenvalues $\det(A-\lambda I)=(4-\lambda)(1-\lambda)+2=\lambda^2-5\lambda+6=(\lambda-2)(\lambda-3)$. Two distinct eigenvalues, so $A$ is diagonalisable (Theorem 7.7).
$P$ and $D$ $$P=\begin{bmatrix}1&2\\1&1\end{bmatrix},\qquad D=\begin{bmatrix}2&0\\0&3\end{bmatrix},\qquad P^{-1}=\frac{1}{1-2}\begin{bmatrix}1&-2\\-1&1\end{bmatrix}=\begin{bmatrix}-1&2\\1&-1\end{bmatrix}.$$
Check $AP=\begin{bmatrix}4-2&8-2\\1+1&2+1\end{bmatrix}=\begin{bmatrix}2&6\\2&3\end{bmatrix}$ and $PD=\begin{bmatrix}2&6\\2&3\end{bmatrix}$ ✓. So $A=PDP^{-1}$.
Note $P$ and $D$ are not unique. Listing the eigenvalues as $3,2$ gives $P=\begin{bmatrix}2&1\\1&1\end{bmatrix}$, $D=\begin{bmatrix}3&0\\0&2\end{bmatrix}$; scaling a column of $P$ (say $(1,1)\to(-5,-5)$) also works. What cannot change is the pairing: column $i$ of $P$ must be an eigenvector for the $i$-th diagonal entry of $D$.
✏️ Example 2 — A $3\times3$ matrix with a repeated eigenvalue
Diagonalise $A=\begin{bmatrix}2&2&-1\\1&3&-1\\-1&-2&2\end{bmatrix}$ if possible.
💡 Show solution▶
Eigenvalues Expanding along the first row,
$$\begin{aligned}\det(A-\lambda I)&=(2-\lambda)\left[(3-\lambda)(2-\lambda)-2\right]-2\left[(2-\lambda)-1\right]-\left[-2+(3-\lambda)\right]\\&=(2-\lambda)(\lambda-1)(\lambda-4)-3(1-\lambda)\\&=(\lambda-1)\left[(2-\lambda)(\lambda-4)+3\right]=-(\lambda-1)(\lambda^2-6\lambda+5)=-(\lambda-1)^2(\lambda-5).\end{aligned}$$ So $\lambda=1$ (algebraic multiplicity $2$) and $\lambda=5$.
$E_1$ $A-I=\begin{bmatrix}1&2&-1\\1&2&-1\\-1&-2&1\end{bmatrix}\to\begin{bmatrix}1&2&-1\\0&0&0\\0&0&0\end{bmatrix}$: $x_1=-2s+t$ with $x_2=s$, $x_3=t$. Basis $\{(-2,1,0),\ (1,0,1)\}$ — geometric multiplicity $2$, equal to the algebraic multiplicity.
$P$ and $D$ $2+1=3$ independent eigenvectors, so $A$ is diagonalisable: $$P=\begin{bmatrix}-2&1&-1\\1&0&-1\\0&1&1\end{bmatrix},\qquad D=\begin{bmatrix}1&0&0\\0&1&0\\0&0&5\end{bmatrix}.$$
Check $A(-2,1,0)=(-2,1,0)$, $A(1,0,1)=(1,0,1)$, $A(-1,-1,1)=(-5,-5,5)$, so $AP=PD$ ✓; and $\det P=-4\ne0$, so $P$ is invertible ✓.
✏️ Example 3 — A matrix that cannot be diagonalised
Show that $A=\begin{bmatrix}1&1\\-1&3\end{bmatrix}$ is not diagonalisable.
💡 Show solution▶
Eigenvalues $\det(A-\lambda I)=(1-\lambda)(3-\lambda)+1=\lambda^2-4\lambda+4=(\lambda-2)^2$: only $\lambda=2$, algebraic multiplicity $2$.
Eigenspace $A-2I=\begin{bmatrix}-1&1\\-1&1\end{bmatrix}$ has rank $1$, so $E_2=\operatorname{span}\{(1,1)\}$ is only a line: geometric multiplicity $1$.
Conclusion All eigenvectors are multiples of $(1,1)$, so there are no two independent eigenvectors and, by Theorem 7.6, $A$ is not diagonalisable. Such a matrix is sometimes called defective. Geometrically, $A$ fixes the direction $(1,1)$ but shears everything else.
✏️ Example 4 — Deciding quickly
Which of these are diagonalisable? (a) $\begin{bmatrix}1&2&0\\0&3&4\\0&0&-1\end{bmatrix}$ (b) $\begin{bmatrix}2&1&3\\0&2&-1\\0&0&5\end{bmatrix}$ (c) $\begin{bmatrix}3&-1&1\\-1&3&-1\\1&-1&3\end{bmatrix}$
💡 Show solution▶
(a) Triangular with diagonal entries $1,3,-1$ — three distinct eigenvalues, so it is diagonalisable (Theorem 7.7), without finding a single eigenvector.
(b) Eigenvalues $2,2,5$. Section 7.1, Example 4 showed $\dim E_2=1$, so $\dim E_2+\dim E_5=2\lt3$: not diagonalisable.
(c) Eigenvalues $2,2,5$ again, but Section 7.1, Example 3 showed $\dim E_2=2$, so $2+1=3$: diagonalisable, with $P=\begin{bmatrix}1&-1&1\\1&0&-1\\0&1&1\end{bmatrix}$, $D=\operatorname{diag}(2,2,5)$.
Lesson A repeated eigenvalue is a warning sign, not a verdict: you must find the dimension of its eigenspace.
⚠️ Common mistakes with diagonalisation
1. Mismatched order. If the first column of $P$ belongs to $\lambda=3$, the first diagonal entry of $D$ must be $3$.
2. “Repeated eigenvalue, so not diagonalisable.” Wrong — see Examples 2 and 4(c). Count the dimensions of the eigenspaces.
3. Diagonalisable and invertible are different ideas. $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ is diagonal (so diagonalisable) but not invertible; $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ is invertible but not diagonalisable. Invertibility is about the eigenvalue $0$; diagonalisability is about having enough eigenvectors.
4. Putting the zero vector, or dependent vectors, in $P$. Then $P$ is not invertible and $A=PDP^{-1}$ is meaningless.
Powers of a Matrix
📐 Theorem 7.9 — Powers of a diagonalisable matrix
If $A=PDP^{-1}$ with $D=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$, then for every positive integer $k$
$$A^k=PD^kP^{-1},\qquad D^k=\operatorname{diag}\left(\lambda_1^k,\dots,\lambda_n^k\right).$$
Equivalently, if $\mathbf{x}=c_1\mathbf{v}_1+\dots+c_n\mathbf{v}_n$ in the eigenvector basis, then $A^k\mathbf{x}=c_1\lambda_1^k\mathbf{v}_1+\dots+c_n\lambda_n^k\mathbf{v}_n$.
Why. $$A^2=(PDP^{-1})(PDP^{-1})=PD(P^{-1}P)DP^{-1}=PD^2P^{-1}:$$ the inner $P^{-1}P$ pairs cancel, and the same happens for every $k$. The second form is Theorem 7.4(2) applied to each eigenvector separately. It shows the main message of this section: after many steps, the term with the strictly largest $|\lambda_i|$ dominates (as long as its coefficient $c_i$ is not zero).
✏️ Example 5 — A formula for $A^k$
For $A=\begin{bmatrix}4&-2\\1&1\end{bmatrix}$ find a formula for $A^k$, then compute $A^5$ and $A^k\mathbf{x}$ for $\mathbf{x}=(3,2)$.
💡 Show solution▶
Use Example 1 $$A^k=PD^kP^{-1}=\begin{bmatrix}1&2\\1&1\end{bmatrix}\begin{bmatrix}2^k&0\\0&3^k\end{bmatrix}\begin{bmatrix}-1&2\\1&-1\end{bmatrix}=\begin{bmatrix}2^k&2\cdot3^k\\2^k&3^k\end{bmatrix}\begin{bmatrix}-1&2\\1&-1\end{bmatrix}$$ $$A^k=\begin{bmatrix}2\cdot3^k-2^k&2^{k+1}-2\cdot3^k\\3^k-2^k&2^{k+1}-3^k\end{bmatrix}.$$
$k=5$ With $2^5=32$, $3^5=243$: $$A^5=\begin{bmatrix}486-32&64-486\\243-32&64-243\end{bmatrix}=\begin{bmatrix}454&-422\\211&-179\end{bmatrix}.$$
$A^k\mathbf{x}$ Write $\mathbf{x}=(3,2)=c_1(1,1)+c_2(2,1)$: $c_1+2c_2=3$, $c_1+c_2=2$, so $c_2=1$, $c_1=1$. Then $$A^k\mathbf{x}=2^k\begin{bmatrix}1\\1\end{bmatrix}+3^k\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}2^k+2\cdot3^k\\2^k+3^k\end{bmatrix}.$$ For large $k$ the $3^k$ term dominates, so $A^k\mathbf{x}$ points more and more nearly along $(2,1)$ and grows by a factor close to $3$ each step. Watch this happen in Interactive 1.
Interactive 1 — $A^k\mathbf{x}$ Split Along the Eigenvectors
Write $\mathbf{x}=c_1\mathbf{v}_1+c_2\mathbf{v}_2$. Then $A^k\mathbf{x}=c_1\lambda_1^k\mathbf{v}_1+c_2\lambda_2^k\mathbf{v}_2$: the blue and green arrows are the two pieces, and the red arrow is their sum. Because the vectors grow (or shrink) fast, the picture is rescaled at each step so that the red arrow has length 1 — watch the piece with the larger $|\lambda|$ take over, and the red arrow swing onto the dominant eigenvector line.
0
$A^k\mathbf{x}$ (actual size)—
$c_1\lambda_1^k$ (along $\mathbf{v}_1$)—
$c_2\lambda_2^k$ (along $\mathbf{v}_2$)—
angle to the dominant line—
✏️ Example 6 — Building a matrix from its eigenvectors
An engineer wants a $2\times2$ matrix that stretches every vector along the line $y=x$ by a factor $3$ and reverses every vector along $y=-x$. Find the matrix and compute $A^{10}$.
💡 Show solution▶
Eigen-data We want $A(1,1)=3(1,1)$ and $A(1,-1)=-(1,-1)$. So $P=\begin{bmatrix}1&1\\1&-1\end{bmatrix}$, $D=\begin{bmatrix}3&0\\0&-1\end{bmatrix}$, and $P^{-1}=\dfrac{1}{-2}\begin{bmatrix}-1&-1\\-1&1\end{bmatrix}=\dfrac12\begin{bmatrix}1&1\\1&-1\end{bmatrix}$.
Power $$A^{k}=P\begin{bmatrix}3^k&0\\0&(-1)^k\end{bmatrix}P^{-1}=\frac12\begin{bmatrix}3^k+(-1)^k&3^k-(-1)^k\\3^k-(-1)^k&3^k+(-1)^k\end{bmatrix}.$$ With $3^{10}=59049$: $A^{10}=\begin{bmatrix}29525&29524\\29524&29525\end{bmatrix}$.
Diagonalisation as a Change of Basis
There is a beautiful way to read $A=PDP^{-1}$. Let $\mathcal B=\{\mathbf{v}_1,\dots,\mathbf{v}_n\}$ be the eigenvector basis formed by the columns of $P$. From Section 5.5, $P$ is the change-of-coordinates matrix from $\mathcal B$ to the standard basis $\mathcal E$, $P=P_{\mathcal E\leftarrow\mathcal B}$: $\mathbf{x}=P[\mathbf{x}]_{\mathcal B}$, so $[\mathbf{x}]_{\mathcal B}=P^{-1}\mathbf{x}$. Reading $A\mathbf{x}=PDP^{-1}\mathbf{x}$ from right to left:
1. $P^{-1}$convert $\mathbf{x}$ into eigen-coordinates $[\mathbf{x}]_{\mathcal B}=(c_1,\dots,c_n)$
2. $D$stretch each coordinate: $c_i\mapsto\lambda_ic_i$
3. $P$convert back to standard coordinates
📐 Theorem 7.10 — The matrix of $\mathbf{x}\mapsto A\mathbf{x}$ in an eigenvector basis
Suppose $A=PDP^{-1}$ with $D$ diagonal, and let $\mathcal B$ be the basis of $\mathbb{R}^n$ formed by the columns of $P$. Then the matrix of the linear transformation $T(\mathbf{x})=A\mathbf{x}$ relative to $\mathcal B$ is $D$: $$[T]_{\mathcal B}=D,\qquad\text{that is,}\qquad[A\mathbf{x}]_{\mathcal B}=D\,[\mathbf{x}]_{\mathcal B}\ \text{ for every }\mathbf{x}.$$
This links back to Section 6.3, where the matrix of a transformation depends on the basis you choose. Diagonalising $A$ means choosing the basis in which the transformation is as simple as possible: in eigen-coordinates it is a pure stretch along each axis. Interactive 2 shows the same transformation through both pairs of eyes.
Interactive 2 — The Same Map Seen in Two Coordinate Systems
The grid is made of lines parallel to the eigenvectors $\mathbf{v}_1$ (blue) and $\mathbf{v}_2$ (green); the shaded cell has corners $\mathbf{0},\mathbf{v}_1,\mathbf{v}_1+\mathbf{v}_2,\mathbf{v}_2$. Press Play to apply $A$ gradually. In standard coordinates the grid is slanted and its cells change shape; switch to eigen-coordinates and the very same motion is a plain stretch of each axis by $\lambda_1$ and $\lambda_2$ — that is $D$.
3 $p(\lambda)=\lambda^2-6\lambda+9=(\lambda-3)^2$. $A-3I=\begin{bmatrix}-1&-1\\1&1\end{bmatrix}$ has rank $1$, so $E_3=\operatorname{span}\{(1,-1)\}$ is one-dimensional: not diagonalisable.
5 $p(\lambda)=\lambda^2+1$ has no real roots, so there are no real eigenvectors: not diagonalisable over $\mathbb{R}$. (It is a $90^\circ$ rotation; over $\mathbb{C}$ it can be diagonalised with $D=\operatorname{diag}(i,-i)$.)
8 $p(\lambda)=-\lambda^3+3\lambda+2=-(\lambda+1)^2(\lambda-2)$. $E_{-1}$: $A+I$ is the all-ones matrix, so $x_1+x_2+x_3=0$: basis $\{(-1,1,0),(-1,0,1)\}$. $E_2$: $(1,1,1)$. $P=\begin{bmatrix}-1&-1&1\\1&0&1\\0&1&1\end{bmatrix}$, $D=\operatorname{diag}(-1,-1,2)$.
9 Block form: $p(\lambda)=\left[(4-\lambda)(2-\lambda)+1\right](3-\lambda)=(\lambda-3)^2(3-\lambda)=-(\lambda-3)^3$. $A-3I=\begin{bmatrix}1&-1&0\\1&-1&0\\0&0&0\end{bmatrix}$ has rank $1$, so $\dim E_3=2\lt3$: not diagonalisable ($E_3$ has basis $\{(1,1,0),(0,0,1)\}$).
10 Triangular: $\lambda=1$ three times. $A-I=\begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix}$ has rank $2$, so $\dim E_1=1\lt3$: not diagonalisable.
II
Powers of matrices.
1 For $A=\begin{bmatrix}1&0\\6&-1\end{bmatrix}$ (Exercise I.1), use the diagonalisation to find $A^{100}$ and $A^{101}$.
2 For $A=\begin{bmatrix}7&2\\-4&1\end{bmatrix}$, find a formula for $A^k$ and use it to compute $A^3$.
3 For the Markov matrix $A=\begin{bmatrix}0.9&0.2\\0.1&0.8\end{bmatrix}$, find $A^k$ and the limit of $A^k$ as $k\to\infty$.
4 Let $P=\begin{bmatrix}1&1\\1&2\end{bmatrix}$ and $D=\begin{bmatrix}2&0\\0&-1\end{bmatrix}$. Compute $A=PDP^{-1}$ and $A^4$.
5 For $A=\begin{bmatrix}4&-2\\1&1\end{bmatrix}$ and $\mathbf{x}=(3,2)$, compute $A^4\mathbf{x}$ using the eigenvector expansion of Example 5, and check by finding $A^2\mathbf{x}$ directly first.
6 For $A=\begin{bmatrix}1&2\\2&1\end{bmatrix}$ and $\mathbf{x}=(1,0)$, find $A^k\mathbf{x}$. What happens to the ratio of its two entries as $k\to\infty$?
💡 Show solutions▶
1 $D^{100}=\operatorname{diag}\left(1^{100},(-1)^{100}\right)=I$, so $A^{100}=PIP^{-1}=I$ and $A^{101}=A^{100}A=A$. (Indeed $A^2=I$: the matrix is an “oblique reflection” that fixes the line through $(1,3)$ and reverses the direction $(0,1)$, so doing it twice changes nothing.)
2 From I.2, $P=\begin{bmatrix}1&1\\-2&-1\end{bmatrix}$, $P^{-1}=\begin{bmatrix}-1&-1\\2&1\end{bmatrix}$: $$A^k=\begin{bmatrix}3^k&5^k\\-2\cdot3^k&-5^k\end{bmatrix}\begin{bmatrix}-1&-1\\2&1\end{bmatrix}=\begin{bmatrix}2\cdot5^k-3^k&5^k-3^k\\2\cdot3^k-2\cdot5^k&2\cdot3^k-5^k\end{bmatrix}.$$ $k=3$: $A^3=\begin{bmatrix}250-27&125-27\\54-250&54-125\end{bmatrix}=\begin{bmatrix}223&98\\-196&-71\end{bmatrix}$.
3 $p(\lambda)=\lambda^2-1.7\lambda+0.7=(\lambda-1)(\lambda-0.7)$; $E_1$: $(2,1)$, $E_{0.7}$: $(1,-1)$. $P=\begin{bmatrix}2&1\\1&-1\end{bmatrix}$, $P^{-1}=\tfrac13\begin{bmatrix}1&1\\1&-2\end{bmatrix}$: $$A^k=\frac13\begin{bmatrix}2+0.7^k&2-2(0.7)^k\\1-0.7^k&1+2(0.7)^k\end{bmatrix}\ \longrightarrow\ \frac13\begin{bmatrix}2&2\\1&1\end{bmatrix}\ \text{ as }k\to\infty,$$ because $0.7^k\to0$. Every column of the limit is the steady state $\left(\tfrac23,\tfrac13\right)$.
6 $(1,0)=\tfrac12(1,1)+\tfrac12(1,-1)$, so $A^k\mathbf{x}=\tfrac12\cdot3^k(1,1)+\tfrac12(-1)^k(1,-1)=\tfrac12\left(3^k+(-1)^k,\ 3^k-(-1)^k\right)$. The ratio of the entries is $\dfrac{3^k+(-1)^k}{3^k-(-1)^k}\to1$: the vector lines up with the dominant eigenvector $(1,1)$, approaching from alternate sides.
III
True or false? Give a reason or a counterexample. (Matrices are $n\times n$.)
1 Every diagonalisable matrix is invertible.
2 Every invertible matrix is diagonalisable.
3 A $4\times4$ matrix with four distinct real eigenvalues is diagonalisable.
4 If $A$ is diagonalisable, then so is $A^2$.
5 Similar matrices have the same eigenvectors.
6 If $A=PDP^{-1}$ with $D$ diagonal, then every column of $P$ is an eigenvector of $A$.
7 A $3\times3$ matrix with eigenvalues $2,2,5$ and $\dim E_2=1$ is diagonalisable.
8 For every real $k$, $\begin{bmatrix}1&k\\0&2\end{bmatrix}$ is diagonalisable, but $\begin{bmatrix}2&k\\0&2\end{bmatrix}$ is diagonalisable only when $k=0$.
9 If $A$ is diagonalisable and its only eigenvalues are $1$ and $-1$, then $A^2=I$.
10 If $A$ and $B$ are similar, then $\det A=\det B$ and $\operatorname{tr}A=\operatorname{tr}B$.
💡 Show solutions▶
1False $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ is diagonal but singular. A diagonalisable matrix may have $0$ as an eigenvalue.
2False $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ has $\det=1$ but only one independent eigenvector.
3True Theorem 7.7: four distinct eigenvalues give four independent eigenvectors.
4True $A^2=PDP^{-1}PDP^{-1}=PD^2P^{-1}$ and $D^2$ is diagonal.
5False They share eigenvalues, not eigenvectors: if $B=P^{-1}AP$ and $A\mathbf{x}=\lambda\mathbf{x}$, then $B(P^{-1}\mathbf{x})=\lambda P^{-1}\mathbf{x}$, so the eigenvector of $B$ is $P^{-1}\mathbf{x}$. E.g. $A=\begin{bmatrix}4&-2\\1&1\end{bmatrix}$ is similar to $D=\operatorname{diag}(2,3)$; $(1,0)$ is an eigenvector of $D$ but not of $A$.
6True $AP=PD$ says $A\mathbf{p}_i=d_i\mathbf{p}_i$ for each column, and no column of an invertible $P$ is $\mathbf{0}$.
7False $\dim E_2+\dim E_5=1+1=2\lt3$ (Theorem 7.8). Section 7.1, Example 4 is such a matrix.
8True The first has distinct eigenvalues $1,2$. For the second, $\lambda=2$ is double and $A-2I=\begin{bmatrix}0&k\\0&0\end{bmatrix}$ has nullity $2$ only when $k=0$.
9True $D^2=I$ because $(\pm1)^2=1$, so $A^2=PD^2P^{-1}=PP^{-1}=I$. (Without diagonalisability it can fail: $\begin{bmatrix}1&1\\0&1\end{bmatrix}^2\ne I$.)
10True By Theorem 7.5 they have the same characteristic polynomial; $\det A=p(0)$ and $\operatorname{tr}A$ is $(-1)^{n-1}$ times the coefficient of $\lambda^{n-1}$, so both agree. (Directly: $\det(P^{-1}AP)=\det A$.)
IV
Applications.
1Migration. Each year $5\%$ of a country's city dwellers move to rural areas and $3\%$ of rural dwellers move to cities. With $\mathbf{x}_k=(\text{city},\text{rural})$ as fractions of the population, $\mathbf{x}_{k+1}=M\mathbf{x}_k$ with $M=\begin{bmatrix}0.95&0.03\\0.05&0.97\end{bmatrix}$. If $\mathbf{x}_0=(0.6,0.4)$, find $\mathbf{x}_k$, the city fraction after $10$ years, and the long-run split.
2Fibonacci numbers. $F_0=0$, $F_1=1$, $F_{k+1}=F_k+F_{k-1}$. Show that $\begin{bmatrix}F_{k+1}\\F_k\end{bmatrix}=A^k\begin{bmatrix}1\\0\end{bmatrix}$ with $A=\begin{bmatrix}1&1\\1&0\end{bmatrix}$, diagonalise $A$, and derive $F_k=\dfrac{\varphi^k-\psi^k}{\sqrt5}$ where $\varphi=\tfrac{1+\sqrt5}{2}$, $\psi=\tfrac{1-\sqrt5}{2}$. Use it to find $F_{20}$.
3 Let $T(\mathbf{x})=A\mathbf{x}$ with $A=\begin{bmatrix}7&2\\-4&1\end{bmatrix}$. Find a basis $\mathcal B$ of $\mathbb{R}^2$ for which $[T]_{\mathcal B}$ is diagonal. Then, for $\mathbf{x}=(2,-3)$, find $[\mathbf{x}]_{\mathcal B}$, $[T(\mathbf{x})]_{\mathcal B}$ and $T(\mathbf{x})$.
4Square root of a matrix. Find a matrix $B$ with positive eigenvalues such that $B^2=\begin{bmatrix}5&4\\4&5\end{bmatrix}$.
💡 Show solutions▶
1 $p(\lambda)=\lambda^2-1.92\lambda+0.92=(\lambda-1)(\lambda-0.92)$. $E_1$: $-0.05x_1+0.03x_2=0$, $\mathbf{v}_1=(3,5)$; $E_{0.92}$: $\mathbf{v}_2=(1,-1)$. Solve $c_1(3,5)+c_2(1,-1)=(0.6,0.4)$: $8c_1=1$, so $c_1=\tfrac18$, $c_2=0.6-\tfrac38=0.225$. $$\mathbf{x}_k=\frac18\begin{bmatrix}3\\5\end{bmatrix}+0.225\,(0.92)^k\begin{bmatrix}1\\-1\end{bmatrix}.$$ After $10$ years the city fraction is $0.375+0.225(0.92)^{10}\approx0.375+0.0977=0.4727$ (about $47.3\%$). Long run: $(0.92)^k\to0$, so $37.5\%$ city and $62.5\%$ rural.
2 $A\begin{bmatrix}F_k\\F_{k-1}\end{bmatrix}=\begin{bmatrix}F_k+F_{k-1}\\F_k\end{bmatrix}=\begin{bmatrix}F_{k+1}\\F_k\end{bmatrix}$, and $(F_1,F_0)=(1,0)$, so induction gives the formula. $p(\lambda)=\lambda^2-\lambda-1$, roots $\varphi,\psi$ with eigenvectors $(\varphi,1)$, $(\psi,1)$ (since $\lambda^2=\lambda+1$). Now $(1,0)=\dfrac{1}{\sqrt5}\left[(\varphi,1)-(\psi,1)\right]$ because $\varphi-\psi=\sqrt5$. So $A^k(1,0)=\dfrac{1}{\sqrt5}\left[\varphi^k(\varphi,1)-\psi^k(\psi,1)\right]$, whose second entry is $F_k=\dfrac{\varphi^k-\psi^k}{\sqrt5}$. $F_{20}=\dfrac{\varphi^{20}-\psi^{20}}{\sqrt5}=6765$ (since $|\psi^{20}|/\sqrt5\approx0.00003$, $F_{20}$ is the integer nearest to $\varphi^{20}/\sqrt5\approx6765.00003$).
3 From I.2, take $\mathcal B=\{(1,-2),(1,-1)\}$; then $[T]_{\mathcal B}=\begin{bmatrix}3&0\\0&5\end{bmatrix}$. $c_1(1,-2)+c_2(1,-1)=(2,-3)$ gives $c_1=1$, $c_2=1$, so $[\mathbf{x}]_{\mathcal B}=(1,1)$, $[T(\mathbf{x})]_{\mathcal B}=(3,5)$ and $T(\mathbf{x})=3(1,-2)+5(1,-1)=(8,-11)$. Check: $A(2,-3)=(14-6,\,-8-3)=(8,-11)$ ✓.
4 $\begin{bmatrix}5&4\\4&5\end{bmatrix}$ has eigenvalues $9$ (eigenvector $(1,1)$) and $1$ (eigenvector $(1,-1)$). Take $B=P\operatorname{diag}(3,1)P^{-1}$ with $P=\begin{bmatrix}1&1\\1&-1\end{bmatrix}$: $B=\begin{bmatrix}3&1\\3&-1\end{bmatrix}\cdot\tfrac12\begin{bmatrix}1&1\\1&-1\end{bmatrix}=\begin{bmatrix}2&1\\1&2\end{bmatrix}$. Check: $B^2=\begin{bmatrix}5&4\\4&5\end{bmatrix}$ ✓.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. How quickly does tomorrow's weather “forget” today's? A two-state weather chain answers this with its second eigenvalue.
Why it matters. Weather generators used in agriculture, hydrology and insurance model wet and dry days as a Markov chain. The long-run fraction of dry days comes from the eigenvalue $1$; how fast the chain settles down is controlled by the other eigenvalue $\lambda_2$, because $A^k=PD^kP^{-1}$ contains $\lambda_2^k$.
What to do.
Sample case: a dry day is followed by a dry day with probability $0.9$; a wet day is followed by a dry day with probability $0.5$. With $\mathbf{x}=(\text{dry},\text{wet})$, $A=\begin{bmatrix}0.9&0.5\\0.1&0.5\end{bmatrix}$.
Diagonalise $A$ and write $A^k$ explicitly. What is the long-run probability of a dry day?
Starting from a wet day, $\mathbf{x}_0=(0,1)$, find $\mathbf{x}_k$ and the first day $k$ on which the probability of a dry day is within $0.01$ of its long-run value.
Collect real data: from a public weather record for your town (for example Muscat in winter or Mumbai in the monsoon), count dry→dry, dry→wet, wet→dry and wet→wet transitions over a month, build your own $A$ and repeat steps 2–3. Compare $|\lambda_2|$ for the two seasons.
Tools: Calculator, spreadsheet; a public daily-rainfall record.
Hand in: A half-page note with your matrices, their eigenvalues, the “memory time” in days, and one reason why real weather might not be a perfect Markov chain.
Going further. The same calculation measures how fast a shuffled deck of cards becomes random, or how quickly a web surfer's position stops depending on the starting page (Section 7.3).
Instructor note: eigenvalues $1$ (eigenvector $(5,1)$) and $0.4$ (eigenvector $(1,-1)$); long-run dry probability $\tfrac56\approx0.833$. From a wet day $P(\text{dry on day }k)=\tfrac56-\tfrac56(0.4)^k$, within $0.01$ once $(0.4)^k\lt0.012$, i.e. from day $k=5$. A season with $|\lambda_2|$ close to $1$ has long dry or wet spells.
🧪 Self-check quiz
0/5 correct
Q1. If $A=PDP^{-1}$ with $D$ diagonal, the columns of $P$ are
Q2. A $3\times3$ matrix has eigenvalues $1$, $2$ and $3$. Then $A$
Q3. The matrix $\begin{bmatrix}2&1\\0&2\end{bmatrix}$ is
Q4. If $A=P\begin{bmatrix}2&0\\0&-1\end{bmatrix}P^{-1}$, then $A^4=$
Q5. If $A=PDP^{-1}$ with $D=\operatorname{diag}(3,-2,1)$, then $\det A=$