WEBVTT

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Multiply a vector by a matrix and,
in general, two things happen to it.

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Its length changes,
and its direction changes.

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Today I want to ask about the second
of those, because for almost every matrix

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there are a few directions
that simply refuse to be turned.

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So here is a piece
of the plane, ruled into squares.

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And here is the matrix
we will apply to it:

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two, one, one, two.

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And let me say what applying it means.

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It means multiplication.

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Take a point on the grid, write
it as a column, put A in front,

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and out comes a different point:

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two

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x plus y, and x plus two y.

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A matrix is a map
that carries one point to another.

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Take one point and try it.

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This one, at one, zero.

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Multiply it by A:

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two times one plus zero is two,

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and one plus two times zero is one.

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So A sends that point over to two, one.

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Now watch every point
do that same sum at once.

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The whole thing stretches
out along one diagonal,

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and our point lands exactly
where the arithmetic said it would.

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Let me put it back, and follow one
single arrow through the same motion.

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This one points straight up,
at zero, one and a half.

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Put it through the same sum
and the answer is one and a half, three.

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Now apply the matrix.

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The arrow gets longer, certainly.

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But look where it ends up. It has swung
over to the right. The matrix turned it.

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Not every arrow does that.

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Let me put the plane
back and try two others,

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one along this diagonal
and one along the other.

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And this time let us do the multiplication
first, before anything moves.

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Read those two answers. The first one
is three times the vector we put in.

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The second one is the vector we put
in, unchanged, which is one times it.

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Now apply the matrix and watch
the arrows do exactly that.

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The red one triples and stays
on the line it was already on.

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The green one does not move at all.

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Those two directions
are the eigenvectors of this matrix.

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Everything else in the plane swings
round toward the stretching direction.

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These two just sit there and get scaled,

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by three and by one, and those two
numbers are their eigenvalues.

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So let us write down
exactly what that means.

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Look again at what just
happened to that red arrow.

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Here is the vector one, one.

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We put it through the matrix,
and out came another vector: three, three.

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And here is the observation
the whole subject is built on.

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That answer is nothing but three
times the vector we started with.

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One vector, one plain number, no matrix.

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Which tells us
something about the picture.

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Both arrows lie on this one
line through the origin.

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So on this line, applying
A is nothing but multiplying by three.

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The arrow got longer, but it
was not turned by so much as a degree.

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Now try an arrow that is not on that line.

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Here is w, pointing straight up.

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A sends it to one, two.

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And one, two is not
a multiple of zero, one.

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No number times zero is ever one.

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So w came back pointing
somewhere else entirely,

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like almost every arrow in the plane.

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The arrows that survive,

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the ones A puts back on their own
line, are the eigenvectors of A.

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And the number that does the stretching,
three in this case, is its eigenvalue.

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Now say that for any
matrix and any vector.

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Here is a general v,
drawn on a general line.

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A vector v is an eigenvector
of A when A times v comes out

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as a plain multiple of v itself.

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Written down, that is A v equals lambda
v, and that is the whole definition.

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That lambda is the eigenvalue.

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And there is one piece of fine print:
v is not allowed to be the zero vector.

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A times zero is zero

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for every matrix there is, so the zero
vector would qualify for everything.

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Read the equation geometrically now.

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It says that when A gets hold of v,

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it puts it back down
somewhere on that same line.

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If lambda is bigger than one,
A stretches the arrow out along it.

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If lambda is between zero and one,
it pulls the arrow in toward the origin.

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Shorter, yes. Turned, no.

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And now let lambda keep
going, down past zero.

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Watch the arrow shrink as it goes,
die away to nothing at the origin,

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and come back out the other
side, pointing backwards.

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That still counts.

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Backwards along a line is the same line.

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Notice what the equation does not say:

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it does not say A leaves v alone,
it says A leaves the line alone.

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So an eigenvector
is a direction the matrix cannot turn,

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and the eigenvalue says what
happens to lengths along it.

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Which leaves the real question:
given a matrix, how do you find them?

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Right. We have a matrix, and we
want the directions it cannot turn.

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So start from the definition
and push the algebra around.

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A v equals lambda

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v.

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Move everything to one side.

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A v minus lambda v equals the zero vector.

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I would love to factor the v out, but I
cannot subtract a number from a matrix,

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so I slip in the identity:

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lambda v is the same thing
as lambda times I times v.

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And now it factors.

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The matrix A minus lambda I,
acting on v, gives the zero vector.

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Look at what that line actually says.

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Something times something comes out zero.

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And one of the two
somethings we already know about:

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v is not the zero vector.

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That was the fine print in the definition.

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So the zero on the right
cannot have come from the vector.

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It has to have come
from the matrix over here.

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A minus lambda I has to take a nonzero
vector and crush it flat to nothing.

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And there is exactly one number
that measures whether a matrix does that.

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If you want to say how much
of a matrix there is, in a single number,

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that number is its determinant.

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A matrix that flattens a nonzero vector
onto the origin has determinant zero.

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Formally, then:

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lambda is an eigenvalue of A exactly when
the determinant of A minus lambda I comes

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out zero. That is one equation in one
unknown, and we can actually solve it.

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Let us do it with our matrix.

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Subtract lambda from each entry

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on the diagonal, take the determinant of
what is left, and set that equal to zero.

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The determinant of a two by two
is read straight off the two diagonals.

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Multiply down this one.

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Then multiply down the other.

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And take the second
product away from the first.

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So that is two minus lambda,
all squared, take away one times one.

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Multiply the square out
and keep every term.

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Lambda squared, minus four lambda,
plus four, minus one, equals zero.

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And four minus one is three, so that
line tidies itself up where it stands.

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Lambda squared minus
four lambda plus three.

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This has a name:

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it is the characteristic polynomial of A,

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and its degree is always
the size of the matrix.

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This one factors on sight:

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lambda minus three, times lambda minus
one. So the eigenvalues are three and one.

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And there is our confirmation,

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because those are exactly the two
numbers we watched on the grid.

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One arrow tripled.

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The other one stayed put.

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We have the eigenvalues.

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Now we want the directions
that go with them.

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Go back to the factored
equation, A minus lambda

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I times v equals zero, and put
the first eigenvalue into it.

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With lambda equal to three, we subtract
three from each entry on the diagonal.

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That gives us minus one, one, one,

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minus one, multiplying the column x, y,
and the answer has to be the zero vector.

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Both rows say the same thing.

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Minus x plus y equals zero,
or in other words y equals x.

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So any vector whose two
entries are equal will do,

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and the tidiest one to write down is one,

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one.

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There is the line of them,
and there is the arrow.

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And notice
that an eigenvector is never alone.

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Every multiple of one,
one is an eigenvector too,

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with the very same eigenvalue three.

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What we have really found is a whole
line the matrix keeps to itself.

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Now the second eigenvalue.

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Lambda equals one, so we subtract
one from the diagonal instead,

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and this time the matrix
comes out as four ones.

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Every row now says x plus y equals zero,

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so y is minus x, and the tidiest
eigenvector to pick is one, minus one.

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There is the second line,
and there is the second arrow.

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Now hold on to both of those arrows,

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and let the matrix loose
on the whole picture again.

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Every grid line
has been dragged off somewhere new.

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The red arrow ran out along its own
line and stopped three times further

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from the origin, which is its eigenvalue.

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The green one did not move at all,
because its eigenvalue is one.

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Those two directions were untouchable.

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One more thing, and then a warning.

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Here is a matrix that rotates
the plane through a quarter turn:

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zero, minus one, one, zero.

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And here are three arrows sitting in it.

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Every arrow in the plane gets
swung round by ninety degrees.

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Watch these three go. Not one of them
finishes on the line it started on.

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A rotation has no favourite direction.

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And the algebra says the same thing.

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Subtract lambda from the diagonal,

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take the determinant, and cross
the two diagonals exactly as before.

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That comes out as lambda squared plus one,

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and over the real numbers
that has no roots at all.

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This matrix has no eigenvectors.

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It is not a defect
in the matrix, it is honest geometry.

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So keep that in your pocket. Eigenvectors
are common, but they are not guaranteed.

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Now let me lift exactly
this rotation up into three dimensions,

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where something rather lovely happens.

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Here is a rotation of space
about the vertical axis,

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and here are two arrows lying
flat in the horizontal plane.

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Watch them turn.

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Every one of them changes direction,

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exactly the way they
did down in the plane.

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But now there is somewhere to stand
that the rotation cannot reach.

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An arrow pointing
straight up along the axis.

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The rotation carries the whole
of space around it and leaves

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that one arrow precisely where it was.

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So that arrow is an eigenvector,
and its eigenvalue is one.

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And this is a general fact
about three dimensional space.

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Every rotation has an axis, and the axis
is the eigenvector with eigenvalue one.

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Euler proved that, and the proof
is essentially this picture.

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So there is the whole idea, in one line.

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A times v equals lambda times v.

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The eigenvectors
are the directions the matrix cannot turn.

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The eigenvalue attached

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to each one says what happens
to lengths along that direction.

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Find those directions,

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and you have found the frame in which your
matrix is nothing but a list of stretches.

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Which is why, from vibration to stability
to the ranking of the whole internet,

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this tends to be the first
question anybody asks about a matrix.