Finding eigenvalues by factoring $\det(A - tI)$ works for a $3 \times 3$ matrix. It is hopeless for the matrices engineers actually meet: a model of a bridge, a circuit board, or the web can have thousands or billions of rows, and a characteristic polynomial of degree one million can't be expanded, let alone factored. What those matrices do allow is multiplying a vector by them, quickly, because almost all of their entries are $0$. The power method finds the dominant eigenvalue and its eigenvector using nothing but that: multiply, rescale, repeat.
Take an arrow, multiply it by a matrix, and shrink the answer back to a manageable size. Do it again. And again. The arrow swings around for a few steps, then stops turning: it has found the matrix's most important eigenvector, and the amount it got stretched by is the eigenvalue. That's the whole method. The matrices on this page are small enough to draw, but the method is built for huge ones, where factoring $\det(A - tI)$ is out of the question and multiplying is all a computer can afford.
On Matrix Powers, the iterates $x_0, Ax_0, A^2x_0, \ldots$ lined up along the eigenvector of the dominant eigenvalue, as long as the start used it. That was a theorem about where the dots go. This page turns it into an algorithm.
Every time the arrow is multiplied by $A$ it grows; every time it's shrunk back to length $1$ it lands on the circle again. The growth factor settles to $3$, and the arrow settles onto the line through $(1, 2)$. Those are the dominant eigenvalue and its eigenvector, found without ever writing down a polynomial.
Why rescale at all? Because $A^kx_0$ grows like $\lambda_1^k$, and after a few hundred steps the numbers overflow (or, if $|\lambda_1| < 1$, underflow to $0$). Dividing by a positive number never changes which way a vector points, so shrinking after every step loses nothing. There are two popular ways to shrink.
The power method for a square matrix $A$ starts from a vector $x_0 \neq 0$ and repeats two moves: multiply, $y_{k+1} = Ax_k$; then rescale, $x_{k+1} = y_{k+1}/m_{k+1}$. The scale factor $m_{k+1}$ is either the entry of $y_{k+1}$ of largest size, sign included (handy by hand: the new vector's largest entry is then $1$), or the length $\|y_{k+1}\|$ (what computers usually do: $x_{k+1}$ is then a unit vector).
Here is the opening figure's matrix by hand, with the largest-entry scaling. It's the example Hefferon uses for this method.
Run the power method on $A = \begin{pmatrix} 3 & 0 \\ 8 & -1 \end{pmatrix}$ from $x_0 = (1, 1)$, dividing by the entry of largest size.
Step $1$: $Ax_0 = (3, 7)$, so $m_1 = 7$ and $x_1 = \left(\frac37, 1\right)$.
Step $2$: $Ax_1 = \left(\frac97, \frac{24}{7} - 1\right) = \left(\frac97, \frac{17}{7}\right)$, so $m_2 = \frac{17}{7} \approx 2.43$ and $x_2 = \left(\frac{9}{17}, 1\right)$.
Step $3$: $Ax_2 = \left(\frac{27}{17}, \frac{55}{17}\right)$, so $m_3 = \frac{55}{17} \approx 3.24$ and $x_3 = \left(\frac{27}{55}, 1\right)$.
Two more steps give $m_4 = \frac{161}{55} \approx 2.93$ and $m_5 = \frac{487}{161} \approx 3.02$. The scale factors zigzag in on $3$, and $x_k$ closes in on $\left(\frac12, 1\right)$. Check: $A$ is triangular, so its eigenvalues are $3$ and $-1$, and $A\left(\frac12, 1\right) = \left(\frac32, 3\right) = 3\left(\frac12, 1\right)$.
Dividing by the largest entry puts every iterate on the edge of the box where $|x| \le 1$ and $|y| \le 1$. Here the iterates walk along its top edge, overshooting back and forth, toward $\left(\frac12, 1\right)$.
Why the zigzag? The other eigenvalue is $-1$, and its part of $x_k$ flips sign every step while it shrinks. The next section makes that precise. First, your turn: in free response, type a number; fractions like 17/7 are fine.
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Rescaling only changes lengths, so $x_k$ always points the same way as $A^kx_0$. And Matrix Powers already knows where $A^kx_0$ points: split $x_0$ along eigenvectors and each piece grows by its own eigenvalue. The dominant piece outgrows the rest, and the leftover shrinks relative to it by $|\lambda_2/\lambda_1|$ every step.
Let $A$ be an $n \times n$ matrix with a basis of eigenvectors $v_1, \ldots, v_n$, $Av_i = \lambda_iv_i$, where $\lambda_1$ is dominant and $\lambda_2$ is an eigenvalue of second-largest size. Run the power method from $x_0 = c_1v_1 + \cdots + c_nv_n$ with $c_1 \neq 0$.
Each $x_k$ is $A^kx_0$ divided by a nonzero number, since $x_{k+1} = Ax_k/m_{k+1}$. By the powers theorem,
$$A^kx_0 = c_1\lambda_1^k\left(v_1 + \delta_k\right), \qquad \delta_k = \sum_{i \ge 2} \frac{c_i}{c_1}\left(\frac{\lambda_i}{\lambda_1}\right)^kv_i,$$and every $|\lambda_i/\lambda_1| \le |\lambda_2/\lambda_1| < 1$, so $\|\delta_k\| \le C'|\lambda_2/\lambda_1|^k$ with $C' = \sum_{i \ge 2} |c_i/c_1|\,\|v_i\|$.
(1) With unit-length scaling, $x_k$ is the unit vector along $A^kx_0$, which is $\pm(v_1 + \delta_k)/\|v_1 + \delta_k\|$, with the sign of $c_1\lambda_1^k$. For nonzero $a$ and $b$, $\left\|\frac{a}{\|a\|} - \frac{b}{\|b\|}\right\| \le \frac{2\|a - b\|}{\|b\|}$ (add and subtract $\frac{a}{\|b\|}$ and use the triangle inequality twice), so $\|e_k\| \le 2\|\delta_k\|/\|v_1\|$, which is the bound with $C = 2C'/\|v_1\|$. Then $\|Ax_k\| = \|A(v_1 + \delta_k)\|/\|v_1 + \delta_k\| \to \|\lambda_1v_1\|/\|v_1\| = |\lambda_1|$.
(2) Suppose the largest-size entry of $v_1$ is entry $i$, and only that one. Since $\delta_k \to 0$, for large $k$ the vector $A^kx_0$ also has entry $i$ as its single largest-size entry. A vector divided by its own largest-size entry has largest entry $1$, and dividing $Ax_k$ by its largest-size entry undoes whatever multiple $x_k$ carried, so for large $k$, $x_k = A^kx_0/(A^kx_0)_i = (v_1 + \delta_k)/(v_1 + \delta_k)_i \to v_1/(v_1)_i$. Finally $m_{k+1}$ is the largest-size entry of $Ax_k = A^{k+1}x_0/(A^kx_0)_i$, which is entry $i$: $m_{k+1} = \lambda_1(v_1 + \delta_{k+1})_i/(v_1 + \delta_k)_i \to \lambda_1$. $\blacksquare$
Let $A$ be a $2 \times 2$ matrix with eigenvectors $v_1, v_2$ on two different lines (or a $3 \times 3$ with eigenvectors $v_1, v_2, v_3$ not all in one plane), where $\lambda_1$ is dominant and $\lambda_2$ comes next in size. If $x_0$ uses $v_1$, then the power method's arrows swing onto the line of $v_1$ and stay there, and each step shrinks what's left off that line by about $|\lambda_2/\lambda_1|$. The lengths $\|Ax_k\|$ settle to $|\lambda_1|$; the largest-entry scale factors settle to $\lambda_1$ itself, as long as $v_1$ has a single entry of largest size.
This is a visual argument, not a formal proof. Split the arrow into two legs, one along each eigenvector. Multiplying by $A$ stretches the main leg by $\lambda_1$ and the leftover leg by $\lambda_2$, so the leftover compared to the main leg changes by $\lambda_2/\lambda_1$. Rescaling shrinks both legs by the same amount, so it doesn't touch that comparison. For the example's $A = \begin{pmatrix} 3 & 0 \\ 8 & -1 \end{pmatrix}$, $\lambda_2/\lambda_1 = -\frac13$: the leftover flips side and shrinks to a third every step, which is exactly the zigzag above.
The full proof is in the n version.
The largest-entry scale factors come with fine print in that theorem, and here's why it's there.
The largest-entry scaling has one catch. If $v_1$ has two entries of the same largest size, like $(1, -1)$, the "largest entry" can trade places from one step to the next, and $x_k$ flips between $(1, -1)$ and $(-1, 1)$ while $m_k$ flips between $\lambda_1$ and $-\lambda_1$. The direction has settled; only the bookkeeping hasn't. Unit-length scaling never has this problem.
Here is the speed claim at work on the opening figure's numbers.
The opening figure ran $A = \begin{pmatrix} 3 & 0 \\ 8 & -1 \end{pmatrix}$ from $x_0 = (1, 1)$ with unit-length scaling. How fast should $\|Ax_k\|$ approach $3$, and does it?
The eigenvalues are $3$ and $-1$, so the theorem predicts errors shrinking by about $\left|\frac{-1}{3}\right| = \frac13$ a step. The lengths were $5.39, 2.53, 3.19, 2.94, 3.02$, with errors $2.39, 0.47, 0.19, 0.06, 0.02$. After the first big jump, each error is about a third of the one before. And the sign of $-1$ shows up too: the lengths land above $3$, below, above, below, above, as the leftover flips sides.
The scale factors are one way to read off $\lambda_1$. A computer usually uses a sharper one: compare $x$ with $Ax$ using a dot product.
For a square matrix $A$ and a vector $x \neq 0$, the Rayleigh quotient is the number $R(x) = \dfrac{x \cdot Ax}{x \cdot x}$.
On an exact eigenvector it gives the eigenvalue on the nose, and it doesn't care how long $x$ is.
If $Av = \lambda v$ with $v \neq 0$, then $R(v) = \lambda$. Also $R(cx) = R(x)$ for every number $c \neq 0$, so $R$ depends only on the line through $x$.
$v \cdot Av = v \cdot (\lambda v) = \lambda(v \cdot v)$, and $v \cdot v = \|v\|^2 \neq 0$, so $R(v) = \lambda$. For the second claim, $(cx) \cdot A(cx) = c^2(x \cdot Ax)$ and $(cx) \cdot (cx) = c^2(x \cdot x)$, and the $c^2$ cancels. $\blacksquare$
If $Av = \lambda v$, the top of the fraction is $v \cdot \lambda v = \lambda(v \cdot v)$, and the $v \cdot v$ cancels with the bottom. Doubling $x$ (or scaling it by any $c$) multiplies the top and the bottom by $c^2$, which also cancels. The full proof is in the n version.
So once $x_k$ is close to the line of $v_1$, $R(x_k)$ is close to $R(v_1) = \lambda_1$. On a symmetric matrix it's closer than you'd expect.
Run the power method on $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ from $x_0 = (1, 0)$, and compare the largest-entry scale factors with the Rayleigh quotients.
Since $R$ ignores scaling, track the unscaled iterates: $(1, 0), (2, 1), (5, 4), (14, 13), (41, 40)$. The largest-entry scale factors are $2, \frac52, \frac{14}{5}, \frac{41}{14}$, that is $2, 2.5, 2.8, 2.93$. The Rayleigh quotients are
$$R(2, 1) = \frac{(2, 1) \cdot (5, 4)}{5} = \frac{14}{5} = 2.8, \quad R(5, 4) = \frac{(5, 4) \cdot (14, 13)}{41} = \frac{122}{41} \approx 2.976, \quad R(14, 13) = \frac{1094}{365} \approx 2.997.$$The eigenvalues are $3$ and $1$ (eigenvectors $(1, 1)$ and $(1, -1)$). The scale factors' errors go $0.5, 0.2, 0.07$, down by about a third each step, as the theorem says ($|\lambda_2/\lambda_1| = \frac13$). The Rayleigh errors go $0.2, 0.024, 0.0027$: down by about a ninth each step.
In the plane, the arrows close in on the diagonal $y = x$ at the same speed either way. The Rayleigh quotient just squeezes a better estimate out of the same arrow.
A third per step for one estimate, a ninth for the other, from the very same arrows. That's worth knowing about.
That ninth is no accident: for a symmetric matrix, the Rayleigh quotient's error shrinks like $|\lambda_2/\lambda_1|^{2k}$, twice as many correct digits per step. (The reason is that a symmetric matrix's eigenvectors are perpendicular, the subject of the Spectral Theorem page later in the course.) For a matrix that isn't symmetric, expect only $|\lambda_2/\lambda_1|^k$.
Your turn. In free response, type a number; fractions like 14/5 are fine.
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The theorem's bound $|\lambda_2/\lambda_1|^k$ is the speedometer. Each step multiplies the error by about $|\lambda_2/\lambda_1|$, so the closer the runner-up is to the winner, the slower the race.
Two matrices have dominant eigenvalue $5$. The first has $4$ as its next eigenvalue, the second has $1$. About how many steps does each need to make the error a thousand times smaller?
The first shrinks the error by $\frac45$ a step. We need $\left(\frac45\right)^k \le \frac{1}{1000}$: $\left(\frac45\right)^{30} \approx 0.0012$ is not quite there, and $\left(\frac45\right)^{31} \approx 0.00099$ is. So about $31$ steps. (With logarithms: $k \ge \frac{\ln 1000}{\ln 1.25} \approx 30.96$.)
The second shrinks it by $\frac15$: $\left(\frac15\right)^4 = 0.0016$ and $\left(\frac15\right)^5 = 0.00032$, so $5$ steps. Same winner, six times the work, all because of the runner-up.
Here are the two runs side by side, with each step's error drawn on a scale where every gridline is ten times smaller than the one above.
On that scale the errors fall along straight lines, and the slope of each line is the shrink factor. Your turn. In free response, type a number; fractions like 3/4 are fine.
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The theorem has three hypotheses: a basis of eigenvectors, one eigenvalue strictly bigger in size than the rest, and a start that uses it. Here's what goes wrong without the last two.
A tie in size. $A = \begin{pmatrix} 0 & 2 \\ 2 & 0 \end{pmatrix}$ has eigenvalues $2$ and $-2$. From $x_0 = (1, 0)$: $Ax_0 = (0, 2)$, rescaled to $(0, 1)$; then $A(0, 1) = (2, 0)$, rescaled to $(1, 0)$, and around it goes. The scale factor is $2$ every single time, which looks settled, but the vectors never are: neither eigenvalue outgrows the other.
A complex pair on top. If the eigenvalues of largest size are $a \pm bi$ with $b \neq 0$, the matrix turns the plane of that pair by an angle every step, like a rotation-scaling in disguise. A real arrow keeps circling. $A = \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$, with eigenvalues $1 \pm i$, turns every arrow $45^\circ$ each step, forever.
Neither failure is an arithmetic mistake: in both, no single eigenvalue is bigger in size than all the others, so the theorem simply doesn't apply.
A start that skips $v_1$. Hefferon asks what happens if $c_1 = 0$. In exact arithmetic, the dominant eigenvalue never shows up: for $A = \begin{pmatrix} 3 & 0 \\ 8 & -1 \end{pmatrix}$, the start $x_0 = (0, 1)$ is an eigenvector for $-1$, and every step gives $A(0, 1) = (0, -1)$, scale factor $-1$, forever. On a real computer, though, round-off error is your friend: the tiniest speck of $v_1$ that sneaks in gets multiplied by $3$ every step while everything else stays the same size.
A speck of size one in a million needs about $13$ steps to catch up, since $3^{13} \approx 1.6$ million. So in practice $c_1 = 0$ costs a few extra steps, not the answer, and a random start makes $c_1 = 0$ essentially impossible anyway.
An algorithm whose official strategy is "keep hitting it until it works," and it's rescued by its own rounding errors. I've never related to an algorithm more.
Try all three. Drag the start and step the method; watch both the arrow and the estimate.
A start near the winning line needs only a couple of steps; a start on the other eigenvector's line never leaves it; and with a tie, only the eigenvectors themselves ever hold still.
Hefferon's last question: how can the power method find the smallest eigenvalue? Turn the matrix inside out. If $Av = \lambda v$, then $v = \lambda A^{-1}v$, so $A^{-1}v = \frac1\lambda v$: the same eigenvectors, with the eigenvalues flipped over. The smallest $|\lambda|$ becomes the biggest $\frac{1}{|\lambda|}$, and the power method on $A^{-1}$ finds it. Slide everything first and you can aim at any eigenvalue you like.
If $Av = \lambda v$ and the number $s$ is not an eigenvalue of $A$, then $(A - sI)^{-1}v = \dfrac{1}{\lambda - s}\,v$. So if $A$ has a basis of eigenvectors, $(A - sI)^{-1}$ has the same eigenvectors with eigenvalues $\frac{1}{\lambda - s}$. The power method on $(A - sI)^{-1}$ finds the eigenvalue of $A$ nearest $s$ (when only one is nearest), and the error shrinks by about $\dfrac{|\lambda_{\text{near}} - s|}{|\lambda_{\text{next}} - s|}$ each step, where $\lambda_{\text{next}}$ is the second-nearest.
$(A - sI)v = Av - sv = (\lambda - s)v$. Since $s$ is not an eigenvalue, $A - sI$ is invertible and $\lambda - s \neq 0$; apply $(A - sI)^{-1}$ to both sides and divide by $\lambda - s$ to get $(A - sI)^{-1}v = \frac{1}{\lambda - s}v$. Applied to each vector of an eigenvector basis of $A$, this gives a basis of eigenvectors of $(A - sI)^{-1}$ with eigenvalues $\frac{1}{\lambda_i - s}$. The biggest in size comes from the smallest $|\lambda_i - s|$, the eigenvalue nearest $s$, and it is dominant exactly when that nearest eigenvalue is the only one at that distance. The runner-up comes from the second-nearest, so by the power method theorem the error shrinks like $\frac{1/|\lambda_{\text{next}} - s|}{1/|\lambda_{\text{near}} - s|} = \frac{|\lambda_{\text{near}} - s|}{|\lambda_{\text{next}} - s|}$ per step. $\blacksquare$
This is a visual argument, not a formal proof. First the check: $(A - sI)v = Av - sv = (\lambda - s)v$, so undoing $A - sI$ divides by $\lambda - s$ instead. Now picture the eigenvalues on a number line with $s$ among them. The new eigenvalues are $\frac{1}{\lambda - s}$: one over the distance to $s$ (with a sign for the side). The eigenvalue sitting right next to $s$ has a tiny distance, so its new eigenvalue is enormous and wins by a mile; the others, farther away, get small. Slide $s$ and the winner changes to whichever eigenvalue $s$ is nearest.
The full proof is in the n version.
Nobody actually computes $(A - sI)^{-1}$, though. Multiplying by it is the same as solving a system, and a system is cheap once you've factored the matrix.
The inverse power method with shift $s$ (a number that is not an eigenvalue of $A$) is the power method on $(A - sI)^{-1}$: each step solves $(A - sI)y_{k+1} = x_k$ for $y_{k+1}$, then rescales it. With $s = 0$ it runs on $A^{-1}$ and finds the eigenvalue of smallest size.
Factor $A - sI = LU$ once (LU Factorization), and every step after that is a forward and a back substitution, about as cheap as one multiplication.
Let $A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 5 & 1 \\ 0 & 0 & 7 \end{pmatrix}$. Which eigenvalue does the inverse power method find with shift $s = 0$? With $s = 6.5$? With $s = 6$? How fast?
$A$ is triangular, so its eigenvalues are $2, 5, 7$, with eigenvectors $(1, 0, 0)$, $(1, 3, 0)$, and $(1, 5, 10)$.
With $s = 0$, the nearest eigenvalue to $0$ is $2$, and the next is $5$: the method finds $2$, shrinking the error by $\frac25$ a step. (The plain power method would find $7$, at speed $\frac57$.)
With $s = 6.5$, the distances are $4.5, 1.5, 0.5$: it finds $7$, at speed $\frac{0.5}{1.5} = \frac13$. Its scale factors settle to $\frac{1}{7 - 6.5} = 2$.
With $s = 6$, the distances are $4, 1, 1$: a tie between $5$ and $7$. The new eigenvalues are $-1$ and $1$, the same size, and the method can't decide. Nudge the shift a little off center and it decides fast.
In space, the same start heads for a different eigenvector depending on the shift: the line of $(1, 0, 0)$ for $s = 0$, the line of $(1, 5, 10)$ for $s = 6.5$.
A good guess for an eigenvalue makes a great shift: the closer $s$ is, the smaller $|\lambda_{\text{near}} - s|$, and the faster the method locks on. Slide the shift and see.
Your turn. In free response, type a number; fractions like -1/4 are fine.
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Three equal masses sit in a row between two walls, joined by four equal springs. Pull them aside and let go, and they jiggle in a messy way that is really a mix of three pure motions, called modes, each with its own frequency. Engineers care most about the lowest one: it's the easiest to set off, and it's the one a bridge or a building will sway in. With $x$ the three displacements, Newton's law says $x'' = -Kx$ (in units where the springs and masses are $1$), with the stiffness matrix
$$K = \begin{pmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{pmatrix}.$$Each middle row says a mass is pulled back by its own two springs ($2x_i$) and tugged by its neighbors. A mode is an eigenvector $Kv = \lambda v$, swinging at frequency $\sqrt{\lambda}$, so the lowest note is the smallest eigenvalue. That's a job for the inverse power method.
Run the inverse power method (no shift) on $K$ from $x_0 = (1, 1, 1)$, scaling by the largest entry.
Solving $Ky = x$ is the same as multiplying by $K^{-1} = \frac14\begin{pmatrix} 3 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 3 \end{pmatrix}$ (check: $K$ times its first column is $\frac14(4, 0, 0)$, and so on).
Step $1$: $K^{-1}(1, 1, 1) = \left(\frac32, 2, \frac32\right)$, so $m_1 = 2$ and $x_1 = \left(\frac34, 1, \frac34\right)$.
Step $2$: $K^{-1}x_1 = \frac14(5, 7, 5)$, so $m_2 = \frac74$ and $x_2 = \left(\frac57, 1, \frac57\right)$.
Step $3$: $K^{-1}x_2 = \frac14\left(\frac{34}{7}, \frac{48}{7}, \frac{34}{7}\right)$, so $m_3 = \frac{12}{7} \approx 1.714$ and $x_3 = \left(\frac{17}{24}, 1, \frac{17}{24}\right)$.
The outer entries close in on $\frac{1}{\sqrt 2} \approx 0.707$ and the scale factors on $\frac{1}{2 - \sqrt 2} \approx 1.707$. So the smallest eigenvalue of $K$ is $\frac{1}{1.707\ldots} = 2 - \sqrt 2 \approx 0.586$, with mode $\left(\frac{1}{\sqrt2}, 1, \frac{1}{\sqrt 2}\right)$, or $(1, \sqrt 2, 1)$: all three masses swing the same way, the middle one $\sqrt 2$ times as far. Check: $K(1, \sqrt2, 1) = (2 - \sqrt 2, \ 2\sqrt 2 - 2, \ 2 - \sqrt 2) = (2 - \sqrt 2)(1, \sqrt 2, 1)$.
Here are those iterates beside the motion they describe.
The other eigenvalues of $K$ are $2$ and $2 + \sqrt 2$, so each step shrank the error by $\frac{2 - \sqrt 2}{2} \approx 0.29$: three steps already gave two correct digits. For a real structure $K$ has thousands of rows but only a few nonzero entries in each, and this is how its lowest modes get computed.
The most famous power method of all runs on a matrix with billions of rows: Google's PageRank. The ranking of the web is the steady-state vector of a huge Markov chain, an eigenvector for the eigenvalue $1$, which is always the dominant one there. The model is the subject of the Markov Chains page; the computation is this page. With Google's usual damping factor of $0.85$, the runner-up eigenvalue is known to have size at most $0.85$ (a theorem beyond this page), so about $85$ steps give six correct digits ($0.85^{85} \approx 10^{-6}$), however big the web gets.
The toolkit for this page: multiply by $A$ and rescale, over and over; the iterates line up with the dominant eigenvector, the scale factors (or $\|Ax_k\|$, or the Rayleigh quotient) settle to $\lambda_1$, and the error shrinks by $|\lambda_2/\lambda_1|$ a step. It fails on ties and complex pairs, and needs a start that uses $v_1$. To aim at a different eigenvalue, run it on $(A - sI)^{-1}$: it finds the eigenvalue nearest $s$.
Every question so far came right after its idea, which is a hint a quiz never gives you. Down here the hint is gone. First, true or false: a claim from anywhere on this page, and you decide whether it holds.
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Next, a mixed set: every question type on this page, shuffled. Type a number when it asks you to.
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Made it this far? Time for something with a little heat: shifting can speed up the plain power method too, if you pick the shift well. In free response, type a number; fractions are fine.
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