FP2 3.1 - Eigenvalues and eigenvectors

FP2 3.1 - Eigenvalues and eigenvectors

Find the directions that a matrix stretches without turning away from their own line. Calculate eigenvalues and eigenvectors of two by two and three by three matrices, including zero, repeated and complex eigenvalues, and give normalised vectors when requested.

Directions that survive a transformation

Can a real matrix have no real direction that survives its action? A quarter-turn seems a promising candidate: every non-zero real vector turns through a right angle. Yet it will still have eigenvalues once we allow complex numbers. That small surprise connects the geometry you know to a much wider algebra.

You already know

Matrix multiplication gives the image of a column vector. An invariant point satisfies Av=v\mathbf{A}\mathbf{v}=\mathbf{v}; an invariant line need not consist of fixed points. A square matrix is singular exactly when its determinant is zero. Use the two by two determinant or cofactor expansion from Core Pure.

For A=(3122)\mathbf{A}=\begin{pmatrix}3&1\\2&2\end{pmatrix}, multiplying (11)\begin{pmatrix}1\\1\end{pmatrix} gives (44)\begin{pmatrix}4\\4\end{pmatrix}. Its length changes but its direction does not. Multiplying (1−2)\begin{pmatrix}1\\-2\end{pmatrix} gives the same vector back. These are especially convenient directions for this particular matrix.

[DIAGRAM: asset_name: Two eigenvector directions under a matrix; asset_slug: 9fm0-fp294-eigendirections; caption: The blue direction is stretched by 4 while the sage direction is fixed; description: Equal-scale coordinate panels show vectors v=(1,1) and w=(1,-2) before transformation and their images (4,4) and (1,-2) under A with rows (3,1) and (2,2). Blue highlights the first direction and sage the second. Grey dashed complete lines distinguish directions from individual vectors. All labels are ink.]
The blue direction is stretched by 4 while the sage direction is fixed

Eigenvalue and eigenvector

For a square matrix A\mathbf{A}, a non-zero vector v\mathbf{v} is an eigenvector with eigenvalue λ\lambda when

Av=λv\mathbf{A}\mathbf{v}=\lambda\mathbf{v}

For real vectors and real λ\lambda, the image remains on the same line through the origin: a negative eigenvalue reverses direction; eigenvalue 11 fixes the vector; eigenvalue 00 sends it to the origin.

The zero vector satisfies the equation for every λ\lambda, so allowing it would tell us nothing. A zero eigenvalue is perfectly valid; a zero eigenvector is not. Any non-zero scalar multiple of an eigenvector is another eigenvector for the same eigenvalue, because A(cv)=cAv=λ(cv)\mathbf{A}(c\mathbf{v})=c\mathbf{A}\mathbf{v}=\lambda(c\mathbf{v}).

Key idea

An eigenvector is a non-zero input whose image is a scalar multiple of itself. Always keep its eigenvalue attached to it.

Neither the eigenvector definition nor the methods in this lesson are supplied in the formula booklet. The next check separates the two different roles of zero.

Find eigenvalues and eigenvectors in two dimensions

Rearrange the defining equation to (A−λI)v=0(\mathbf{A}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}. A non-zero solution exists only when the coefficient matrix is singular. Its determinant therefore gives the equation for λ\lambda. Subtract λ\lambda from diagonal entries only.

Characteristic equation

For an nn by nn matrix,

det⁡(A−λI)=0\det(\mathbf{A}-\lambda\mathbf{I})=0

is the characteristic equation. For A=(abcd)\mathbf{A}=\begin{pmatrix}a&b\\c&d\end{pmatrix}, it is

λ2−(a+d)λ+ad−bc=0\lambda^2-(a+d)\lambda+ad-bc=0

For each root, solve (A−λI)v=0(\mathbf{A}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0} and choose a non-zero solution. The equivalent equation det⁡(λI−A)=0\det(\lambda\mathbf{I}-\mathbf{A})=0 has the same roots.

Not in the formula booklet: learn it.

The determinant is the route to the eigenvalues, not the final eigenvectors. Those come from simultaneous linear equations.

Worked example 1: two distinct eigenvalues

Question. Find the eigenvalues and a corresponding eigenvector for each for A=(3122)\mathbf{A}=\begin{pmatrix}3&1\\2&2\end{pmatrix}.

Plan. Solve the characteristic equation, then substitute each eigenvalue into the vector equation.

det⁡(A−λI)=(3−λ)(2−λ)−2diagonal subtraction=λ2−5λ+4=(λ−4)(λ−1)expand and factorλ=4, 1set determinant to zero\begin{aligned} \det(\mathbf{A}-\lambda\mathbf{I})&=(3-\lambda)(2-\lambda)-2 && \text{diagonal subtraction}\\ &=\lambda^2-5\lambda+4=(\lambda-4)(\lambda-1) && \text{expand and factor}\\ \lambda&=4,\ 1 && \text{set determinant to zero} \end{aligned}

For v=(xy)\mathbf{v}=\begin{pmatrix}x\\y\end{pmatrix}, the first row gives

λ=4:−x+y=0⟹v=(11)choose x=1λ=1:2x+y=0⟹v=(1−2)choose x=1\begin{aligned} \lambda=4:&\quad -x+y=0\quad\Longrightarrow\quad \mathbf{v}=\begin{pmatrix}1\\1\end{pmatrix} && \text{choose }x=1\\ \lambda=1:&\quad 2x+y=0\quad\Longrightarrow\quad \mathbf{v}=\begin{pmatrix}1\\-2\end{pmatrix} && \text{choose }x=1 \end{aligned}

Why this step? Each singular system has a free scale. Choosing a convenient non-zero value gives one representative, not a unique vector.

Answer. The eigenpairs are 4,(11)4,\begin{pmatrix}1\\1\end{pmatrix} and 1,(1−2)1,\begin{pmatrix}1\\-2\end{pmatrix}, allowing non-zero multiples of either vector.

Check. The products are (44)\begin{pmatrix}4\\4\end{pmatrix} and (1−2)\begin{pmatrix}1\\-2\end{pmatrix}, agreeing with both eigenvalues.

Change only the lower-left entry, from 22 to 66. One eigenvalue becomes zero; retain it.

Worked example 2: a singular matrix still has eigenvectors

Question. Find the eigenvalues and corresponding eigenvectors of B=(3162)\mathbf{B}=\begin{pmatrix}3&1\\6&2\end{pmatrix}.

Plan. Use the same determinant and first-row equations, checking the second row afterwards.

det⁡(B−λI)=(3−λ)(2−λ)−6changed entry=λ2−5λ=λ(λ−5)factorλ=0, 5both roots countλ=0:3x+y=0⟹v=(1−3)non-zero inputλ=5:−2x+y=0⟹v=(12)choose x=1\begin{aligned} \det(\mathbf{B}-\lambda\mathbf{I})&=(3-\lambda)(2-\lambda)-6 && \text{changed entry}\\ &=\lambda^2-5\lambda=\lambda(\lambda-5) && \text{factor}\\ \lambda&=0,\ 5 && \text{both roots count}\\ \lambda=0:&\quad 3x+y=0\quad\Longrightarrow\quad\mathbf{v}=\begin{pmatrix}1\\-3\end{pmatrix} && \text{non-zero input}\\ \lambda=5:&\quad -2x+y=0\quad\Longrightarrow\quad\mathbf{v}=\begin{pmatrix}1\\2\end{pmatrix} && \text{choose }x=1 \end{aligned}

Why this step? The matrix is singular, so some non-zero inputs must collapse to zero. They supply the eigenvectors for eigenvalue zero.

Answer. Eigenvalue 00 has eigenvectors c(1−3)c\begin{pmatrix}1\\-3\end{pmatrix}; eigenvalue 55 has eigenvectors c(12)c\begin{pmatrix}1\\2\end{pmatrix}, with c≠0c\ne0 in each case.

Check. B(1−3)=0\mathbf{B}\begin{pmatrix}1\\-3\end{pmatrix}=\mathbf{0} and B(12)=(510)\mathbf{B}\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}5\\10\end{pmatrix}.

The equation must subtract the scalar times the identity, not the scalar from every entry.

Watch out

Wrong: replace (3162)\begin{pmatrix}3&1\\6&2\end{pmatrix} by (3−λ1−λ6−λ2−λ)\begin{pmatrix}3-\lambda&1-\lambda\\6-\lambda&2-\lambda\end{pmatrix}. Right: the off-diagonal entries stay 11 and 66, because the identity has zeros there.

Change the lower-left entry once more, keeping the same method.

Work systematically with three coordinates

For three by three matrices, the characteristic equation is cubic. A zero entry can simplify its expansion. A supplied eigenvalue gives a known factor; substitute it into the polynomial before dividing. Once a root is known, use enough component equations to establish a vector and check all three rows.

Worked example 3: use a simple root of a cubic

Question. Find the eigenvalues and corresponding eigenvectors of T=(210131012)\mathbf{T}=\begin{pmatrix}2&1&0\\1&3&1\\0&1&2\end{pmatrix}.

Plan. Expand along the first row, factor out 2−λ2-\lambda, then solve the component equations.

det⁡(T−λI)=(2−λ)((3−λ)(2−λ)−1)−(2−λ)two non-zero terms=(2−λ)((3−λ)(2−λ)−2)common factor=(2−λ)(λ−1)(λ−4)factor the quadratic\begin{aligned} \det(\mathbf{T}-\lambda\mathbf{I}) &=(2-\lambda)\big((3-\lambda)(2-\lambda)-1\big)-(2-\lambda) && \text{two non-zero terms}\\ &=(2-\lambda)\big((3-\lambda)(2-\lambda)-2\big) && \text{common factor}\\ &=(2-\lambda)(\lambda-1)(\lambda-4) && \text{factor the quadratic} \end{aligned}

Thus λ=1,2,4\lambda=1,2,4. For λ=2\lambda=2, the outer rows give y=0y=0, and the middle row gives x+z=0x+z=0. For the other roots, the outer rows give y=(λ−2)x=(λ−2)zy=(\lambda-2)x=(\lambda-2)z, so z=xz=x.

λ=1:v1=(1−11)take x=z=1λ=2:v2=(10−1)use the separate caseλ=4:v4=(121)take x=z=1\begin{aligned} \lambda=1:&\quad\mathbf{v}_1=\begin{pmatrix}1\\-1\\1\end{pmatrix} && \text{take }x=z=1\\ \lambda=2:&\quad\mathbf{v}_2=\begin{pmatrix}1\\0\\-1\end{pmatrix} && \text{use the separate case}\\ \lambda=4:&\quad\mathbf{v}_4=\begin{pmatrix}1\\2\\1\end{pmatrix} && \text{take }x=z=1 \end{aligned}

Why this step? Dividing by λ−2\lambda-2 would lose the root 22. Handle it before using that non-zero factor.

Answer. The eigenvalues are 1,2,41,2,4, with corresponding eigenvectors v1,v2,v4\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_4 above.

Check. Their images are (1,−1,1)T(1,-1,1)^{\mathrm T}, (2,0,−2)T(2,0,-2)^{\mathrm T} and (4,8,4)T(4,8,4)^{\mathrm T}, respectively.

Change only the central entry from 33 to 11. The factorisation strategy survives, but the other two eigenvalues change.

Worked example 4: change one entry and retain a zero eigenvalue

Question. Find the eigenvalues and corresponding eigenvectors of U=(210111012)\mathbf{U}=\begin{pmatrix}2&1&0\\1&1&1\\0&1&2\end{pmatrix}.

Plan. Repeat the first-row expansion and separate the root 22 when solving for vectors.

det⁡(U−λI)=(2−λ)((1−λ)(2−λ)−2)same expansion=(2−λ)λ(λ−3)factorλ=0, 2, 3all three rootsλ=0:v0=(1−21)y=−2x, z=xλ=2:v2=(10−1)y=0, z=−xλ=3:v3=(111)y=x, z=x\begin{aligned} \det(\mathbf{U}-\lambda\mathbf{I}) &=(2-\lambda)\big((1-\lambda)(2-\lambda)-2\big) && \text{same expansion}\\ &=(2-\lambda)\lambda(\lambda-3) && \text{factor}\\ \lambda&=0,\ 2,\ 3 && \text{all three roots}\\ \lambda=0:&\quad\mathbf{v}_0=\begin{pmatrix}1\\-2\\1\end{pmatrix} && y=-2x,\ z=x\\ \lambda=2:&\quad\mathbf{v}_2=\begin{pmatrix}1\\0\\-1\end{pmatrix} && y=0,\ z=-x\\ \lambda=3:&\quad\mathbf{v}_3=\begin{pmatrix}1\\1\\1\end{pmatrix} && y=x,\ z=x \end{aligned}

Why this step? The outer rows determine the same relationships as before, but the changed middle row determines which eigenvalues make them consistent.

Answer. The eigenvalues are 0,2,30,2,3, with the corresponding vectors shown.

Check. Multiplication gives 0\mathbf{0}, (2,0,−2)T(2,0,-2)^{\mathrm T} and (3,3,3)T(3,3,3)^{\mathrm T}, in that order.

Any vector found from one equation must satisfy the remaining equations too. The determinant guarantees existence, not the correctness of your chosen components.

Watch out

Wrong: for λ=2\lambda=2, cancel λ−2\lambda-2 and conclude x=zx=z. Right: substitute λ=2\lambda=2 first; the middle equation gives z=−xz=-x. Dividing by zero would replace the correct direction by a wrong one.

One more change to the central entry introduces exact surds rather than new methods.

Continue this lesson.

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